Subject CT6. CMP Upgrade 2013/14. CMP Upgrade

Size: px
Start display at page:

Download "Subject CT6. CMP Upgrade 2013/14. CMP Upgrade"

Transcription

1 CT6: CMP Upgrade 013/14 Page 1 Subject CT6 CMP Upgrade 013/14 CMP Upgrade This CMP Upgrade lists all significant changes to the Core Reading and the ActEd material since last year so that you can manually amend your 013 study material to make it suitable for study for the 014 exams. It includes replacement pages and additional pages where appropriate. Alternatively, you can buy a full replacement set of up-to-date Course Notes at a significantly reduced price if you have previously bought the full price Course Notes in this subject. Please see our 014 Student Brochure for more details. This CMP Upgrade contains: All non-trivial changes to the Syllabus objectives and Core Reading. Changes to the ActEd Course Notes, Question and Answer Bank, and Series X Assignments that will make them suitable for study for the 014 exams. IFE: 014 Examinations

2 Page CT6: CMP Upgrade 013/14 1 Changes to the Syllabus objectives and Core Reading 1.1 Syllabus objectives There have been no changes to the syllabus. 1. Core Reading Chapter Page 10 The following paragraph has been deleted. It is often convenient to express this result in terms of the value of a statistic, such as X, rather than the sample values X. So, for example, f( qω X) = f( XΩq) f( q). f( X) In practice these are equivalent. Chapter 3 Page 4 In Section 1.1, where the exponential distribution is defined, it has been specified that l > 0 and x > 0. Page 5 In Section 1., where the gamma distribution is defined, it has been specified that a > 0, l > 0 and x > 0. Page 7 In Section.1, where the Pareto distribution is defined, it has been specified that a > 0, l > 0 and x > 0. IFE: 014 Examinations

3 CT6: CMP Upgrade 013/14 Page 3 Page 10 Under the section on Method of Moments, x i has been redefined: x i = the i th value in the sample Page 11 In Section.3, where the Weibull distribution is defined, it has been specified that c > 0, g > 0 and x > 0. Page 16 The section on MLE has been modified to include formulae for the discrete case. A replacement page is provided. Page 19 In the first paragraph of Core Reading, references to estimates have been changed to estimators. Page 0 In the first paragraph of Core Reading, references to estimates have been changed to estimators. In Section 3.6, the word n-denominator has been inserted before sample variance. Pages 7 to 30 Various changes have been made to the section on mixture distributions, to help improve the flow. Replacement pages are provided. IFE: 014 Examinations

4 Page 4 CT6: CMP Upgrade 013/14 Chapter 4 Pages 6 and 7 There have been several modifications to Pages 6 and 7, including a deletion of the discussion on complete and incomplete integrals. Replacement pages are provided. Page 8 Some alterations have been made to the notation used in the discussion of the reinsurer s conditional claims distribution. A replacement page is provided. Page 10 The second and third paragraphs of Core Reading have been altered. A replacement page is provided. Pages 19 and 0 The following passage has been deleted: The same device as the one used to obtain (1.3) can be used to convert (3.1) to a complete integral. (3.1) can be written as E(Y) = Ú kx f ( x) dx - Ú kx f ( x) dx + M Ú f ( x) dx 0 M/ k M/ k = k E(X) k Ú ( - / ) ( ). / M k x M k f x dx The new mean amount paid by the insurer is E(Y) = - Ú + 0 kex ( ( ) yf( y M/ k) dy ). (3.) Note that the integral in (3.) has the same form as that in (1.3). Page 0 The paragraph of Core Reading in the middle of the page has been replaced with: One important general point that can be made is that the new mean claim amount paid by the insurer is not k times the mean claim amount paid by the insurer without inflation. IFE: 014 Examinations

5 CT6: CMP Upgrade 013/14 Page 5 Page 4 Some alterations have been made to the discussion on estimation. A replacement page is provided. Chapter 7 Page 18 At the bottom of the page, the following fragment of a sentence has been deleted: and the distribution of S follows that of N. Pages 19 to 4 Some alterations have been made to the discussion of the compound Poisson distribution and the result concerning the sum of compound Poisson distributions. Replacement pages are provided. Pages 4 to 6 In the discussion on the compound binomial distribution, the distribution of the number of claims has been changed from N ~ bin(n, q) to the more usual N ~ bin(n, p) so that all of the q s are now p s and vice-versa. Replacement pages are provided. Chapter 8 Page 3 The second paragraph has been rewritten as: If, for example, N ~ Poi( l ), S I has a compound Poisson distribution with Poisson parameter, and the i th individual claim amount is Y i. Similarly, S R has a compound Poisson distribution with Poisson parameter, and the individual claim amount is Z i. th i IFE: 014 Examinations

6 Page 6 CT6: CMP Upgrade 013/14 Pages 5 and 6 Some small changes have been made to the example in the Core Reading to improve the flow. Replacement pages are provided. Pages 6 and 8 Some changes have been made to the notation used for the PDF of the reinsurer s conditional claims distribution, to tie in with the changes made to the notation in Chapter 4. Replacement pages are provided. Page 11 Two-thirds of the way down the page, a sentence has been rewritten as: Thus, the distribution of Y j is compound binomial, with individual claim amount random variable X j. At the bottom of the page, the following sentence has been deleted: It was seen in Section 3.5 of Chapter 7 that there is no general result about the distribution of such a sum. Page 13 The first part of Section 3. has been altered to: Consider a portfolio consisting of n independent policies. The aggregate claims from the i-th policy are denoted by the random variable S i, where S i has a compound Poisson distribution with parameters i, and the CDF of the individual claim amounts distribution is F(x). Notice that, for simplicity, the CDF of the distribution of individual claim amounts, F(x), is assumed to be identical for all the policies. In this example the distribution of individual claim amounts, ie F(x), is assumed to be known but the values of the Poisson parameters, ie the i s, are not known. IFE: 014 Examinations

7 CT6: CMP Upgrade 013/14 Page 7 Page 17 The first part of Section 3.4 has been altered to: Now a different example is considered. Suppose, as before, there is a portfolio of n policies. The aggregate claims from a single policy have a compound Poisson distribution with parameters, and the CDF of the individual claim amounts random variable is F(x). The Poisson parameters are the same for all policies in the portfolio. Chapter 11 The origin years in the run-off triangles have been brought more up-to-date. This doesn t affect the understanding of the material in anyway so we have not provided replacement pages. IFE: 014 Examinations

8 Page 8 CT6: CMP Upgrade 013/14 Changes to the ActEd Course Notes All Chapters The Summary and Formulae Pages have been updated for all chapters in an attempt to better reflect the key material to be revised for the exam. Replacement pages are provided for all chapters. Chapter Page 10 The following paragraph has been deleted: The notation can get a little confusing here. Remember that X is a sample mean, whereas X refers to the whole group of values of X ie a vector containing all the values in the sample. Chapter 3 Page 17 The 1 st step in the description of how to find an MLE has been modified to take account of the discrete case: (1) Write down the likelihood function for the available data. If the likelihood is based on a set of known values x 1, x,, x n, then the likelihood function will take the form f ( x1) f( x) f( x n ), where f ( x) is the PDF of X where X is a continuous random variable (or P( X = x1) P( X = x) P( X = xn ) in the case where X is a discrete random variable). Chapter 4 Pages 6 and 7 There have been several modifications to Pages 6 and 7 to try and improve the clarity, including the deletion of Questions 4.3 and 4.4. Replacement pages are provided. IFE: 014 Examinations

9 CT6: CMP Upgrade 013/14 Page 9 Pages 8 to 10 Some alterations have been made to the notation used in the discussion of the reinsurer s conditional claims distribution. Replacement pages are provided. Page 36 Solutions 4.3 and 4.4 have been deleted to correspond with the deletion of Question 4.3. Page 37 In Solution 4.5, the z s have been replaced with w s to tie in with the altered notation relating to the reinsurer s conditional claims distribution. In Solution 4.6, the z s have been replaced with w s to tie in with the altered notation relating to the reinsurer s conditional claims distribution. Pages 44 and 45 In Solution 4.16, the y s have been replaced with w s to tie in with the altered notation relating to the reinsurer s conditional claims distribution. Pages 45 and 46 Solution 4.17 has been modified, to improve clarity, as follows: Without the excess, the average amount paid by the insurer in respect of each loss is the mean of a Pa ( a, l ) distribution: l a -1 With an excess L, the average amount paid by the insurer in respect of each loss is: a l Ú ( x - L) a dx a 1 L ( l + + x) Using the substitution y = x- L, this is: a l Ú ya a 1 0 ( l + + L+ y) dy IFE: 014 Examinations

10 Page 10 CT6: CMP Upgrade 013/14 We now transform the integrand into the mean of a Pa(, L) a l+ distribution: a a -a-1 Ê l ˆ Ú ( l + L) yal ( l + L+ y) dy = Á ya dy Ë l + L Ú a + 1 ( l + L+ y) 0 0 a Ê l ˆ l + L = Á Ë l + L a -1 a So, the ratio of the risk premiums will be: a a Ê l ˆ l + L l Ê l ˆ Ê 5,000 ˆ Á = = = Ë l L a 1 a 1 Á Ël + L Á Ë5, ie a reduction of.9%. Chapter 5 Page 17 The first sentence of the last paragraph has been modified to say: In fact, what is required is ( ) = E l x. E X x ( ) Chapter 7 Page 41 In Solution 7.9, the q s have been changed p s and vice-versa to tie in with the change to the Core Reading, where N is now presented has having a binomial distribution with parameters n and p. Chapter 8 Page 7 At the top of Page 7, the PDF of the reinsurer s conditional claim amount distribution is now defined as: g W f X ( w+ M) ( w) = 1 - F ( M) X IFE: 014 Examinations

11 CT6: CMP Upgrade 013/14 Page 11 Chapter 10 Significant changes have been made to the ActEd notes for this chapter to improve the explanations and examples. Replacement pages are provided for the whole chapter. Chapter 11 The origin years in the run-off triangles have been brought more up-to-date. This doesn t affect the understanding of the material in anyway so we have not provided replacement pages. Page 11 We have added the following paragraphs to the top of Page 11, before Question 11.7 to help explain the general statistical model: Note that some of the terminology above is being used in a different context to previously. The development factors r j in this general statistical model are defined differently to the development factors that we have met previously. The development factors that we met previously were used to project forward cumulative claims data. The development factors in the general statistical model are being used to model incremental data. They are defined above as the proportion of claims from a particular accident year that are paid in the j th development year. As such, they are a set of factors that add up to 1. A question may help you to see what is going on. IFE: 014 Examinations

12 Page 1 CT6: CMP Upgrade 013/14 3 Changes to the Q&A Bank Part Questions Question.10 The headings in the table have been corrected to j, Y j and P j. (Previously these were incorrectly labelled as i, Y i and P i.) Solution.1 Two formulae were incorrectly labelled E( Y ). These have now been changed to E( Z ). IFE: 014 Examinations

13 CT6: CMP Upgrade 013/14 Page 13 4 Changes to the X Assignments This section provides details of changes that have been made to the 013 X Assignments, so that you can continue to use these for the 014 exams. However, if you are having your attempts marked by ActEd, you will need to use the 014 version of the X Assignments. Assignment X Solutions We have corrected a typo in Solution X.6(i): = - = EY ( ) (85 35 ) 3,808 (Previously, the answer was quoted incorrectly as 3,803 1.) 3 Assignment X3 Questions Question X3. has been reworded to say: Define a model for each incremental entry, C ij, in general terms and explain each element of the formula. Question X3.7 now refers to Policy Year not Accident Year in the run-off triangle. Assignment X3 Solutions Solution X3.7 now refers to Policy Year not Accident Year. We have corrected a typo in Solution X3.11(i)(c): dr (100 1,300 )( 3) (1 3 )(100,600 ) d (100 1,300 ) 300 3, , ,800 (100 1,300 ), 600 3, (100 1,300 ) (Previously, the denominator was not squared. The final answer is unaffected.) IFE: 014 Examinations

14 Page 14 CT6: CMP Upgrade 013/14 Assignment X4 Solutions We have corrected a typo in the notes (in italics) to solution X4.: n > z (Previously, the subscript of the z was 0.01.) We have added the following new statement (worth [1] mark) to Solution X4.4(i): This reproducibility removes the variability of using different sets of random numbers, which is helpful for comparing different models. Any two of the three statements are now needed to score a maximum of [] marks on this part. IFE: 014 Examinations

15 CT6: CMP Upgrade 013/14 Page 15 5 Other tuition services In addition to this CMP Upgrade you might find the following services helpful with your study. 5.1 Study material We offer the following study material in Subject CT6: ASET (ActEd Solutions with Exam Technique) and Mini-ASET Flashcards Mock Exam A and Additional Mock Pack (AMP) Revision Notes Online Classroom. For further details on ActEd s study materials, please refer to the 014 Student Brochure, which is available from the ActEd website at 5. Tutorials We offer the following tutorials in Subject CT6: a set of Regular Tutorials (lasting two or three full days) a Block Tutorial (lasting two or three full days) a Revision Day (lasting one full day) a Taught Course (lasting four full days) a series of webinars (lasting approximately an hour and a half each) CT6 Online Classroom. For further details on ActEd s tutorials, please refer to our latest Tuition Bulletin, which is available from the ActEd website at IFE: 014 Examinations

16 Page 16 CT6: CMP Upgrade 013/ Marking You can have your attempts at any of our assignments or mock exams marked by ActEd. When marking your scripts, we aim to provide specific advice to improve your chances of success in the exam and to return your scripts as quickly as possible. For further details on ActEd s marking services, please refer to the 014 Student Brochure, which is available from the ActEd website at IFE: 014 Examinations

17 CT6: CMP Upgrade 013/14 Page 17 6 Feedback on the study material ActEd is always pleased to get feedback from students about any aspect of our study programmes. Please let us know if you have any specific comments (eg about certain sections of the notes or particular questions) or general suggestions about how we can improve the study material. We will incorporate as many of your suggestions as we can when we update the course material each year. If you have any comments on this course please send them by to CT6@bpp.com or by fax to IFE: 014 Examinations

18 All study material produced by ActEd is copyright and is sold for the exclusive use of the purchaser. The copyright is owned by Institute and Faculty Education Limited, a subsidiary of the Institute and Faculty of Actuaries. Unless prior authority is granted by ActEd, you may not hire out, lend, give out, sell, store or transmit electronically or photocopy any part of the study material. You must take care of your study material to ensure that it is not used or copied by anybody else. Legal action will be taken if these terms are infringed. In addition, we may seek to take disciplinary action through the profession or through your employer. These conditions remain in force after you have finished using the course. IFE: 014 Examinations

19 CT6-01: Decision theory Page 5 Chapter 1 Summary Zero-sum games In this chapter we study zero-sum two-player games. If we call our players (who are in conflict) Alice and Bob then the term zero-sum tells us that whatever Alice loses, Bob must win; there are no payments to or receipts from third parties. Both Alice and Bob have a number of different strategies open to them. The payoffs from each combination of strategies can be represented in a matrix. The payoffs associated with Alice s available strategies (labelled I, II, III, etc) determine the columns of the matrix. The payoffs associated with Bob s available strategies (labelled 1,, 3, etc) determine the rows of the payoff matrix. One strategy is said to dominate another if the first strategy is at least as good as the second and in some cases better. Dominated strategies can always be discarded. Problems in Decision Theory usually involve the determination of optimum strategies and the corresponding payoff, or value, of the game. Two criteria that are often used to determine optimum strategies are the minimax criterion and the Bayes criterion. Under the minimax criterion, all players will adopt the strategy that minimises their maximum loss (or maximises their minimum gain). The minimax criterion can be thought of as a best-of-all-evils approach. A saddle point is the name given to an entry of a payoff matrix that is the largest in its column and the smallest in its row. If a saddle point exists then each player will adopt the pure strategy, with the options that correspond to the saddle point being chosen. If there is no saddle point, a randomised strategy can be adopted to enable a player to minimise his/her maximum expected loss. This means that the player will vary his or her choice of strategy in a random fashion but in accordance with some fixed set of probabilities. Statistical games A statistical game can be regarded as a game between nature (which controls the relevant features of a population) and the statistician (who is trying to make a decision about the population).

20 Page 6 CT6-01: Decision theory An example of a statistical game is where a statistician wishes to determine whether a coin is fair (F) or biased (B) towards tails, with nature acting as his/her opponent. A decision function sets out the action for the statistician to take based on each outcome of an event, eg the event might be a single toss of the coin and one example of a decision function is: F if the coin toss results in heads and B if the coin toss results in tails. A risk function sets out the expected loss for a particular decision function and a given actual state of nature. In order to calculate the expected loss, probabilities must be assigned to each state of nature. The statistician can then use the minimax criterion to find the decision function that minimises the maximum value of risk function or the Bayes criterion to find the decision function that minimises the expected value of the risk function.

21 CT6-0: Bayesian statistics Page 7 Chapter Summary Bayesian estimation vs classical estimation A common problem in statistics is to estimate the value of some unknown parameter. The classical approach to this problem is to treat as a fixed, but unknown, constant and use sample data to estimate its value. For example, if represents some population mean then its value may be estimated by a sample mean. The Bayesian approach is to treat as a random variable. Prior distribution of θ The prior distribution of represents the knowledge available about the possible values of before the collection of any sample data. The prior PDF of q is denoted f ( q ). Likelihood function of the sample data A likelihood function is then determined, based on a set of observations x = ( x1, x,..., xn ). The likelihood function is just the same as the joint density (or, in the discrete case, the joint probability) of X 1, X,, X n. However, the likelihood is considered to be a function of rather than one of x 1, x,, x n. Since we are assuming that the random variables X 1,, X n are independent, the joint density function f X ( x1, x,, xn) is equal to the product of the individual density functions f X x i ( i). The likelihood function is therefore: L( x q ) = fx q ( xi ) n i= 1 Posterior distribution of θ i The PDF (or probability function) of the prior distribution and the likelihood function are then combined to obtain the PDF (or probability function) of the posterior distribution for.

22 Page 8 CT6-0: Bayesian statistics The relationship is given by Bayes Formula as: Posterior PDF μ Prior PDF Likelihood Loss functions ( q ) μ ( q) ( q) f x f L x A loss function, such as quadratic (or squared) error loss, absolute error loss or zero-one error loss gives a measure of the loss incurred when ˆ q is used as an estimator of the true value of q. In other words, it measures the seriousness of an incorrect estimator. Under squared error loss, the Bayesian estimator that minimises the expected loss function is the mean of the posterior distribution. Under absolute error loss, it is the median of the posterior distribution that minimises the expected loss function. Under zero-one error loss, the mode of the posterior distribution minimises the expected loss function. Conjugate prior For a given likelihood, if the prior distribution leads to a posterior distribution belonging to the same family as the prior distribution, then this prior is called the X l Poi l then a conjugate conjugate prior for this likelihood. For example if ( ) prior for l is a gamma distribution, since this leads to a gamma posterior for l. Uninformative prior In the absence of any other information as to the prior distribution of a parameter, it is useful to use an uninformative prior, which assumes that the parameter is equally likely X l Poi l then an uninformative prior to take any possible value. For example if ( ) for l is a U ( 0, ) distribution.

23 CT6-0: Bayesian statistics Page 8a Chapter Formulae Bayes Formula PAB ( i) PB ( i) PB ( i A) = Â PAB ( ) PB ( ) i i ( q) PA ( q) f f( q A) = Ú PA ( q) f( q) dq i (discrete form) (continuous form) Posterior distribution Posterior PDF μ Prior PDF Likelihood Loss function Loss Bayesian estimator Squared error ( ) mean Absolute error median Zero-one 0 if ˆ q= q 1 if ˆ q π q mode

24 Page 8b CT6-0: Bayesian statistics This page has been left blank so that you can keep the chapter summaries together as a revision tool.

25 CT6-03: Loss distributions Page 15 3 Estimation The methods of maximum likelihood, moments and percentiles can be used to fit distributions to sets of data. The fit of the distribution can be tested formally by using a test. The method of percentiles is outlined in Section 3.7; the other methods and the test have been covered in Subject CT3, Probability and Mathematical Statistics. The formulae for the densities, the moments and the moment generating functions (where they exist) for the distributions discussed in this chapter are given in the Formulae and Tables for Actuarial Examinations. We will give a brief summary of the method of moments and of maximum likelihood estimation for those students who have not taken Subject CT3 recently. An example of using the chi square distribution will also be given, in case you have forgotten how to carry out this test. 3.1 The method of moments For a distribution with r parameters, the moments are as follows: m j = 1 Â n j x n i i = 1 j = 1, r where m j = E(X j ), a function of the unknown parameter,, being estimated n = the sample size x i = the i th value in the sample The estimate for the parameter,, can be determined by solving the equation above. Where there is more than one parameter, they can be determined by solving the simultaneous equations for each m j. To obtain a method of moments estimator for a parameter, we equate the corresponding sample and population non-central moments. So, for example, if we were trying to estimate the value of a single parameter, and we had a sample of n claims whose sizes were x 1, x,, x n, we would solve the equation: n 1 EX ( ) = Â x n i = 1 i ie we would equate the first non-central moments for the population and the sample.

26 Page 16 CT6-03: Loss distributions If we were trying to find estimates for two parameters (for example if we were fitting a gamma distribution and needed to find estimates for both parameters), we would solve the simultaneous equations: n 1 EX ( ) = Â x and n i = 1 i EX ( ) n 1 = Â xi n i = 1 In fact in the two parameter case, estimates are usually obtained by equating sample and population means and variances. If we define the sample variance to have a denominator n, this will give the same estimates as would be obtained by equating the first two noncentral moments. n k 1 More generally, we use as many equations of the form EX ( ) = Â x, k 1,, as are needed to determine estimates of the relevant parameters. n i = 1 k i 3. Maximum Likelihood Estimation The likelihood function of a random variable, X, is the probability (or PDF) of observing what was observed given a hypothetical value of the parameter,. The maximum likelihood estimate (MLE) is the one that yields the highest probability (or PDF), ie that maximises the likelihood function. For the sample in Section 3.1 above, the likelihood function L() can be expressed as: n L( q) = P( X = xi q ) for a discrete random variable, X, or i = 1 n L( q) = f ( xi q ) for a continuous random variable, X. i = 1 To determine the MLE the likelihood function needs to be maximised. Often it is practical to consider the loglikelihood function: n l( q) = logl( q) = Âlog P( X = xi q ) for a discrete random variable, X, or i = 1 ( q) log ( q) log ( q) l L f x n = =Â i = 1 i for a continuous random variable, X.

27 CT6-03: Loss distributions Page 17 If l() can be differentiated with respect to, the MLE, expressed as ˆ q, satisfies the expression: d l( qˆ ) dq = 0 Where there is more than one parameter, the MLEs for each parameter can be determined by taking partial derivatives of the loglikelihood function and setting each to zero. The determination of MLEs when the data is incomplete is covered in Chapter 4. The steps involved in finding a maximum likelihood estimate (MLE) are as follows: (1) Write down the likelihood function for the available data. If the likelihood is based on a set of known values x 1, x,, x n, then the likelihood function will take the form f ( x1) f( x) f( x n ), where f ( x) is the PDF of X where X is a continuous random variable (or P( X = x1) P( X = x) P( X = xn ) in the case where X is a discrete random variable). () Take natural logs. This will usually simplify the algebra. (3) Maximise the log-likelihood function. This usually involves differentiating the log-likelihood function with respect to each of the unknown parameters, and setting the resulting expression(s) equal to zero. (4) Solve the resulting equation(s) to find the MLEs of the parameters. Check that the values you have found do maximise the likelihood function. This will usually involve differentiating a second time. Note that where there are two or more parameters to estimate, checking for a maximum is complicated and is very unlikely to be required for the Subject CT6 exam. We will now look at the distributions described above and consider how the parameters can be estimated in each case. 3.3 The exponential and gamma distributions It is possible to use the method of maximum likelihood (ML) or the method of moments to estimate the parameter of the exponential distribution.

28 Page 18 CT6-03: Loss distributions Example An insurance company uses an exponential distribution to model the cost of repairing insured vehicles that are involved in accidents. Find the maximum likelihood estimate of the mean cost, given that the average cost of repairing a sample of 1,000 vehicles was,00. Solution Let X 1, X,, X n denote the individual repair costs (where n 1, 000 ). The likelihood of obtaining these values for the costs, if they come from an exponential distribution with parameter is: n xi i L e e e i1 1 (where x n n x i i1 n x n nx denotes the average claim amount). To find the MLE, we need to find the value of that maximises the likelihood, or, alternatively, the value that maximises the log-likelihood: log Lnlog nx Differentiating to look for stationary points: n log L nx Setting this to zero gives: 1/ x ie 1/ 00, The second derivative is n log L 0, which shows that this is a maximum. Alternatively you can note that the likelihood function is continuous and is always n nx e positive (by necessity) and that point that we find must be a maximum. 0 as 0 or. So any stationary

29 CT6-03: Loss distributions Page 7 5 Mixture distributions The exponential distribution is one of the simplest models for insurance losses. Suppose that each individual in a large insurance portfolio incurs losses according to an exponential distribution. What we re thinking of here is that the amounts of Mr Ferrari s insurance claims over a period of years can be assumed to have an exponential distribution with a certain value of. Practical knowledge of almost any insurance portfolio reveals that the means of these various distributions will differ among the policyholders. Thus the description of the losses in the portfolio is that each loss follows its own exponential distribution, ie the exponential distributions have means that differ from individual to individual. So, although Mr Trabant s claims also have an exponential distribution, the value of is different for him. So rather than having identically distributed claim amounts X ~ Exp( ), what we have are claim amounts whose distributions are conditional on the value of, ie X ~ Exp( ). A description of the variation among the individual means must now be found. One way to do this is to assume that the exponential means themselves follow a distribution. In the exponential case, it is convenient to make the following assumption. Let i = 1/ i be the reciprocal of the mean loss for the i-th policyholder. Assume that the variation among the i can be described by a known gamma distribution Ga(, ), ie assume that ~ Ga (, ) where f ( ) exp( ) ( ) 1, 0. Take particular note that this is a PDF in with known values of and. (There are no x s in the above PDF.) This formulation has much in common with that used in Bayesian estimation. Indeed, the fundamental idea in Bayesian estimation is that the parameter of interest (here, ) can be treated as a random variable with a known distribution. Notice, however, that the purpose here is not to estimate the individual i, but to describe the aggregate losses over the whole portfolio.

30 Page 8 CT6-03: Loss distributions Estimation of the individual i can be treated by Bayesian estimation, when the Ga(, ) distribution would be referred to as a prior distribution. In this problem of describing the losses over the whole portfolio, the Ga(, ) distribution is used to average the exponential distributions; the Ga(, ) distribution is referred to as the mixing distribution and the resulting loss distribution as a mixture distribution. So we are trying to achieve something different from what we achieved in Chapter. There we were looking for a point estimate of the parameter value, using information from a prior distribution as well as from our random sample. Here we are trying to find the overall distribution of the actual claim amounts, assuming the given make-up of the different values for different policies in the portfolio. The random variable X represents the amount of a single randomly selected claim and E( X ) represents the mean claim amount for all risks in the portfolio. To find the overall distribution of claim amounts, we need to work out the marginal distribution of X. This is obtained by integrating the joint density function f X, over all possible values of. The PDF of the mixture (or marginal) distribution of X is fx ( x ) = Ú fx, l ( x, l) dl 0 = Ú fl ( ) f ( x ) d 0 XΩl l l l = = a d Ú G( a ) 0 a -1 exp( - ) exp( - x) d l dl l l l a d a Ú exp{ -( x + ) } d G( a ) 0 l d l l We now make the integrand look like the PDF of a Ga ( + 1, x + ) a d distribution: fx ( x ) = a d G( a + 1) G( a ) a + 1 ( x + d ) a ( x + d ) G ( a + 1) + 1 Ú 0 a l exp{ -( x + d) l} dl

31 CT6-03: Loss distributions Page 9 Integrating the PDF over all possible values of l will give us 1, so that a fx ( x ) d G( a + 1) = G( a ) a + 1 ( x + d ) a ad =, x > 0 a + 1 ( x + d ) which can be recognised as the PDF of the Pareto distribution, Pa(, ). This result gives a very nice interpretation of the Pareto distribution: Pa(, ) arises when exponentially distributed losses are averaged using a Ga(, ) mixing distribution. In order to see what is going on here, you might like to compare this with the corresponding process for finding an unconditional probability from a conditional one in the discrete case. In order to calculate an unconditional probability P( X ), we can write: P( X x) P( X xy y ) P( Y y ) P( X xy y ) P( Y y ) 1 1 where the summation is taken over all possible values of Y. Here we carry out the same process, except that probabilities are replaced by probability density functions, and summation is replaced by integration. The parameter can take any value in a continuous range of possibilities, so we integrate over all possible values. Example Another generalisation of the Pareto distribution uses the idea of a mixture distribution discussed earlier. We ve seen that if losses are exponential with mean 1/ l and l ~ Ga( a, d ), then the mixture distribution of losses is Pa( ad, ). This can be generalised if it is assumed that the losses are Ga( k, l ) and l ~ Ga( a, d ). If k = 1 then losses are exponential with mean 1/ l since Ga ( 1, l) Exp ( 1 / l ) and the Pa( ad, ) mixture distribution is obtained exactly as before.

32 Page 30 CT6-03: Loss distributions For general k, the PDF of the mixture distribution of the loss, X, is f X ( x ) = Ú fl ( l) f 0 Ω ( x l) dl X l = a d Ú0 k a -1 l k -1 l exp( - dl) x exp( -lx) dl G( a ) G( k) = a k - 1 d x G( a ) G( k) Ú0 a + k -1 exp( -( + x) ) d l d l l We now make the integrand look like the PDF of a Ga ( + k, + x) a d distribution: fx ( x ) = G ( a k) a + ( d x) a k - 1 d x + + G( a) G( k) k Ú + 0 G a + ( d x) ( a + k) k a + k -1 l exp( -( d + x) l) dl Integrating the PDF over all possible values of l will give us 1, so that fx ( x ) = a G( a + k) d x G( a ) G( k) ( d + x) k -1 a + k, x > 0 which can be recognised as the PDF of the generalised Pareto distribution, Pa a, d, k. ( ) In the Tables the parameter is called. This is the situation referred to in Section.1. If we regard the generalised Pareto as the result of mixing together Gamma( k, ) distributions whose parameters come from a Gamma(, ) distribution, then we can find the mean of the generalised Pareto by calculating: k 1 a a-1 -d l EX ( ) = Ú d l e dl l ( a) 0 G where k / is the mean of the Gamma( k, ) distribution. If you follow the algebra through (use the usual procedure of making the integral look like another gamma distribution), you should find that you get the formula k 1 as required.

33 CT6-03: Loss distributions Page 3 Question 3.15 The annual number of claims from an individual policy in a portfolio has a Poisson( ) distribution. The variability in among policies is modelled by assuming that over the portfolio, individual values of have a Gamma(, ) distribution. Derive the mixture distribution for the annual number of claims from each policy in the portfolio. Question 3.16 Claim numbers from individual policies in a portfolio have a Bin( n, p) distribution. The parameter p varies over the portfolio with a Beta(, ) distribution. Find the mixture distribution.

34 Page 30b CT6-03: Loss Distributions This page has been deliberately left blank

35 CT6-03: Loss distributions Page 33 Chapter 3 Summary Loss distributions Individual claim amounts can be modelled using a loss distribution. Loss distributions are often positively skewed and long-tailed. The (cumulative) distribution function of X is denoted by FX ( x). It is defined by the equation: FX ( x) P( X x). The (probability) density function of X is denoted by f X ( x). It is defined by the equation: f ( x) F ( x), wherever this derivative exists. X X Distributions such as the exponential, normal, lognormal, gamma, Pareto, generalised Pareto, Burr or Weibull distribution are commonly used to model individual claim amounts. Once the form of the loss distribution has been decided upon, the values of the parameters must be estimated. This may be done using the method of maximum likelihood, the method of moments or the method of percentiles. Goodness of fit can then be checked using a chi square test. Method of maximum likelihood The steps involved in finding a maximum likelihood estimate (MLE) are as follows: (1) Write down the likelihood function for the available data. () Take natural logs. This will usually simplify the algebra. (3) Maximise the log-likelihood function. This usually involves differentiating the log-likelihood function with respect to each of the unknown parameters, and setting the resulting expression(s) equal to zero. (4) Solve the resulting equation(s) to find the MLEs of the parameters. (5) Differentiating the log-likelihood function a second time to check that the estimates are indeed maxima.

36 Page 34 CT6-03: Loss distributions Method of moments The method of moments involves equating population and sample moments to solve for the unknown parameter values. For example, if there is just one parameter to estimate, we could equate the population mean with the sample mean. If there are two parameters to estimate, we could equate the first two non-central population moments with the equivalent non-central sample moments. Equivalently, we could equate the first two central population moments with the equivalent central sample moments, noting that (for equivalence) we would need to use the n-denominator sample variance. Method of percentiles The method of percentiles involves equating population and sample percentiles to solve for the unknown parameter values. For example, if there is just one parameter to estimate, we could equate the population median with the sample median. If there are two parameters to estimate, we could equate the population lower and upper quartiles with the sample lower and upper quartiles. Mixture distributions Let X be a random variable representing losses on an insurance portfolio. Suppose that the distribution of X depends on the value of an unknown parameter l where l is itself a random variable. For example l may vary by policyholder. The mixture distribution of X is also known as the marginal or unconditional distribution of X. It represents the overall distribution of losses, once the effects of the different l s have been averaged out.

37 CT6-03: Loss distributions Page 34a Chapter 3 Formulae Moment generating function of X M () t E( e ) X tx PDF of a mixture (or marginal) distribution f ( x ) = f ( l) f ( x l) dl X Ú all l l Gamma distribution Parameters: 0, 0 1 fx ( x) x e ( ) XΩl x, x 0 X ~ Gamma(, ) X ~ Note that the exponential distribution is a special case of the gamma distribution where a = 1. Lognormal distribution Parameters: - < m <, s > 0 f X 1 È 1 Êlogx - m ˆ ( x) = expí- Á xs p Í Ë s Î, x > 0 Pareto distribution Parameters: a > 0, l > 0 fx a al ( x) = ( l + x) a + 1 Ê l ˆ, x > 0 FX ( x) = 1- Á Ë l + x a

38 Page 34b CT6-03: Loss distributions Generalised Pareto distribution Parameters: a > 0, l > 0, k > 0 f X a k-1 G ( a + k) l x ( x) =, x > 0 G a + k ( a) G ( k)( l + x) Burr distribution Parameters: a > 0, l > 0, g > 0 f X a g -1 agl x ( x) = g ( l + x ) a+ 1 Ê l ˆ, x > 0 FX ( x) = 1- Á Ë g l + x a Weibull distribution Parameters: c > 0, g > 0 g g -1 -cx f ( x) = cg x e, x > 0 F ( x) = 1- e X X g -cx

39 CT6-04: Reinsurance Page 5 So the graph looks like this: We are now in a position to consider the statistical calculations relating to reinsurance arrangements. 1.1 Excess of loss reinsurance In excess of loss reinsurance, the insurer will pay any claim in full up to an amount M, the retention level; any amount above M will be borne by the reinsurer. In this section the company refers to the direct writer. The excess of loss reinsurance arrangement can be written in the following way: if the claim is for amount X then the insurer will pay Y where: Y = X if X M Y = M if X M. The reinsurer pays the amount Z X Y. Question 4.1 Write down an expression for Y if only a layer between M and M is reinsured.

40 Page 6 CT6-04: Reinsurance The insurer s liability is affected in two obvious ways by reinsurance: (i) (ii) the mean amount paid is reduced; the variance of the amount paid is reduced. Both these conclusions are simple consequences of the fact that excess of loss reinsurance puts an upper limit on large claims. The mean amounts paid by the insurer and the reinsurer under excess of loss reinsurance can now be obtained. Observe that the mean amount paid by the insurer without reinsurance is E(X) = Ú x f( x) dx (1.1) 0 where f( x ) is the PDF of the claim amount X. With a retention level of M the mean amount paid by the insurer becomes M Ú 0 EY ( ) = xf( x) dx+ MP( X> M ). (1.) This is because M EY ( ) = xf( xdx ) + Mf( xdx ) Ú Ú, from the definition of Y. 0 M Similarly, we can calculate EY ( ) using: ( ) M Ú EY = x f( xdx ) + M f( xdx ) 0 M Ú 0 M ( ) = x f( x) dx+ M P X > M Then ( ) ( ) È ( ) var Y = E Y - Î E Y. Ú More generally, the moment generating function of Y, the amount paid by the insurer, is M tx tm MY () ty t = Ee ( ) = Ú e f( x) dx + e P( X > M ). 0

41 CT6-04: Reinsurance Page 7 Question 4. Find E( Y ) when X has a Pareto distribution with parameters 00 and 6, and M 80. Under excess of loss reinsurance, the reinsurer will pay Z where: Z = 0 if X M Z = X M if X > M. With a retention level of M the mean amount paid by the reinsurer becomes: EZ ( ) = Ú ( x-m) f( xdx ), from the definition of Z (1.3) M Similarly, we can calculate E( Z ) using: E( Z ) = ( x-m) f( x) dx Then var ( Z) E( Z ) ÈE( Z) Ú M = - Î. More generally, the moment generating function of Z, the amount paid by the reinsurer, is: tz M t0 tx ( -M) () = ( ) = + M t E e e f( x) dx e f( x) dx Z Ú Ú 0 M

42 Page 8 CT6-04: Reinsurance 1. The reinsurer s conditional claims distribution Now consider reinsurance (once again) from the point of view of the reinsurer. The reinsurer may have a record only of claims that are greater than M. If a claim is for less than M the reinsurer may not even know a claim has occurred. The reinsurer thus has the problem of estimating the underlying claims distribution when only those claims greater than M are observed. The statistical terminology is to say that the reinsurer observes claims from a truncated distribution. In this case the values observed by the reinsurer relate to a conditional distribution, since the numbers are conditional on the original claim amount exceeding the retention limit. Let W be the random variable with this truncated distribution. Then: W = X - M X > M Suppose that the underlying claim amounts have PDF f ( x) and CDF F( x). Suppose that the reinsurer is only informed of claims greater than the retention M and has a record of w = x M. What is the PDF g(w) of the amount, w, paid by the reinsurer? The argument goes as follows: PW ( < w) = P( X< w+ M X> M) w+ M f( x) = Ú dx (using Bayes' Formula) M 1 - FM ( ) = Fw ( + M) - FM ( ) 1 - FM ( ) (using the definition of the CDF) Differentiating w.r.t. w, the PDF of the reinsurer s claims is fw ( + M) gw ( ) =, w > 0. (1.4) 1 - FM ( ) Note that this is just the original PDF applied to the gross amount w+ M, divided by the probability that the claim exceeds M. Question 4. Using the notation above, if X is Exp( ), find the distribution of W.

43 CT6-04: Reinsurance Page 9 Question 4. What if X Pa( a, l)? It is also possible that the reinsurer is aware of all the claims that are made under the original policies. In this case the distribution of the reinsurer s outgo may include those claims on which it does not in fact pay anything. In this case we are looking at the unconditional distribution of the reinsurer s outgo Z, where: Z 0 X M Z X M X M The calculation of the mean and variance of the reinsurer s outgo in this case will be similar to the corresponding calculations for the insurer. Example If claims from a portfolio have a N ( 500, 400 ) distribution, and there is a retention limit of M 550, find the mean amount paid by the reinsurer on all claims. Solution We want 550 Ú X Ú X, where f x EZ ( ) = 0 f ( x) dx+ ( x-550) f ( x) dx X ( ) is the PDF of x the N ( 500, 400 ) distribution. The first integral equals zero. Substituting u in the second integral, we get: EZ ( ) = u (0u 50) e Ú - du p.5 È u = 0 Í- e -50[ 1 -F(.5)] = Í p Î The first term in square brackets can be evaluated directly..5 We will give some general formulae later in this chapter which will enable us to calculate these types of integrals relatively quickly.

44 Page 10 CT6-04: Reinsurance If we now use W to denote the amount payable by the reinsurer on claims in which it is involved, ie W Z Z 0 X M X M, then: EW ( ) EZ ( ) PZ ( 0) EZ ( ) P( X M) This follows from equation (1.4). 1.3 Proportional reinsurance In proportional reinsurance the insurer pays a fixed proportion of the claim, whatever the size of the claim. Using the same notation as above, the proportional reinsurance arrangement can be written as follows: if the claim is for an amount X then the company will pay Y where Y X 0 1. The parameter is known as the retained proportion or retention level; note that the term retention level is used in both excess of loss and proportional reinsurance though it means different things. As the amount paid by the insurer on a claim X is Y = a X and the amount paid by the reinsurer is Z = ( 1-a ) X, the distribution of both of these amounts can be found by a simple change of variable. Question 4. Claims occur as a generalised Pareto distribution with parameters 6, 00 and k 4. A proportional reinsurance arrangement is in force with a retained proportion of 80%. Find the mean and variance of the amount paid by the insurer and the reinsurer on an individual claim.

45 CT6-04: Reinsurance Page 3 And for Year : 110 Ê110ˆ 1710 EY ( ) = - Á = Ë So the percentage increase from Year 0 to Year 1 is 7.%, and the percentage increase from Year 1 to Year is 6.9%. Note that these figures are less than 10%, as expected. Question 4.1 What is the limit of EY ( n) as n tends to infinity?

46 Page 4 CT6-04: Reinsurance 4 Estimation Consider the problem of estimation in the presence of excess of loss reinsurance. Suppose that the claims record shows only the net claims paid by the insurer. A typical claims record might be x1, x, M, x3, M, x4, x5,... (4.1) and an estimate of the underlying gross claims distribution is required. As before, we wish to estimate the parameters for the distribution we ve assumed for the claims. The method of moments is not available since even the mean claim amount cannot be computed. On the other hand, it may be possible to use the method of percentiles without alteration; this would happen if the retention level M is high and only the higher sample percentiles were affected by the (few) reinsurance claims. The statistical terminology for a sample of the form (4.1) is censored. In general, a censored sample occurs when some values are recorded exactly and the remaining values are known only to exceed a particular value, here the retention level M. Maximum likelihood can be applied to censored samples. The likelihood function is made up of two parts. If the values of x 1, x,..., x n are recorded exactly these contribute a factor of n L1( q) = Pf( x i ; q) 1 If a further m claims are referred to the reinsurer, then the insurer records a payment of M for each of these claims. These censored values then contribute a factor m L ( q ) = P Pr( X > M) ie [Pr( X > M)] m 1 The complete likelihood function is n L( q) = P f( x ; q) [1- F( M; q)] 1 i where F(.; ) is the CDF of the claims distribution. m

47 CT6-04: Reinsurance Page 31 Chapter 4 Summary Reinsurance Reinsurance is insurance for insurance companies. By using reinsurance, the insurer seeks to protect itself from large claims. The mean amount paid by the insurer is reduced, and the variance of the amount paid by the insurer is reduced. Reinsurance may be proportional or non-proportional (ie individual excess of loss). Proportional reinsurance Under proportional reinsurance, the insurer and the reinsurer split the claim in pre-defined proportions. For a claim amount X, the amount paid by the insurer is Y = a X and the amount paid by the reinsurer is Z ( 1 a ) proportion or retention level, 0< a < 1. Non-proportional reinsurance (individual excess of loss) = - X where a is known as the retained Under individual excess of loss, the insurer will pay any claim in full up to an amount M, the retention level; any amount above M will be met by the reinsurer. For a claim amount, X, the amount paid by the insurer is: Y Ï X if X M = Ì ÓM if X > M The amount paid by the reinsurer is: Z Ï0 if X M = Ì ÓX - M if X > M

48 Page 3 CT6-04: Reinsurance Reinsurer s conditional claims distribution It may be the case that the reinsurer is only informed of claims greater than the retention level M. In this case, the reinsurer observes claims from a truncated (or conditional) distribution. Let W be the random variable associated with this distribution, then: W = Z Z > 0 = X - M X > M The PDF of the reinsurer s conditional distribution is given by: g W Excesses f X ( w+ M) ( w) = 1 - F ( M) X When a policy excess applies, the policyholder pays for the first part of each loss up to an excess level L ; any amount greater than L will be met by the insurer. The positions of the policyholder and the insurer as far as losses are concerned are the same as those of the insurer and the reinsurer respectively under individual excess of loss reinsurance. When a policy excess applies, the insurer s conditional distribution takes the same form as that of the reinsurer s conditional distribution above. Inflation and individual excess of loss reinsurance If claims are inflated by a factor of k but the retention level remains fixed at M then the amount paid by the insurer is: Y ÏkX if kx M = Ì Ó M if kx > M The amount paid by the reinsurer is: Z Ï0 if kx M = Ì ÓkX - M if kx > M

Practice Exam 1. (A) (B) (C) (D) (E) You are given the following data on loss sizes:

Practice Exam 1. (A) (B) (C) (D) (E) You are given the following data on loss sizes: Practice Exam 1 1. Losses for an insurance coverage have the following cumulative distribution function: F(0) = 0 F(1,000) = 0.2 F(5,000) = 0.4 F(10,000) = 0.9 F(100,000) = 1 with linear interpolation

More information

3 Continuous Random Variables

3 Continuous Random Variables Jinguo Lian Math437 Notes January 15, 016 3 Continuous Random Variables Remember that discrete random variables can take only a countable number of possible values. On the other hand, a continuous random

More information

ASM Study Manual for Exam P, First Edition By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA Errata

ASM Study Manual for Exam P, First Edition By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA Errata ASM Study Manual for Exam P, First Edition By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA (krzysio@krzysio.net) Errata Effective July 5, 3, only the latest edition of this manual will have its errata

More information

This exam is closed book and closed notes. (You will have access to a copy of the Table of Common Distributions given in the back of the text.

This exam is closed book and closed notes. (You will have access to a copy of the Table of Common Distributions given in the back of the text. TEST #3 STA 5326 December 4, 214 Name: Please read the following directions. DO NOT TURN THE PAGE UNTIL INSTRUCTED TO DO SO Directions This exam is closed book and closed notes. (You will have access to

More information

Errata for the ASM Study Manual for Exam P, Fourth Edition By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA

Errata for the ASM Study Manual for Exam P, Fourth Edition By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA Errata for the ASM Study Manual for Exam P, Fourth Edition By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA (krzysio@krzysio.net) Effective July 5, 3, only the latest edition of this manual will have its

More information

MATH/STAT 3360, Probability Sample Final Examination Model Solutions

MATH/STAT 3360, Probability Sample Final Examination Model Solutions MATH/STAT 3360, Probability Sample Final Examination Model Solutions This Sample examination has more questions than the actual final, in order to cover a wider range of questions. Estimated times are

More information

Parameter Estimation

Parameter Estimation Parameter Estimation Chapters 13-15 Stat 477 - Loss Models Chapters 13-15 (Stat 477) Parameter Estimation Brian Hartman - BYU 1 / 23 Methods for parameter estimation Methods for parameter estimation Methods

More information

CONTINUOUS RANDOM VARIABLES

CONTINUOUS RANDOM VARIABLES the Further Mathematics network www.fmnetwork.org.uk V 07 REVISION SHEET STATISTICS (AQA) CONTINUOUS RANDOM VARIABLES The main ideas are: Properties of Continuous Random Variables Mean, Median and Mode

More information

ASM Study Manual for Exam P, Second Edition By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA Errata

ASM Study Manual for Exam P, Second Edition By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA Errata ASM Study Manual for Exam P, Second Edition By Dr. Krzysztof M. Ostaszewski, FSA, CFA, MAAA (krzysio@krzysio.net) Errata Effective July 5, 3, only the latest edition of this manual will have its errata

More information

Course 1 Solutions November 2001 Exams

Course 1 Solutions November 2001 Exams Course Solutions November Exams . A For i =,, let R = event that a red ball is drawn form urn i i B = event that a blue ball is drawn from urn i. i Then if x is the number of blue balls in urn, ( R R)

More information

Joint Probability Distributions and Random Samples (Devore Chapter Five)

Joint Probability Distributions and Random Samples (Devore Chapter Five) Joint Probability Distributions and Random Samples (Devore Chapter Five) 1016-345-01: Probability and Statistics for Engineers Spring 2013 Contents 1 Joint Probability Distributions 2 1.1 Two Discrete

More information

This exam contains 13 pages (including this cover page) and 10 questions. A Formulae sheet is provided with the exam.

This exam contains 13 pages (including this cover page) and 10 questions. A Formulae sheet is provided with the exam. Probability and Statistics FS 2017 Session Exam 22.08.2017 Time Limit: 180 Minutes Name: Student ID: This exam contains 13 pages (including this cover page) and 10 questions. A Formulae sheet is provided

More information

Mixture distributions in Exams MLC/3L and C/4

Mixture distributions in Exams MLC/3L and C/4 Making sense of... Mixture distributions in Exams MLC/3L and C/4 James W. Daniel Jim Daniel s Actuarial Seminars www.actuarialseminars.com February 1, 2012 c Copyright 2012 by James W. Daniel; reproduction

More information

SPRING 2007 EXAM C SOLUTIONS

SPRING 2007 EXAM C SOLUTIONS SPRING 007 EXAM C SOLUTIONS Question #1 The data are already shifted (have had the policy limit and the deductible of 50 applied). The two 350 payments are censored. Thus the likelihood function is L =

More information

Exam P Review Sheet. for a > 0. ln(a) i=0 ari = a. (1 r) 2. (Note that the A i s form a partition)

Exam P Review Sheet. for a > 0. ln(a) i=0 ari = a. (1 r) 2. (Note that the A i s form a partition) Exam P Review Sheet log b (b x ) = x log b (y k ) = k log b (y) log b (y) = ln(y) ln(b) log b (yz) = log b (y) + log b (z) log b (y/z) = log b (y) log b (z) ln(e x ) = x e ln(y) = y for y > 0. d dx ax

More information

Notes for Math 324, Part 20

Notes for Math 324, Part 20 7 Notes for Math 34, Part Chapter Conditional epectations, variances, etc.. Conditional probability Given two events, the conditional probability of A given B is defined by P[A B] = P[A B]. P[B] P[A B]

More information

Question Points Score Total: 76

Question Points Score Total: 76 Math 447 Test 2 March 17, Spring 216 No books, no notes, only SOA-approved calculators. true/false or fill-in-the-blank question. You must show work, unless the question is a Name: Question Points Score

More information

Solutions to the Spring 2015 CAS Exam ST

Solutions to the Spring 2015 CAS Exam ST Solutions to the Spring 2015 CAS Exam ST (updated to include the CAS Final Answer Key of July 15) There were 25 questions in total, of equal value, on this 2.5 hour exam. There was a 10 minute reading

More information

Course 4 Solutions November 2001 Exams

Course 4 Solutions November 2001 Exams Course 4 Solutions November 001 Exams November, 001 Society of Actuaries Question #1 From the Yule-Walker equations: ρ φ + ρφ 1 1 1. 1 1+ ρ ρφ φ Substituting the given quantities yields: 0.53 φ + 0.53φ

More information

1: PROBABILITY REVIEW

1: PROBABILITY REVIEW 1: PROBABILITY REVIEW Marek Rutkowski School of Mathematics and Statistics University of Sydney Semester 2, 2016 M. Rutkowski (USydney) Slides 1: Probability Review 1 / 56 Outline We will review the following

More information

Lecture 2: Repetition of probability theory and statistics

Lecture 2: Repetition of probability theory and statistics Algorithms for Uncertainty Quantification SS8, IN2345 Tobias Neckel Scientific Computing in Computer Science TUM Lecture 2: Repetition of probability theory and statistics Concept of Building Block: Prerequisites:

More information

Department of Statistical Science FIRST YEAR EXAM - SPRING 2017

Department of Statistical Science FIRST YEAR EXAM - SPRING 2017 Department of Statistical Science Duke University FIRST YEAR EXAM - SPRING 017 Monday May 8th 017, 9:00 AM 1:00 PM NOTES: PLEASE READ CAREFULLY BEFORE BEGINNING EXAM! 1. Do not write solutions on the exam;

More information

ACTEX CAS EXAM 3 STUDY GUIDE FOR MATHEMATICAL STATISTICS

ACTEX CAS EXAM 3 STUDY GUIDE FOR MATHEMATICAL STATISTICS ACTEX CAS EXAM 3 STUDY GUIDE FOR MATHEMATICAL STATISTICS TABLE OF CONTENTS INTRODUCTORY NOTE NOTES AND PROBLEM SETS Section 1 - Point Estimation 1 Problem Set 1 15 Section 2 - Confidence Intervals and

More information

Math 416 Lecture 3. The average or mean or expected value of x 1, x 2, x 3,..., x n is

Math 416 Lecture 3. The average or mean or expected value of x 1, x 2, x 3,..., x n is Math 416 Lecture 3 Expected values The average or mean or expected value of x 1, x 2, x 3,..., x n is x 1 x 2... x n n x 1 1 n x 2 1 n... x n 1 n 1 n x i p x i where p x i 1 n is the probability of x i

More information

Part (A): Review of Probability [Statistics I revision]

Part (A): Review of Probability [Statistics I revision] Part (A): Review of Probability [Statistics I revision] 1 Definition of Probability 1.1 Experiment An experiment is any procedure whose outcome is uncertain ffl toss a coin ffl throw a die ffl buy a lottery

More information

Course: ESO-209 Home Work: 1 Instructor: Debasis Kundu

Course: ESO-209 Home Work: 1 Instructor: Debasis Kundu Home Work: 1 1. Describe the sample space when a coin is tossed (a) once, (b) three times, (c) n times, (d) an infinite number of times. 2. A coin is tossed until for the first time the same result appear

More information

1 Introduction. P (n = 1 red ball drawn) =

1 Introduction. P (n = 1 red ball drawn) = Introduction Exercises and outline solutions. Y has a pack of 4 cards (Ace and Queen of clubs, Ace and Queen of Hearts) from which he deals a random of selection 2 to player X. What is the probability

More information

MAT 271E Probability and Statistics

MAT 271E Probability and Statistics MAT 271E Probability and Statistics Spring 2011 Instructor : Class Meets : Office Hours : Textbook : Supp. Text : İlker Bayram EEB 1103 ibayram@itu.edu.tr 13.30 16.30, Wednesday EEB? 10.00 12.00, Wednesday

More information

Notes for Math 324, Part 19

Notes for Math 324, Part 19 48 Notes for Math 324, Part 9 Chapter 9 Multivariate distributions, covariance Often, we need to consider several random variables at the same time. We have a sample space S and r.v. s X, Y,..., which

More information

Creating New Distributions

Creating New Distributions Creating New Distributions Section 5.2 Stat 477 - Loss Models Section 5.2 (Stat 477) Creating New Distributions Brian Hartman - BYU 1 / 18 Generating new distributions Some methods to generate new distributions

More information

Practice Problems Section Problems

Practice Problems Section Problems Practice Problems Section 4-4-3 4-4 4-5 4-6 4-7 4-8 4-10 Supplemental Problems 4-1 to 4-9 4-13, 14, 15, 17, 19, 0 4-3, 34, 36, 38 4-47, 49, 5, 54, 55 4-59, 60, 63 4-66, 68, 69, 70, 74 4-79, 81, 84 4-85,

More information

Errata and updates for ASM Exam C/Exam 4 Manual (Sixth Edition) sorted by date

Errata and updates for ASM Exam C/Exam 4 Manual (Sixth Edition) sorted by date Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Date 1 Errata and updates for ASM Exam C/Exam 4 Manual (Sixth Edition) sorted by date Please note that the SOA has announced that when using

More information

EXAMINATIONS OF THE ROYAL STATISTICAL SOCIETY

EXAMINATIONS OF THE ROYAL STATISTICAL SOCIETY EXAMINATIONS OF THE ROYAL STATISTICAL SOCIETY GRADUATE DIPLOMA, 2015 MODULE 1 : Probability distributions Time allowed: Three hours Candidates should answer FIVE questions. All questions carry equal marks.

More information

Random Variables. Random variables. A numerically valued map X of an outcome ω from a sample space Ω to the real line R

Random Variables. Random variables. A numerically valued map X of an outcome ω from a sample space Ω to the real line R In probabilistic models, a random variable is a variable whose possible values are numerical outcomes of a random phenomenon. As a function or a map, it maps from an element (or an outcome) of a sample

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6684/0 Edexcel GCE Statistics S Silver Level S Time: hour 30 minutes Materials required for examination papers Mathematical Formulae (Green) Items included with question Nil Candidates

More information

Distributions of Functions of Random Variables. 5.1 Functions of One Random Variable

Distributions of Functions of Random Variables. 5.1 Functions of One Random Variable Distributions of Functions of Random Variables 5.1 Functions of One Random Variable 5.2 Transformations of Two Random Variables 5.3 Several Random Variables 5.4 The Moment-Generating Function Technique

More information

STAT 414: Introduction to Probability Theory

STAT 414: Introduction to Probability Theory STAT 414: Introduction to Probability Theory Spring 2016; Homework Assignments Latest updated on April 29, 2016 HW1 (Due on Jan. 21) Chapter 1 Problems 1, 8, 9, 10, 11, 18, 19, 26, 28, 30 Theoretical Exercises

More information

Lecture Notes for BUSINESS STATISTICS - BMGT 571. Chapters 1 through 6. Professor Ahmadi, Ph.D. Department of Management

Lecture Notes for BUSINESS STATISTICS - BMGT 571. Chapters 1 through 6. Professor Ahmadi, Ph.D. Department of Management Lecture Notes for BUSINESS STATISTICS - BMGT 571 Chapters 1 through 6 Professor Ahmadi, Ph.D. Department of Management Revised May 005 Glossary of Terms: Statistics Chapter 1 Data Data Set Elements Variable

More information

STA2603/205/1/2014 /2014. ry II. Tutorial letter 205/1/

STA2603/205/1/2014 /2014. ry II. Tutorial letter 205/1/ STA263/25//24 Tutorial letter 25// /24 Distribution Theor ry II STA263 Semester Department of Statistics CONTENTS: Examination preparation tutorial letterr Solutions to Assignment 6 2 Dear Student, This

More information

Algorithms for Uncertainty Quantification

Algorithms for Uncertainty Quantification Algorithms for Uncertainty Quantification Tobias Neckel, Ionuț-Gabriel Farcaș Lehrstuhl Informatik V Summer Semester 2017 Lecture 2: Repetition of probability theory and statistics Example: coin flip Example

More information

5. Conditional Distributions

5. Conditional Distributions 1 of 12 7/16/2009 5:36 AM Virtual Laboratories > 3. Distributions > 1 2 3 4 5 6 7 8 5. Conditional Distributions Basic Theory As usual, we start with a random experiment with probability measure P on an

More information

Errata and updates for ASM Exam C/Exam 4 Manual (Sixth Edition) sorted by page

Errata and updates for ASM Exam C/Exam 4 Manual (Sixth Edition) sorted by page Errata for ASM Exam C/4 Study Manual (Sixth Edition) Sorted by Page Errata and updates for ASM Exam C/Exam 4 Manual (Sixth Edition) sorted by page Please note that the SOA has announced that when using

More information

EECS 126 Probability and Random Processes University of California, Berkeley: Spring 2015 Abhay Parekh February 17, 2015.

EECS 126 Probability and Random Processes University of California, Berkeley: Spring 2015 Abhay Parekh February 17, 2015. EECS 126 Probability and Random Processes University of California, Berkeley: Spring 2015 Abhay Parekh February 17, 2015 Midterm Exam Last name First name SID Rules. You have 80 mins (5:10pm - 6:30pm)

More information

9 Bayesian inference. 9.1 Subjective probability

9 Bayesian inference. 9.1 Subjective probability 9 Bayesian inference 1702-1761 9.1 Subjective probability This is probability regarded as degree of belief. A subjective probability of an event A is assessed as p if you are prepared to stake pm to win

More information

Midterm Exam 1 (Solutions)

Midterm Exam 1 (Solutions) EECS 6 Probability and Random Processes University of California, Berkeley: Spring 07 Kannan Ramchandran February 3, 07 Midterm Exam (Solutions) Last name First name SID Name of student on your left: Name

More information

Eco517 Fall 2004 C. Sims MIDTERM EXAM

Eco517 Fall 2004 C. Sims MIDTERM EXAM Eco517 Fall 2004 C. Sims MIDTERM EXAM Answer all four questions. Each is worth 23 points. Do not devote disproportionate time to any one question unless you have answered all the others. (1) We are considering

More information

STATISTICS 1 REVISION NOTES

STATISTICS 1 REVISION NOTES STATISTICS 1 REVISION NOTES Statistical Model Representing and summarising Sample Data Key words: Quantitative Data This is data in NUMERICAL FORM such as shoe size, height etc. Qualitative Data This is

More information

STA 256: Statistics and Probability I

STA 256: Statistics and Probability I Al Nosedal. University of Toronto. Fall 2017 My momma always said: Life was like a box of chocolates. You never know what you re gonna get. Forrest Gump. There are situations where one might be interested

More information

Summary of basic probability theory Math 218, Mathematical Statistics D Joyce, Spring 2016

Summary of basic probability theory Math 218, Mathematical Statistics D Joyce, Spring 2016 8. For any two events E and F, P (E) = P (E F ) + P (E F c ). Summary of basic probability theory Math 218, Mathematical Statistics D Joyce, Spring 2016 Sample space. A sample space consists of a underlying

More information

Estimation of Quantiles

Estimation of Quantiles 9 Estimation of Quantiles The notion of quantiles was introduced in Section 3.2: recall that a quantile x α for an r.v. X is a constant such that P(X x α )=1 α. (9.1) In this chapter we examine quantiles

More information

Outline PMF, CDF and PDF Mean, Variance and Percentiles Some Common Distributions. Week 5 Random Variables and Their Distributions

Outline PMF, CDF and PDF Mean, Variance and Percentiles Some Common Distributions. Week 5 Random Variables and Their Distributions Week 5 Random Variables and Their Distributions Week 5 Objectives This week we give more general definitions of mean value, variance and percentiles, and introduce the first probability models for discrete

More information

IEOR 3106: Introduction to Operations Research: Stochastic Models. Professor Whitt. SOLUTIONS to Homework Assignment 2

IEOR 3106: Introduction to Operations Research: Stochastic Models. Professor Whitt. SOLUTIONS to Homework Assignment 2 IEOR 316: Introduction to Operations Research: Stochastic Models Professor Whitt SOLUTIONS to Homework Assignment 2 More Probability Review: In the Ross textbook, Introduction to Probability Models, read

More information

MATH Notebook 5 Fall 2018/2019

MATH Notebook 5 Fall 2018/2019 MATH442601 2 Notebook 5 Fall 2018/2019 prepared by Professor Jenny Baglivo c Copyright 2004-2019 by Jenny A. Baglivo. All Rights Reserved. 5 MATH442601 2 Notebook 5 3 5.1 Sequences of IID Random Variables.............................

More information

November 2000 Course 1. Society of Actuaries/Casualty Actuarial Society

November 2000 Course 1. Society of Actuaries/Casualty Actuarial Society November 2000 Course 1 Society of Actuaries/Casualty Actuarial Society 1. A recent study indicates that the annual cost of maintaining and repairing a car in a town in Ontario averages 200 with a variance

More information

EXAMINATIONS OF THE ROYAL STATISTICAL SOCIETY (formerly the Examinations of the Institute of Statisticians) HIGHER CERTIFICATE IN STATISTICS, 1996

EXAMINATIONS OF THE ROYAL STATISTICAL SOCIETY (formerly the Examinations of the Institute of Statisticians) HIGHER CERTIFICATE IN STATISTICS, 1996 EXAMINATIONS OF THE ROAL STATISTICAL SOCIET (formerly the Examinations of the Institute of Statisticians) HIGHER CERTIFICATE IN STATISTICS, 996 Paper I : Statistical Theory Time Allowed: Three Hours Candidates

More information

Lecture 2: Review of Probability

Lecture 2: Review of Probability Lecture 2: Review of Probability Zheng Tian Contents 1 Random Variables and Probability Distributions 2 1.1 Defining probabilities and random variables..................... 2 1.2 Probability distributions................................

More information

EDUCATION AND EXAMINATION COMMITTEE OF THE SOCIETY OF ACTUARIES ACTUARIAL MODELS LIFE CONTINGENCIES SEGMENT POISSON PROCESSES

EDUCATION AND EXAMINATION COMMITTEE OF THE SOCIETY OF ACTUARIES ACTUARIAL MODELS LIFE CONTINGENCIES SEGMENT POISSON PROCESSES EDUCATION AND EXAMINATION COMMITTEE OF THE SOCIETY OF ACTUARIES ACTUARIAL MODELS LIFE CONTINGENCIES SEGMENT POISSON PROCESSES (and mixture distributions) by James W. Daniel Copyright 2007 by James W. Daniel;

More information

Recursive Estimation

Recursive Estimation Recursive Estimation Raffaello D Andrea Spring 08 Problem Set : Bayes Theorem and Bayesian Tracking Last updated: March, 08 Notes: Notation: Unless otherwise noted, x, y, and z denote random variables,

More information

INSTITUTE OF ACTUARIES OF INDIA

INSTITUTE OF ACTUARIES OF INDIA INSTITUTE OF ACTUARIES OF INDIA EXAMINATIONS 13 th May 2008 Subject CT3 Probability and Mathematical Statistics Time allowed: Three Hours (10.00 13.00 Hrs) Total Marks: 100 INSTRUCTIONS TO THE CANDIDATES

More information

Probability Theory. Introduction to Probability Theory. Principles of Counting Examples. Principles of Counting. Probability spaces.

Probability Theory. Introduction to Probability Theory. Principles of Counting Examples. Principles of Counting. Probability spaces. Probability Theory To start out the course, we need to know something about statistics and probability Introduction to Probability Theory L645 Advanced NLP Autumn 2009 This is only an introduction; for

More information

Week 2: Review of probability and statistics

Week 2: Review of probability and statistics Week 2: Review of probability and statistics Marcelo Coca Perraillon University of Colorado Anschutz Medical Campus Health Services Research Methods I HSMP 7607 2017 c 2017 PERRAILLON ALL RIGHTS RESERVED

More information

Conditional densities, mass functions, and expectations

Conditional densities, mass functions, and expectations Conditional densities, mass functions, and expectations Jason Swanson April 22, 27 1 Discrete random variables Suppose that X is a discrete random variable with range {x 1, x 2, x 3,...}, and that Y is

More information

Exam C Solutions Spring 2005

Exam C Solutions Spring 2005 Exam C Solutions Spring 005 Question # The CDF is F( x) = 4 ( + x) Observation (x) F(x) compare to: Maximum difference 0. 0.58 0, 0. 0.58 0.7 0.880 0., 0.4 0.680 0.9 0.93 0.4, 0.6 0.53. 0.949 0.6, 0.8

More information

Actuarial Science Exam 1/P

Actuarial Science Exam 1/P Actuarial Science Exam /P Ville A. Satopää December 5, 2009 Contents Review of Algebra and Calculus 2 2 Basic Probability Concepts 3 3 Conditional Probability and Independence 4 4 Combinatorial Principles,

More information

Problem Set 1. MAS 622J/1.126J: Pattern Recognition and Analysis. Due: 5:00 p.m. on September 20

Problem Set 1. MAS 622J/1.126J: Pattern Recognition and Analysis. Due: 5:00 p.m. on September 20 Problem Set MAS 6J/.6J: Pattern Recognition and Analysis Due: 5:00 p.m. on September 0 [Note: All instructions to plot data or write a program should be carried out using Matlab. In order to maintain a

More information

6.1 Moment Generating and Characteristic Functions

6.1 Moment Generating and Characteristic Functions Chapter 6 Limit Theorems The power statistics can mostly be seen when there is a large collection of data points and we are interested in understanding the macro state of the system, e.g., the average,

More information

Dependence. Practitioner Course: Portfolio Optimization. John Dodson. September 10, Dependence. John Dodson. Outline.

Dependence. Practitioner Course: Portfolio Optimization. John Dodson. September 10, Dependence. John Dodson. Outline. Practitioner Course: Portfolio Optimization September 10, 2008 Before we define dependence, it is useful to define Random variables X and Y are independent iff For all x, y. In particular, F (X,Y ) (x,

More information

Problem Points S C O R E Total: 120

Problem Points S C O R E Total: 120 PSTAT 160 A Final Exam Solution December 10, 2015 Name Student ID # Problem Points S C O R E 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 11 10 12 10 Total: 120 1. (10 points) Take a Markov chain

More information

M378K In-Class Assignment #1

M378K In-Class Assignment #1 The following problems are a review of M6K. M7K In-Class Assignment # Problem.. Complete the definition of mutual exclusivity of events below: Events A, B Ω are said to be mutually exclusive if A B =.

More information

Extreme Value Theory.

Extreme Value Theory. Bank of England Centre for Central Banking Studies CEMLA 2013 Extreme Value Theory. David G. Barr November 21, 2013 Any views expressed are those of the author and not necessarily those of the Bank of

More information

TABLE OF CONTENTS - VOLUME 1

TABLE OF CONTENTS - VOLUME 1 TABLE OF CONTENTS - VOLUME 1 INTRODUCTORY COMMENTS MODELING SECTION 1 - PROBABILITY REVIEW PROBLEM SET 1 LM-1 LM-9 SECTION 2 - REVIEW OF RANDOM VARIABLES - PART I PROBLEM SET 2 LM-19 LM-29 SECTION 3 -

More information

2 (Statistics) Random variables

2 (Statistics) Random variables 2 (Statistics) Random variables References: DeGroot and Schervish, chapters 3, 4 and 5; Stirzaker, chapters 4, 5 and 6 We will now study the main tools use for modeling experiments with unknown outcomes

More information

Counting principles, including permutations and combinations.

Counting principles, including permutations and combinations. 1 Counting principles, including permutations and combinations. The binomial theorem: expansion of a + b n, n ε N. THE PRODUCT RULE If there are m different ways of performing an operation and for each

More information

Probability and Estimation. Alan Moses

Probability and Estimation. Alan Moses Probability and Estimation Alan Moses Random variables and probability A random variable is like a variable in algebra (e.g., y=e x ), but where at least part of the variability is taken to be stochastic.

More information

Sampling Distributions

Sampling Distributions Sampling Distributions In statistics, a random sample is a collection of independent and identically distributed (iid) random variables, and a sampling distribution is the distribution of a function of

More information

Continuous random variables

Continuous random variables Continuous random variables Continuous r.v. s take an uncountably infinite number of possible values. Examples: Heights of people Weights of apples Diameters of bolts Life lengths of light-bulbs We cannot

More information

Time: 1 hour 30 minutes

Time: 1 hour 30 minutes Paper Reference(s) 6684/01 Edexcel GCE Statistics S Silver Level S4 Time: 1 hour 30 minutes Materials required for examination papers Mathematical Formulae (Green) Items included with question Nil Candidates

More information

The Random Variable for Probabilities Chris Piech CS109, Stanford University

The Random Variable for Probabilities Chris Piech CS109, Stanford University The Random Variable for Probabilities Chris Piech CS109, Stanford University Assignment Grades 10 20 30 40 50 60 70 80 90 100 10 20 30 40 50 60 70 80 90 100 Frequency Frequency 10 20 30 40 50 60 70 80

More information

Statistics for scientists and engineers

Statistics for scientists and engineers Statistics for scientists and engineers February 0, 006 Contents Introduction. Motivation - why study statistics?................................... Examples..................................................3

More information

Sampling Distributions

Sampling Distributions In statistics, a random sample is a collection of independent and identically distributed (iid) random variables, and a sampling distribution is the distribution of a function of random sample. For example,

More information

Preliminary Statistics Lecture 2: Probability Theory (Outline) prelimsoas.webs.com

Preliminary Statistics Lecture 2: Probability Theory (Outline) prelimsoas.webs.com 1 School of Oriental and African Studies September 2015 Department of Economics Preliminary Statistics Lecture 2: Probability Theory (Outline) prelimsoas.webs.com Gujarati D. Basic Econometrics, Appendix

More information

Probability. Table of contents

Probability. Table of contents Probability Table of contents 1. Important definitions 2. Distributions 3. Discrete distributions 4. Continuous distributions 5. The Normal distribution 6. Multivariate random variables 7. Other continuous

More information

ELEG 3143 Probability & Stochastic Process Ch. 2 Discrete Random Variables

ELEG 3143 Probability & Stochastic Process Ch. 2 Discrete Random Variables Department of Electrical Engineering University of Arkansas ELEG 3143 Probability & Stochastic Process Ch. 2 Discrete Random Variables Dr. Jingxian Wu wuj@uark.edu OUTLINE 2 Random Variable Discrete Random

More information

STAT 285 Fall Assignment 1 Solutions

STAT 285 Fall Assignment 1 Solutions STAT 285 Fall 2014 Assignment 1 Solutions 1. An environmental agency sets a standard of 200 ppb for the concentration of cadmium in a lake. The concentration of cadmium in one lake is measured 17 times.

More information

Probability Distributions for Continuous Variables. Probability Distributions for Continuous Variables

Probability Distributions for Continuous Variables. Probability Distributions for Continuous Variables Probability Distributions for Continuous Variables Probability Distributions for Continuous Variables Let X = lake depth at a randomly chosen point on lake surface If we draw the histogram so that the

More information

Data Modeling & Analysis Techniques. Probability & Statistics. Manfred Huber

Data Modeling & Analysis Techniques. Probability & Statistics. Manfred Huber Data Modeling & Analysis Techniques Probability & Statistics Manfred Huber 2017 1 Probability and Statistics Probability and statistics are often used interchangeably but are different, related fields

More information

MATH/STAT 3360, Probability

MATH/STAT 3360, Probability MATH/STAT 3360, Probability Sample Final Examination This Sample examination has more questions than the actual final, in order to cover a wider range of questions. Estimated times are provided after each

More information

Math Review Sheet, Fall 2008

Math Review Sheet, Fall 2008 1 Descriptive Statistics Math 3070-5 Review Sheet, Fall 2008 First we need to know about the relationship among Population Samples Objects The distribution of the population can be given in one of the

More information

Statistics 100A Homework 5 Solutions

Statistics 100A Homework 5 Solutions Chapter 5 Statistics 1A Homework 5 Solutions Ryan Rosario 1. Let X be a random variable with probability density function a What is the value of c? fx { c1 x 1 < x < 1 otherwise We know that for fx to

More information

Brief Review of Probability

Brief Review of Probability Brief Review of Probability Nuno Vasconcelos (Ken Kreutz-Delgado) ECE Department, UCSD Probability Probability theory is a mathematical language to deal with processes or experiments that are non-deterministic

More information

RMSC 2001 Introduction to Risk Management

RMSC 2001 Introduction to Risk Management RMSC 2001 Introduction to Risk Management Tutorial 4 (2011/12) 1 February 20, 2012 Outline: 1. Failure Time 2. Loss Frequency 3. Loss Severity 4. Aggregate Claim ====================================================

More information

University of Chicago Graduate School of Business. Business 41901: Probability Final Exam Solutions

University of Chicago Graduate School of Business. Business 41901: Probability Final Exam Solutions Name: University of Chicago Graduate School of Business Business 490: Probability Final Exam Solutions Special Notes:. This is a closed-book exam. You may use an 8 piece of paper for the formulas.. Throughout

More information

EXAMINATIONS OF THE ROYAL STATISTICAL SOCIETY

EXAMINATIONS OF THE ROYAL STATISTICAL SOCIETY EXAMINATIONS OF THE ROYAL STATISTICAL SOCIETY GRADUATE DIPLOMA, 2016 MODULE 1 : Probability distributions Time allowed: Three hours Candidates should answer FIVE questions. All questions carry equal marks.

More information

RMSC 2001 Introduction to Risk Management

RMSC 2001 Introduction to Risk Management RMSC 2001 Introduction to Risk Management Tutorial 4 (2011/12) 1 February 20, 2012 Outline: 1. Failure Time 2. Loss Frequency 3. Loss Severity 4. Aggregate Claim ====================================================

More information

Chapter 8: An Introduction to Probability and Statistics

Chapter 8: An Introduction to Probability and Statistics Course S3, 200 07 Chapter 8: An Introduction to Probability and Statistics This material is covered in the book: Erwin Kreyszig, Advanced Engineering Mathematics (9th edition) Chapter 24 (not including

More information

Exam 1 Review With Solutions Instructor: Brian Powers

Exam 1 Review With Solutions Instructor: Brian Powers Exam Review With Solutions Instructor: Brian Powers STAT 8, Spr5 Chapter. In how many ways can 5 different trees be planted in a row? 5P 5 = 5! =. ( How many subsets of S = {,,,..., } contain elements?

More information

Chapter 5. Means and Variances

Chapter 5. Means and Variances 1 Chapter 5 Means and Variances Our discussion of probability has taken us from a simple classical view of counting successes relative to total outcomes and has brought us to the idea of a probability

More information

Arkansas Tech University MATH 3513: Applied Statistics I Dr. Marcel B. Finan

Arkansas Tech University MATH 3513: Applied Statistics I Dr. Marcel B. Finan 2.4 Random Variables Arkansas Tech University MATH 3513: Applied Statistics I Dr. Marcel B. Finan By definition, a random variable X is a function with domain the sample space and range a subset of the

More information

Mathematics and Further Mathematics Pre-U June 2010

Mathematics and Further Mathematics Pre-U June 2010 Mathematics and Further Mathematics Pre-U June 2010 The following question papers for Mathematics and Further Mathematics are the first papers to be taken by Pre-U students at the end of the two-year course.

More information

Chapter 4: An Introduction to Probability and Statistics

Chapter 4: An Introduction to Probability and Statistics Chapter 4: An Introduction to Probability and Statistics 4. Probability The simplest kinds of probabilities to understand are reflected in everyday ideas like these: (i) if you toss a coin, the probability

More information