ECO227: Term Test 2 (Solutions and Marking Procedure)

Transcription

1 ECO7: Term Test (Solutions and Marking Procedure) January 6, 9 Question 1 Random variables X and have the joint pdf f X, (x, y) e x y, x > and y > Determine whether or not X and are independent. [1 marks] Let W X +. Find the pdf for W. [1 marks] Solution We rst nd the marginal pdfs of X and f X (x) f X, (x, y)dy e x y dy e x e y dy e x ( e y ) e x ( lim y e y e ) e x ( 1) e x, x > f (y) f X, (x, y)dx e x y dx e y e x dx e y ( e x ) e y ( lim x e x e () ) e y ( 1) e y, y > Then, f X (x)f (y) (e x )(e y ) e x y f X, (x, y), so X and are independent. Alternatively, f X, (x, y) e x y (e x )(e y ) g(x)h(y), so X and are independent. 1

2 Version 1 Given W X +, < X < W. Dene V. Then according to Theorem 3..3 (p.171), f V (v) 1 f ( v ) 1 e v, v > Now, W X + X + V and we can apply the formula for the sum of two random variables (Theorem , p.). We can substitute either V W X or X W V, i.e., f W (w) e w e w f X (x) f V (w x) dx e x 1 w x e dx e 3 x dx ( ) (e 3 x ) w 3 or f W (w) 3 e w (e 3 w e ) 3 e w 3 e w, w > f X (w v) f V (v) dv e (w v) 1 e v dv e w e 3 v dv ( ) e w (e 3 v ) w 3 3 e w (e 3 w e ) 3 e w 3 e w, w > Version It can be shown that the formula in Theorem (p.) is applicable directly when we substitute X W, whilst it is not applicable when substituting (W X)/. If students feel uncertain about why this is the case, it is better to use the method outlined above.

3 Given W + X, X W >, so W > < W/. Then, f W (w) f X (w y) f (y) dy e (w y) e y dy e w e 3y dy ( ) 1 e w e 3y w 3 3 e w (e 3 w e ) 3 e w 3 e w, w > Marking Most students got this question right, and were awarded full marks. 1 mark was deducted for arithmetic errors. Partial marks were awarded for deriving the marginal pdfs of X and, and for stating the denition of independence in this context. It should be noted that it is not strictly necessary to derive the marginal pdfs here but that they are needed for part. This was a dicult question. Full marks were awarded for any correct solution. Up to 5 marks were awarded for incorrectly substituting (W X)/. 1 mark was deducted for arithmetic errors. Partial marks were awarded when (somewhat) managing to state a (somewhat) correct integral. Question is an exponential random variable with pdf f (y) λe λy, y > Derive the moment-generation function for. [1 marks] Use the moment-generating function to nd the expected value of. [1 marks] Solution By the denition for a moment-generating function (p.57) M (t) E(e t ) λ e ty λe λy dy e (λ t)y dy λ (λ t) (e (λ t)y ) If λ t >, e (λ t)y as y, so M (t) λ (λ t) ( e ) λ λ t If λ t <, e (λ t)y as y, so M (t) does not exist. 3

4 If λ t, M (t) λ (λ t) (e e ). Hence, M (t) exists and is non-zero only when t < λ. Then M (t) λ λ t. Rewrite M (t) λ λ t λ(λ t) 1. Then, dierentiate consecutively using the chain rule to obtain M () (t) Marking M (1) (t) λ(λ t) ( 1) λ(λ t) (t) λ(λ t) 3 ( 1) λ(λ t) 3 M () M (3) (t) 6λ(λ t) ( 1) 6λ(λ t) M () (t) λ(λ t) 5 ( 1) λ(λ t) 5 λ (λ t) 5 Then, E( ) M () λ () (λ ) 5 λ Many students solved this question correctly. When the integral was set up and integrated correctly, but could not be evaluated (e.g., confusion about whether or not it converged), 5 λ marks were awarded. A common incorrect solution was t λ, which obtained 7 marks. It is desirable to state the range of t for which the moment generating function exists and is non-zero (i.e., t < λ), but no marks were deducted for failing to do so. The crucial thing in this question was to realize that we should evaluate M () (t) at t. If E( ) M () () was given and some attempt att dierentiating M (t) four times was made, 5 marks were awarded. Merely stating the denition with no attempt at dierentiating obtained 3 marks (some students failed to realize that the () in M () (t) referred to the fourth-order derivative). 1 mark was deducted for very minor mistakes in dierentiation and for arithmetic errors, and marks were deducted for more serious mistakes in dierentiation. Students should pay attention to that t only occurs once in M (t), implying that this function can be dierentiated using the chain rule. Where M () (t) had been found but was not evaluated at zero, 6 marks were given. When had been solved incorrectly, but was solved correctly, full marks were given for. For example, if one had found λ t λ in, full marks in were given only for λ. Such solutions are marked ECF (Error Carried Forward ). No marks were given when methods other from using the moment generating function were used. Question 3 A random variable is normally distributed with an expected value of 65. (c) If P ( < 5).9, nd the standard devation of. [5 marks] Assuming that the variance of is 16, nd P (6 < < 8). [1 marks] Assuming that the variance of is 16, nd the value y s such that P ( > y s ).15 [5 marks]

5 Solution Since N(65, σ ), P ( < 5) P ( < ) P (Z < ).9, where Z N(, 1) σ σ σ From the standard normal table, P (Z <.81) Hence,.81, so σ 15 σ (c) Marking Since var( ) σ 16, σ. Thus, we have 6 65 P (6 < < 8) P ( < 65 < P ( 1.5 < Z < 3.75) 8 65 ) P (Z < 3.75) P (Z < 1.5) P ( > y s ) P ( 65 From the standard normal table, P (Z < 1.8).9 Hence, y s , so y s 1.8() Common problems included > y s 65 ).1 P (Z > y s 65 ).1 1 P (Z < y s 65 ).1 P (Z < y s 65 ).9 (i) (ii) Mistaking the variance for the standard deviation, e.g., dividing by 16 rather than 16 in the part. Students should be very condent with the terminology (variance, standard deviation, etc.) since it will be crucial for managing the second part of the course. marks were deducted for this mistake. Performing a continuity correction, e.g., calculating P (59.5 < < 8.5) in the part. Students should be reminded that continuity corrections are only carried out when we approximate a discrete random variable (e.g., binomial, poisson) with a continuous distribution (e.g., normal). Here, is exactly normally (so continuously) distributed: we are not approximating anything. 3 marks were deducted for this mistake. Students were only penalized once for mistakes of type (i) and (ii) above. That is, if marks had been deducted in the part, and the same mistake led to incorrect answers on the part, full marks could still be obtained for the part. Such solutions were marked ECF (for Error Carried Forward ). 1 marks was deducted for arithmetic error. Many students evaluated P (Z < 3.75) 1 rather than No marks were deducted for this mistake. 5

6 Some students wrote y s in the (c) part. From reading the standard normal table, we know P (Z < 1.8).8997 and P (Z < 1.9).915. Since.8997 is a lot closer to.9 than.915 is, 1.8 is a better answer than 1.9 (the number is in fact about ). 1 mark was deducted for 1.9, whilst no marks were deducted for A number of students failed to read the standard normal table correctly: whilst most were condent in obtaining probabilities from z-values, many failed to nd z-values corresponding to particular probabilities. Drawing a diagram of the normal curve (with z-values on the horizontal axis) and probabilities as shaded areas under the curve might be helpful to avoid this type of mistakes. Question Suppose that vehicles arrive at a toll-booth independently at a rate of 6 cars per hpur and 3 trucks per hour. (c) Find the probability that at least cars will arrive at the toll-booth within the next 3 minutes. [1 marks] Find the probability that exactly 1 car and no trucks will arrive within 1 minute. [5 marks] Assuming that a vehicle has just arrived, nd the probability that the next vehicle will arrive in less than 1 minute. [5 marks] Solution Let X C,1 be the random variable denoting the number of cars arriving within the next minute. Then X C,1 is poisson distributed with parameter λ C,1 6 cars 1 hour 6 cars 1 car per minute 6 min Let X C,3 be the random variable denoting the number of cars arriving within the next 3 minutes. Then X C,3 is poisson distributed with parameter λ C,3 3λ C,1 3 cars per 3 minutes Now, the probability that at least 3 cars arrive within the next 3 minutes is given by P (X C,3 ) 1 P (X C,3 < ) 1 P (X C,3 1) 1 P (X C,3 ) P (X C,3 1) 1 e 3 3! 1 e 3.81 e ! Let X T,1 be the random variable denoting the number of trucks arriving within the next minute. Then X T,1 is poisson distributed with parameter λ T,1 3 trucks 1 hour 3 trucks 6 min.5 truck per minute 6

7 Since the arrivals of trucks and cars are independent, P (X C,1 1 X T,1 ) P (X C,1 1) P (X T,1 ) ( e ) ( e.5.5 ) 1!! e (c) Let X V,1 be the random variable denoting the number of vehicles arriving within the next minute. Then X V,1 is poisson distributed with parameter λ V,1 6 cars+3 trucks 1 hour 9 vehicles 6 min 1.5 vehicles per minute Let be the random variable denoting the time interval between two arrivals (i.e., the waiting time). Then is exponentially distributed with pdf f (y) λ V,1 e λv,1y 1.5e 1.5y, y > Then the probability that the next arrival is within the next minute (i.e., that the waiting time is less than 1 minute) is given by P ( < 1) ˆ 1 1.5e 1.5y dy e 1.5y 1 (e 1.5 e ) 1 e Marking Many students did well on this question. Some students failed to analyze the question correctly and made use of the wrong distribution. A small number of students solved the question by approximating the poisson probabilities using the normal distribution via the central limit theorem. When this was done correctly (including an appropriate continuity correction), up to 6% of the total marks were awarded. It seemed that some students were unaware which calculations represented exact probability calculations and which represented approximations. Generally, it is important to identify the underlying distribution, and then decide whether to perform the calculation (this should be done whenever feasible) or an appropriate approximation. Marks were awarded as follows: 1 mark was awarded for noting that we had to use a poisson distribution, even if the formula was not correctly recalled. 3 marks were awarded for noting that there were 3 cars per 3 minutes, i.e., that the poisson parameter (λ) was 3. marks were awarded for correctly analyzing the probability scenario (i.e., P (X ) 1 P (X ) P (X 1). marks were awarded for correct recall and application of the poisson formula (calculation). Partial marks were awarded in some of these categories. 1 mark was deducted for arithmetic errors. marks were awarded for noting that the probabilities were independent and multiplying them together. 1 mark was awarded for noting that the probabilities were poisson distributed, and 1 mark per probability correctly calculated. 1 mark was deducted for arithmetic errors. (c) 1 mark was awarded for calculating / writing down the correct poisson parameter (1.5). marks were awarded for using the exponential distribution. marks were awarded for correctly evaluating the integral. 1 mark was deducted for arithmetic errors. Question 5 A normal random variable with expected value µ and variance σ has the following momentgenerating function M (t) exp(µt + σ t /), where exp(x) e x 7

8 Suppose that there are two normal random variables 1 N(µ 1, σ 1) and N(µ, σ ), and let s 1 +. Show that s is also a normal random variable. [1 marks] Use the central limit theorem to prove that a binomial distribution can be approximated by a normal distribution with the same expected value and variance. [1 marks] Solution Assume that 1 and are independent. Then, by Theorem (p.66), M s (t) M 1 (t) M (t) exp(µ 1 t + σ 1t ) exp(µ t + σ t ) exp(µ 1 t + σ 1t + µ t + σ t ) exp(µ 1 t + µ t + σ 1t + σ t ) exp((µ 1 + µ )t + (σ 1 + σ1)t ) Hence, s N(µ 1 + µ, σ 1 + σ 1), i.e., s is normal with mean µ 1 + µ and variance σ 1 + σ 1. Let X i be an independent random variable with a pdf given by P (X i 1) p P (X i ) 1 p implying that E(X i ) 1p + (1 p) p var(x) (1 p + (1 p)) p p(1 p) Let X X 1 + X X n. Then, since X 1, X,..., X n are independent, X follows a binomial distribution with E(X) E(X 1 + X X n ) E(X 1 ) + E(X ) E(X n ) np var(x) var(x 1 + X X n ) var(x 1 ) + var(x ) var(x n ) np(1 p) Since X 1, X,..., X n are independent random variables from the same distribution (with mean µ and variance σ ) then by the central limit theorem (p.3) ( lim a X ) 1 + X X n nµ b P (a Z b) n nσ Here, µ p and σ p(1 p). Thus, lim n lim n ( ( a X 1 + X X n np n p(1 p) a X 1 + X X n np np(1 p) b b ) ) P (a Z b) P (a Z b) Given N(np, np(1 p)), Z approximated by (the normal random variable). np. Hence, X (the binomial random variable) can be np(1 p) 8

9 Marking 5 marks were awarded for stating the correct formula. 3 marks were awarded for rewriting the formula. marks were awarded for correctly stating the conclusion, i.e., that the formula implies s being normal with mean µ 1 + µ and variance σ 1 + σ 1. When failing to recall the correct formula, 1 mark could awarded for stating that s had mean µ 1 + µ and 1 mark for stating that the variance was σ 1 + σ 1. This can be shown without using the moment-generating function, but does not show that s has a normal distribution. Thus, it is incorrect to derive the variance and mean, plug these into the given moment-generating function for a normal, and claim that s is normal (this is a case of assuming the conclusion). Only a small number of students actually attempted proving the statement. The quality of responses was varying. For solutions which did not actually attempt the proof (or approached it incorrectly), marks were awarded for correctly stating the central limit theorem, and 1 mark each for stating that a binomial variable has a mean of np and a variance of np(1 p). The key point in this question was to realize that a binomial random variable X is itself a sum of (Bernoulli) random variables X 1, X,..., X n. The fact that the X i 's are independent draws from the same distribution allows direct application of the central limit theorem to their sum (X). Overall Performance Although the average raw mark was fairly low (58.7), a number of students did very well and many students improved their performance from the rst term test. 9

10 Appealing your mark If you did not do as well as you had expected, remember that only the better two of your three term tests count towards your nal grade. If you are unsatised with how your test was marked, it is possible to make an appeal. The procedure is as follows: 1. Make sure that you understand the solutions and how the test was marked (as described in this document).. Write (on a piece of paper) what questions and what parts of these questions that you wish to appeal. Explain in writing why you believe that these questions were marked incorrectly (i.e., dierently from what was outlined in this document). If your marks have been summed incorrectly, just write which are the questions concerned. 3. Attach your written appeal to your term test and hand it back to me (Rebecca, the TA) at the beginning of the tutorial on January 15. To ensure consistency and fairness, all appeals must be made as outlined above. There will be no ad hoc mark amendments during tutorials. About the second problem set Students who attempted most or almost all problems were awarded one mark (tick on the cover). Students who failed to attempt a substantial number of the assigned problems did not obtain this mark (crossed tick on the cover). 1