Homework 4 Solutions

Transcription

1 Homework 4 Solutions February 10, Random variables X and Y have the joint PMF { c(x 2 + y 2 ) if x {1, 2, 4} and y {1, 3} p X,Y (x, y) = 0 otherwise (1) (a) What is the value of the constant c? (b) What is P[Y < X]? (c) What is P[Y > X]? (d) What is P[Y = X]? (e) What is P[Y = 3]? (f) Find the marginal PMFs p X (x) and p Y (y) (g) Find the expectations E[X], E[Y ], and E[XY ] (h) Find the covariances var(x), var(y ), and var(x + Y ) (i) Let A denote the event X Y. Find E[X A] and var(x A) (a) From the joint PMF, there are six (x, y) coordinate pairs with nonzero probabilities of occurring. These pairs are (1,1), (1,3), (2,1), (2,3), (4,1), and (4,3). The probability of a pair is proportional to the sum of the squares of the coordinates of the pair, x 2 + y 2. Because the probability of the entire sample space must equal to 1, we have: (1 + 1)c + (1 + 9)c + (4 + 1)c + (4 + 9)c + (16 + 1)c + (16 + 9)c = 1 (2) Solving for c, we get, c = 1 72 (3) (b) There are three sample points for which y < x: P(Y < X) = P({2, 1}) + P({4, 1}) + P({4, 3}) = = (4) 1

2 (c) There are two sample points for which y > x: P(Y > X) = P({1, 3}) + P({2, 3}) = = (5) (d) There is only one sample point for which y = x: P(Y = X) = P({1, 1}) = 2 72 (6) Notice that, using the above two parts, as expected. P(Y < X) + P(Y > X) + P(Y = X) = = 1 (7) (e) There are three sample points for which y = 3: P(Y = 3) = P({1, 3}) + P({2, 3}) + P({4, 3}) = = (8) (f) In general, for two discrete random variable X and Y for which a joint PMF is defined, we have, p X (x) = p X,Y (x, y) y= p Y (y) = x= p X,Y (x, y) (9) In this problem the ranges of X and Y are quite restricted so we can determine the marginal PMFs by enumeration. For example, p X (2) = P({(2, 1)}) + P({(2, 3)}) = 18 (10) 72 Overall, we get: and, 12/72 if x = 1, 18/72 if x = 2, p X (x) = 42/72 if x = 4, 24/72 if y = 1, p Y (y) = 48/72 if y = 3, (g) In general, the expected value of any discrete random variable X equals E[X] = 2 x= (11) (12) x p X (x) (13)

3 For this problem, E[X] = = 3 E[Y ] = = 7 3 To compute E[XY ], note that p X,Y (x, y) p X (x)p Y (y). Therefore, X and Y are not independent and we can not assume E[XY ] = E[X]E[Y ]. Thus, we have E[XY ] = x = 1 xy p X,Y (x, y) y = 61 9 (14) (15) (h) The variance of a random variable X can be computed as E[X 2 ] E[X] 2 or as E[(X E[X]) 2 ]. We use the second approach here because X and Y take on such limited ranges. We have, var(x) = (1 3) (2 3)2 + (4 3) = 3 2 var(x) = ( ) ( ) = 8 9 X and Y are not independent, so we can not assume var(x + Y ) = var(x) + var(y ). The variance of X+Y will be computed using var(x + Y ) = E[(X + Y ) 2 ] (E[X + Y ]) 2. Therefore, we have (16) Therefore, E[(X + Y ) 2 ] = = (E[X + Y ]) 2 = (E[X] + E[Y ]) 2 = ( )2 = var(x + Y ) = = (17) (18) (19) (i) There are four (x,y) coordinate pairs in A : (1,1), (2,1), (4,1), and (4,3). Therefore, P(A) = ( )/72 = 49/72. To find E[X A] and var(x A), p X A (x) must be calculated. We have 2/49 if x = 1, 5/49 if x = 2, p X A (x) = 42/49 if x = 4, 0 otherwise, 2 E[X A] = = E[X 2 A] = = var(x A) = E[X 2 A] (E[X A]) 2 = ( )2 = , (23) (20) (21) (22) 3

4 2. Suppose that X, Y, and Z are independent random variables such that each is equal to 0 with probability 0.5 and 1 with probability 0.5 (a) Compute the conditional probability P[X + Y + Z = 1 X Y = 0] (b) Are the events {X = Y }, {Y = Z} and {X = Z} independent? Are they pairwise independent? Explain. (a) Both events occur if and only if both X = Y = 0 and Z = 1. So, P(X + Y + Z = 1 X Y = 0) = P(X + Y + Z = 1, X Y = 0) P(X Y = 0) = (1/8)/(1/2) = 1/4 (24) (b) Not independent. Each event has probability 1/2 but probability all events occur is 1/4 (1/2) 3. Are pairwise independent, since probability of any two occurring is (1/2) 2 = 1/4 4

5 3. Jill sends her resume to 1000 companies she finds on monster.com. Each company responds with probability 3/1000 (independent of what all other companies do). Let R be the number of companies that respond (a) Compute E[R] (b) Computer var[r] (c) Use a Poisson random variable approximation to estimate the probability P[R = 3] (a) R is binomial random variable with n = 1000 and p = 3/1000, therefore, E[R] = np = 3 (25) (b) R is binomial random variable with n = 1000 and p = 3/1000, therefore, var(r) = np(1 p) = 3(1 3/1000) (26) (c) R is approximately Poisson with λ = 3. So, P{R = 3} e λ λ k /k! = e /3! = 9e 3 /2 (27) 5

6 4. The number of calls arriving at a Police Station follows a Poisson distribution with rate 4.6/hour. (a) What is the probability that exactly six calls will come between 8:00 PM and 9:00 PM? (b) Find the probability that exactly seven calls will come between 9:00 PM and 10:30 PM. (a) Let X be the random number arriving in this one-hour time period. We ll use λ = 4.6 and then find P[X = 6] = e 4.6 (4.6) 6 /6! (28) (b) Let Y be the random number arriving during this 90-minute period. The Poisson rate parameter expands and contracts appropriately, so the relevant value of λ is λ = = 6.9, hence P[Y = 7] = e 6.9 (6.9) 7 /7! (29) 6

7 5. Let f(x) = cx 2 for x = 1, 2, 3. Determine the constant c so that function f(x) satisfies the conditions of being a probability mass function. For f(x) to be a probability mass function, we should have, f(x) = c + 4c + 9c = 1 x c = 1/14 (30) 7

8 6. A stock market trader buys 100 shares of stock A and 200 shares of stock B. Let X and Y be the price changes of stock A and B, respectively, over a certain time period, and assume that the Joint PMF of X and Y is uniform over the set of integers x and y satisfying (a) Find the marginal PMFs and the means of X and Y (b) Find the mean of the traders profit 2 x 4 1 y x 1 (31) (a) There are 21 integer pairs (x,y) in the region R = {(x, y) 2 x 4, 1 y x 1} (32) so that the joint PMF of X and Y is p X,Y (x, y) = { 1/21 if (x, y) is in R, (33) For each x in the range [ 2, 4], there are three possible values of Y. Thus, we have, { 3/21 if x = 2, 1, 0, 1, 2, 3, 4, p X (x) = (34) The mean of X is the midpoint of the range [ 2, 4]: E[X] = 1 (35) The marginal PMF of Y is obtained by using the tabular method. We have, 1/21 if y = 3 2/21 if y = 2 3/21 if y = 1, 0, 1, 2, 3 p Y (y) = 2/21 if y = 4 1/21 if y = 5 (36) The mean of Y is E[Y ] = 1 21 ( 3 + 5) ( 2 + 4) + 3 ( ) = 1 (37) 21 (b) The profit is given by so that, P = 100X + 200Y (38) E[P ] = 100 E[X] E[Y ] = = 300 (39) 8