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 tutorial letter intends to help you prepare for the examination. It contains the following information:. About the May/June 24 examinations. 2. Solutions to a previous year s examination paper. 3. Solutions to Assignment 6.. ABOUT THE MAY/JUNE 24 EXAMINATIONS What you are expected to be able to do in the exam You are expected to know the following distributions, remember their density functions or probability mass functions, recognize them, know when they apply, and be able to calculate probabilities from them: The Bernoulli distribution The Binomial distribution (the probability mass function will be given in the paper) The Poisson distribution The uniform distribution The normal distribution You may be asked to work with any other distributions discussed in the study guide as well, in which case the distribution will be given in the exam paper. In the exam you are given the formulas for the Binomial probability mass function, the density function of the general Gamma distribution, the density function of the k-th order statistic from a sample of n from a given distribution, and the series expansion of the exponential function. Further formulas will be given as needed. You will also be given cumulative standard normal distribution tables. The next page contains a brief summary of what you need to be able to do in the exam.
3 STA263/25 Random variables Given a discrete or continuous distribution of a random variable (defined via probability mass function/density function or cumulative distribution function), Bivariate distributions Given a discrete or continuous joint distribution of two random variables (defined via joint probability mass function/ joint density function or joint cumulative distribution function), find: find: cumulative distribution function from probability mass function, probability mass function from cumulative distribution function, probabilities of events distributions of functions of the random variable quartiles, quantiles, medians joint cumulative distribution function from joint probability mass function and vice versa, joint density from joint cumulative distribution function and vice versa, probabilities of events marginal distributions (densities and/or cumulative distribution functions) conditional distributions test for independence of the random variables find joint distribution Given a joint distribution, Given independence and marginal distributions, Work with joint distributions, conditional distributions and marginal distributions (whether deriving density/probability mass functions, cumulative distributions or probabilities of events) Given a joint distribution, find distributions of functions of the random variables, both by the distribution function method and the transformation method. Order statistics Use the formula (given in the exam paper) for the density of k th order statistics to find distributions for various order statistics Expected values Given distribution or joint distribution, find expected values of random variables or expected values of functions of the random variables conditional expectations of random variables and functions of random variables variances and standard deviations covariances and correlation coefficients moments of the distribution Moment generating functions Given a distribution (density or probability mass function), find the corresponding moment generating function calculate moments by differentiating the moment generating function Inequalities Apply Chebyshev s inequality (formula will be given in the exam paper when needed) Limit theorems Apply the central limit theorem Distributions derived from the normal distribution Recognize linear combinations of, sums of squares and quotients of independent normally distributed random variables as coming from normal, chi square, F or t distributions, and specify the parameters and degrees of freedom; including the distributions of the sample mean and sample variance.
4 2. SOLUTIONS TO A PREVIOUS YEAR S EXAMINATION PAPER Your examination paper for this semester will be approximately similar to previous years examination papers. One such paper (October/November 22) formed the basis for Assignment 6 for this semester. We will give you detailed solutions to that question paper here. In the following, we give first all the questions, and then the solutions to all the questions. More examination papers from previous years are available on the myunisa website of this module. We will not provide solutions to them, but you are welcome to contact with your lecturer if you are struggling with a particular question. We have also included the tables and the set of equations, exactly as they will be given to you in your examination paper. The equations are there just to aid your memory you will still need to know exactly which equation to use in which situation! Make sure that you know how to use the given equations and the tables, and how to use your calculator! We wish you all the best in the examination. Please do contact us for help, if you have any queries! With regards, Dr E Rapoo THE QUESTIONS QUESTION Two discrete random variables X and Y have a joint probability mass function given in the table below. x p XY x y 2 3 8 4 y 2 3 8 4 8 8. Calculate the marginal probability mass functions p X x and p Y y of X and Y 4.2 Are the X and Y independent? Justify your answer! 3.3 Calculate the following: (a) P X 2 Y 2 (b) E Y X 2 (c) P X 2 Y [7]
5 STA263/25 QUESTION 2 Let and 4x 2y f Y X y x 4x x y elsewhere 4x x f X x 3 elsewhere 2. Find the marginal density function of Y (6) 2.2 Calculate P 2 X Y (6) 2.3 Determine PY X (6) QUESTION 3 Suppose X and Y are two independent random variables of the exponential distribution with parameter 2 3. Find the distribution function of the random variable U X Y (8) 3.2 Find the moment generating function of X (6) 3.3 Calculate E X 5 by differentiating the moment generating function of X (4) 3.4 Find the moment generating function of the random variable W 2XY (6) [8] [24] QUESTION 4 4. Let X and X 2 be two independent random variables so that the variances of X and X 2 are 2 k and 2 2 2 respectively. Given that the variance of Y 3X 2X 2 is 35 find the value of k (4) 4.2 Find the mean and the variance of the sum Y of the items of a random sample of size 5 from the distribution with probability density function 6x x x f X x elsewhere (8) [2]
6 QUESTION 5 5. Suppose that the number of insurance claims (N) filed during a period of one year is Poisson distributed with EN Use the normal approximation to the Poisson distribution to approximate PN EN 5 (8) 5.2 State if each of the following statements is true or false. If the statement is false, explain why it is false. (a) If two random variables come from the same distribution with the same parameters, then they cannot be independent. (b) Two discrete random variables are independent if their marginal probability mass functions both sum up to. (c) If f Y y f XY x y for all possible values x and y then X and Y are independent. (d) Only continuous random variables have moment generating functions. (e) If the joint distribution function F XY x y is known then the marginal distribution of X can be found by integrating F XY over all values of Y () QUESTION 6 Let X X 2 X be a sample from a N2 25 distribution, independent of the sample Y Y 2 Y 2 coming from a N2 distribution. Let X X i Y 2 Y j 2 i j [8] SX 2 i X 9 ix 2 SY 2 9 2 X i X i 2 2 2 Y 2 2 j 2 j Y j Y 2 Y j 2 2 6. Explain why Q 6 Y j 2 2 does not follow a 2 distribution. (3) j 6.2 Write down the name of the distribution of U 9S2 Y 2 and give the value(s) of the parameter(s). (3) 6.3 Define V A 2 X 2 Y For what value of the constant A does V follow an F distribution? What are the degrees of freedom of this F distribution? Justify your answer! (5) [] TOTAL: []
7 STA263/25 FORMULAE f n x p X k n k p k p nk f X x x e x n! k!n k! [F Xx] k [ F X x] nk f X x e tx tx tx2 2! tx3 3!
8 Table : Cumulative Standardized Normal Probabilities
9 STA263/25 Table (continued)
THE SOLUTIONS QUESTION. The marginal probability mass function for X is found by summing up over all y-values, over each column; and the marginal probability mass function for Y is found by summing up over all x-values, over rows. We get: 8 8 4 4 8 2 x p X x 4 8 3 8 x 2 8 8 x 3 p Y x.2 The random variables are not independent. For instance 8 4 3 8 y 8 8 2 8 4 y 2 4 8 3 8 y 3 p XY 3 p X 3 p Y 8 3 8.3 (a) PX 2 Y 2 P X 2 and Y 2 P Y 2 p XY 2 p XY 2 2 p Y 2 8 8 4 (b) We need to first find the conditional probability distribution of Y given that X 2. We get It follows that PY yx 2 P X 2 Y y P X 2 4 3 8 2 3 y 8 3 8 3 y 2 y 3 p XY 2 y p X 2 EY X 2 PY X 2 2 PY 2X 2 3 PY 3X 2 2 3 2 3 3 4 3 (c) PX 2 Y PX 2 since Y isalwaystrue PX 2 or X 3 p X 2 p X 3 2
STA263/25 QUESTION 2 2. The marginal density function of Y is found by integrating the joint density function over all x-values. Here we are not given the joint density function directly, but it can always be found from the given function as follows: f XY x y f Y X yx f X x Therefore, we have f Y y f XY x ydx f Y X yx f X x dx and elsewhere. 4x 2y 4x 4x dx y 3 4x 2y 4 dx 3 6 x2 2 3 yx 4 6 2 3 y 2 2y for y 3 2.2 To find P 2 x Y we need to first find the conditional density of X given Y.Thisisgivenby f XY xy f XY x y 4x2y 22y 4x2y 3 3 22y 2xy y x f Y y elsewhere. Therefore, and P 2 f XY xy 2x 2x x 2 2 x 2 X Y x dx 2 2 x2 2 x 2 8 5 4 8 2 2 2 4 2 2 2.3 P Y X 3 yx 4xy y 2 x f XY x y dxdy x 4x 2y dy dx 3 dx 4x 2 x 2 dx 3 3 5x2 dx 3 5 x 3 3 3 5 3 5 9
2 QUESTION 3 X and Y are both from the exponential distribution with parameter 2, which means that their density functions are f X x e x x f Y y e y y and the joint density function, since X and Y are independent, is f XY x y e x e y 3. Let U X Y. Then the distribution function of U is F U u PU u PX Y u u ux xyu e y dy e x dx for u f XY x y dxdy u e ux e x dx u e x e u dx e u ue u u zero elsewhere where 2 Alternatively, we can use the transformation method here. If we take U X Y V X then u g x y x y g 2 x y x with inverse transformations x y u The Jacobian is Jx y g x g 2 x g y g 2 y Therefore, f UV u f XY u e e u 2 e u for u
3 STA263/25 and therefore f U u and F U u f UV u d u u f U a da 2 e u du u 2 e u u u a 2 e a da u which gives the same result as before, but the calculations are much longer! 3.2 3.3 From We get M X t Ee tx t e tx e x dx t Mt t e tx dx (assuming t t e tx t M t t 2 t 2 M t 2 t 3 2 t 3 M t 32 t 4 6 t 4 M i t 46 t 4 24 t 5 M t 524 t 6 2 t 6 and therefore EX 5 M 2 5 3.4 For W 2XY, the moment generating function is M W t E e tw E e t2xy e 2txy f XY x y dxdy e 2txy e x e y dxdy QUESTION 4 4. Since X and X 2 are independent, VarY Var3X 2X 2 Var3X Var2X 2 9VarX 4VarX 2 9 k 4 2 9k 8 so to have VarY 35, we must have 9k 8 35 9k 35 8 27 k 3
4 4.2 For a random variable X from the distribution f X x 6x x x elsewhere we get EX E X 2 x 6x x dx 2 x 2 6x x dx 3 and therefore VarX EX 2 EX 2 3 2 2 2 Now, Y X X 2 X 3 X 4 X 5 where the X i are all independent of each other and have the same distribution as X above, i 5 Therefore QUESTION 5 EY 5 Ex 5 2 5 2 VarY 5VarX 5 2 4 5. The normal approximation to the Poisson distribution states that if N has the Poisson distribution, then Z N E N VarN is approximately a standard normal random variable when the Poisson parameter is large enough. Here, EN VarN. Therefore, we get N E N P N E N 5 P VarN 5 VarN P N E N VarN 5 P Z 5 VarN P Z 5 5.2 (a) False. The variables may or may not be independent. (b) False. The marginal distribution must always sum up to, whether the random variables are independent or not.
5 STA263/25 (c) False. Random variables X and Y would be independent if we had f Y y f Y X yx for all x y values. (d) False. Moment generating function is defined also for discrete random variables. (e) False. What is true is that the marginal density function of X can be found by integrating f XY,thejoint density function, over all values of y. QUESTION 6 6. Q is the sum of six independent squared standard normal variables, so it has the 2 6 distribution, not the 2 distribution. 6.2 Since each Y j N2,wehave Y j 2 N and But Y j 2 2 j N Y j 2 Y j 2 2 2 2 2 j 9 9 Y j 2 2 2 j 2 j y j 2 2 2 2 Y j 2 2 9 2 U R Therefore, R has the 2 2 distribution. 6.3 We have so 2 X j 2 5 Xi 2 N Y j 2 N i 5 2 2 2 Y j 2 and j Xi 2 2 2 2 2 W 5 i 2 X j 2 2 F 2 2 Here, and therefore, if we have then V has the F 2 distribution. A j V A 25 W 25 A 25 4
6 3. SOLUTIONS TO ASSIGNMENT 6 The correct answers to the multiple choice questions are given below. Please see Section 2 in this tutorial letter for the correct solutions to the examination questions that these questions are based on!. The correct answer is 2.. The correct answer is. 2. The correct answer is 2. 2. The correct answer is 2. 3. The correct answer is 4. 3. The correct answer is 2. 4. The correct answer is 4. 4. The correct answer is 2. 5. The correct answer is 3. 5. The correct answer is 2. 6. The correct answer is 4. 6. The correct answer is 2. 7. The correct answer is. 7. The correct answer is 2. 8. The correct answer is 2. 8. The correct answer is 2. 9. The correct answer is 4. 9 The correct answer is 4.. The correct answer is 3. 2 The correct answer is 4.