Math 510 midterm 3 answers

Transcription

1 Math 51 midterm 3 answers Problem 1 (1 pts) Suppose X and Y are independent exponential random variables both with parameter λ 1. Find the probability that Y < 7X. P (Y < 7X) 7x 7x f(x, y) dy dx e x e y dy dx e x e y 7x dx ( e x e 7x + e x) dx ( e 8x + e x) dx 1 8 e 8x e x Problem 2 (1 pts) Suppose X and Y have the mass distributions below. Find the joint mass distribution function for X and Y so that X and Y are independent. X Prob Y Prob Since X and Y are independent, P (X i, Y j) P (X i)p (Y j), and therefore we can make the following table for the joint mass distribution function. X 1 X 2 Y Y Y

2 Problem 3 (1 pts) Suppose X and Y are independent random variables, and suppose X is binomial with n 1 and p.4; while Y is binomial with n 12 and p.2. Find the expected value and variance of 2X + 3Y. (Recall that a binomial random variable has expected value np and variance npq.) Expected Value: E[2X + 3Y ] 2E[X] + 3E[Y ] Variance: V ar(2x + 3Y ) 4 V ar(x) + 9 V ar(y ) Problem 4 (1 pts) True or false: TRUE The correlation ρ(x, Y ) is always a number in the range [ 1, 1]. FALSE For any random variables X and Y, we have V ar(x + Y ) V ar(x) + V ar(y ). FALSE If E[XY ] E[X]E[Y ], then X and Y are independent. TRUE If X and Y are independent, then M X+Y (t) M X (t) M Y (t). TRUE If V ar(x + Y ) V ar(x) + V ar(y ), then ρ(x, Y ). FALSE If E[XY Z] E[X Z]E[Y Z], then E[XY ] E[X]E[Y ]. TRUE If the moment generating function for a random variable X is e t2, then X must be normal. FALSE Let X and Y be continuous random variables. The density function for X and the density function for Y determine uniquely the joint density function for (X, Y ). FALSE If X and Y are independent, then SD(X + Y ) SD(X) + SD(Y ). FALSE If E[X + Y ] E[X] + E[Y ], then X and Y are independent. 2

3 Problem 5 (5 pts) Find the moment generating function M(t) for the random variable given in the table below: X Prob M(t) E[e tx ] e t +.2 e 2t Problem 6 (5 pts) Find an example of two random variables X and Y so that ρ(x, Y ) 1. Let X be a non-constant random variable (say, a standard normal random variable) and let Y X. Problem 7 (5 pts) Suppose a random variable X had the moment generating function given by the series below. Find the third moment of X. M X (t) 1 + 5t + 9t 2 + 8t 3 + 3t 4 + Since M X (t) 1 + E[X]t + E[X2 ] t 2 + E[X3 ] t 3 + 2! 3! we can match terms to find E[X 3 ] 8 3! and therefore E[X 3 ] 3! 8 48 Problem 8 (1 pts) Suppose X and Y are normal random variables, both with mean 4 and standard deviation 1. Suppose ρ(x, Y ) 1 5. Find the variance of X + Y. Since ρ(x, Y ) Cov(X, Y ) SD(X)SD(Y ) we have 1 5 Cov(X, Y ) 1 1 and therefore Cov(X, Y ) Now V ar(x + Y ) Cov(X + Y, X + Y ) Cov(X, X) + 2Cov(X, Y ) + V ar(y, Y )

4 Problem 9 (1 pts) Two people are playing a game. A fair die is rolled. If it turns up even, player A pays player B a number of dollars equal to the amount shown on the die. If it turns up odd, player B pays $3 to player A. The procedure is repeated until the die comes up 1, 2, 4, or 5, at which point the game ends. Find the expected winnings of player A. Example: The die is rolled, and comes up 6. Then player A pays $6 to player B. The die is rolled again, and comes up 1. Then player B pays $3 to player A, and the game is over. Player A s winnings here is then dollars. Let X be player A s winnings at the end of the game. Let Y be the first roll of the die. Then E[X] E[E[X Y ]] and we consider the following table: Therefore E[X] E[E[X Y ]] Y prob. E[X Y y] 1 1/ / /6 3 + E[X] 4 1/ / /6 6 + E[X] ( 2) (3 + E[X]) ( 4) ( 6 + E[X]) E[X] E[X] E[X] 1 2 E[X] 3 4 Problem 1 (5 pts) 5 people go to a fancy dinner, where there are 5 seats. The guests seat themselves randomly, and only later do they realize there are name cards, indicating that the seating was actually assigned beforehand. What is the expected number of people who are sitting at their correct seat? Let { 1 if the ith guest sits at the correct seat X i otherwise Then the number of people who sit at the correct seat is N X X 5 4

5 and its expected value is E[N] E[X 1 ] + + E[X 5 ] Since E[X i ] is the probability that the ith guest sits at the correct seat, and this is 1 5, we have 1 E[N] Problem 11 (1 pts) Prove that Cov(X + Y, Z) Cov(X, Z) + Cov(Y, Z). Cov(X + Y, Z) E[(X + Y E[X + Y ])(Z E[Z])] E[(X + Y E[X] E[Y ])(Z E[Z])] E[((X E[X]) + (Y E[Y ]))(Z E[Z])] E[(X E[X])(Z E[Z]) + (Y E[Y ])(Z E[Z])] E[(X E[X])(Z E[Z])] + E[(Y E[Y ])(Z E[Z])] Cov(X, Z) + Cov(Y, Z) Problem 12 (5 pts) Let X be exponential with parameter 1 and Y be exponential with parameter 2. Suppose X and Y are independent. Find the density function for X + Y. f X+Y (t) f X (s)f Y (t s) ds ({ e s 2 e 2(t s), if s > and t s >, otherwise ) ds So we are really integrating over the values of s where s > and s < t. This only occurs when t >, and then s ranges in the interval [, t]. f X+Y { t es 2 e 2(t s) ds if t > if t { t 2 e2t e s ds if t > if t Now t t 2 e 2t s ds 2 e 2t e s ds 2 e 2t (1 e t ) 2 e 2t 2 e t 5

6 So f X+Y (t) { 2 e 2t 2 e t t >, t Problem 13 (5 pts) Prove that ρ(x, Y ) 1. ( X V ar Y ) σ X σ Y ( ) ( X X V ar 2 Cov, Y ) ( ) Y + V ar σ X σ X σ Y σ Y 1 σ 2 V ar(x) 2 1 Cov(X, Y ) + 1 X σ X σ Y σ 2 V ar(y ) Y 1 2ρ(X, Y ) + 1 2(1 ρ(x, Y )) 1 ρ(x, Y ) Therefore ρ(x, Y ) 1 6