STAT/MATH 395 A - Winter Quarter 17 - Midterm - February 17, 17 Name: Student ID Number: Problem #1 # #3 #4 Total Points /5 /7 /8 /4 /4 Directions. Read directions carefully and show all your work. Define the events and random variables you are considering. You do not need to simplify or evaluate unless indicated. If your answer involves an integral, you only need to provide an antiderivative with the appropriate bounds. Partial credit will be assigned based upon the correctness, completeness and clarity of your answers. Correct answers without proper justification will not receive full credit. The exam is closed book, closed notes. Calculators are permitted but not needed to answer questions.
Problem 1.[5 points] Records over the past few decades show that the annual numbers of tornadoes N A and N B in counties A and B of a given state are jointly distributed as follows: Annual number of tornadoes in county B 1 3 p NA (x) Annual number.1.6.5..5 of tornadoes 1.13.15.1.3.43 in county A.5.15.1..3 p NB (y).3.36.7.7 (a) [ points] Fill in the table with the marginal probability mass function values of N A and N B. You do not need to justify your answer for this question. (b) [ points] What is the probability that there are at least twice as many tornadoes in county B as in county A next year? Answer. The event of interest is {N B N A. The corresponding set of couples is R {(, ), (, 1), (, ), (, 3), (1, ), (1, 3). Therefore, P(N B N A ) P((N A, N B ) R) p NA N B (, ) + p NA N B (, 1) + p NA N B (, ) + p NA N B (, 3) + p NA N B (1, ) + p NA N B (1, 3).1 +.6 +.5 +. +.1 +.3 (c) [1 point] The correlation of N A and N B is ρ NA N B.16. What can you say about the relationship between the annual numbers of tornadoes in counties A and B? Answer. The annual numbers of tornadoes in counties A and B are positively linearly correlated. Therefore the numbers of tornadoes tend to be simultaneously greater than, or simultaneously less than, their respective means. However, the small value for the correlation indicates a slight linear association between the two random variables.
Problem.[7 points] The return of a certain financial shock is modeled by a continuous random variable X with probability density function where a > is a constant. f X (x) a e a x, for x R, (a) [ points] Show that M X (t) the moment generating function of X is Answer. M X (t) is given by M X (t) a a, where a < t < a. t M X (t) E[e tx ] e tx f X (x) e tx a e a x a { e (t+a)x + e (t a)x { a 1 t + a e(t+a)x + 1 t a e(t a)x (b) [ points] Show that the expected return is using M X. Answer. The expected return can be found by evaluating the first derivative of M X (t) at t. We have that M X(t) Therefore, the expected return is E[X] M X (). a t (a t ). (c) [3 points] Financial analysts want to know if the distribution of the returns is skewed. Compute the third central moment of X using M X. Answer. The third central moment of X is E[(X E[X]) 3 ]. Since E[X], the calculation simplifies to E[X 3 ]. This can be found by evaluating the third derivative of M X (t) at t. The second derivative is M X (t) a (a + 3t )/(a t ) 3 and d 3 dt 3 M X(t) 4a t(a + t ) (a t ) 4. Therefore, E[X 3 ] d3 dt M 3 X (t). t
Problem 3.[8 points] Let X represent the age (in years) of an insured automobile involved in an accident. Let Y represent the length of time (in years) the owner has insured the automobile at the time of the accident. X and Y have joint probability density function with S {(x, y) x 1, y 1. f XY (x, y) 1 64 (1 xy ) 1 S (x, y) (a) [ points] Compute f X the marginal probability density function of X. Answer. First, for x [, 1], f X (x). Second, for x [, 1], f X (x) ( ) Hence, f X (x) 1 64 1 x 3 1[,1] (x). 1 64 (1 xy ) dy 1 64 ] y1 [1y xy3 1 ( 1 x ). 3 y 64 3 (b) [ points] Show that the mean age of an insured automobile involved in an accident is approximately 5.78 years. Answer. This amounts to computing E[X]: E[X] xf X (x) x ( 1 x ) 1 64 3 64 ] 1 [5x x3. 9 (c) [ points] The marginal probability density function of Y is f Y (y).5(5 3y )1 [,1] (y). Are X and Y independent? Answer. They are not independent because for instance, 5 3 f XY (3, ) f X (3)f Y () 9 64 5 4. (d) [ points] The mean length of time an owner has insured the automobile at the time of the accident is.4375 year. Compute the covariance of X and Y. Answer. The covariance of X and Y is given by Cov(X, Y ) E[XY ] E[X] E[Y ]. Since E[X] and E[Y ] are known, we only need to compute E[XY ]. We have E[XY ] xyf XY (x, y) dy R 1 { xy(1 xy ) dy 64 1 ) (5x x 64 4 1 [ ] 1 5 64 x x3 1 59 4. Hence, Cov(X, Y ) 59 4 5.78.4375.7.
Problem 4.[4 points] An insurance company insures a large number of drivers. Let X be the random variable representing the company s losses under collision insurance, and let Y represent the company s losses under liability insurance. X and Y have joint probability density function f XY (x, y) (x + y)1 S(x, y), with S {(x, y) < x < 1, < y <. What is the probability that the total loss is at least 1? Hint. Consider first the probability of the complement, that is the total loss is less than 1. Answer. The probability of that is the total loss is less than 1 is P(X + Y < 1). The event {X + Y < 1 corresponds to the subregion of R, R {(x, y) < x < 1, < y < 1 x. Therefore, P(X + Y < 1) P((X, Y ) R) x 7 4. { x f XY (x, y) dy (x + y) dy {(x + )(1 x) { 5 x + x + 3 [ 56 x3 + x + 3 x ] 1 (1 x) Hence the desired probability is P(X + Y 1) 1 P(X + Y < 1) 1 7/4.