Solutions to Assignment #8 Math 501 1, Spring 2006 University of Utah

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Muriel Clementine Tate
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1 Solutions to Assignment #8 Math 5, Spring 26 University of Utah Problems:. A man and a woman agree to meet at a certain location at about 2:3 p.m. If the man arrives at a time uniformly distributed between 2:5 and 2:45, and if the woman independently arrives at a time uniformly distributed between 2: and p.m., then find the probability that the first to arrive waits longer than 7 minutes. What is the probability that the man arrives first? Solution: Let M denote the number of minutes the man arrives after 2:. Therefore, f M (m 3, if 5 m 45,, otherwise. Similarly, define W to be the number of minutes the woman arrives after 2:. Thus, f W (w 6, if w 6,., otherwise. By independence, f M,W (m, w 8, if m 45 and w 6,, otherwise. We are interested in finding P {M < W 7 or W < M 7}. Now f M,W is zero off the rectangle R : {(m, w : 5 m 45, w 6}. Let A denote the area, in R, that is bounded between the two lines w m + 7 and w m 7. Then, P {M < W 7 or W < M 7} Area(A Area(R Area(A 8. But A is a parallelopiped; so the area is base height. By the Pythagorean rule, base(a Also, the height is described by sin(45 height(a/4. Because sin(45 / 2, this yields height(a 4/ 2. Therefore, area(a , and P {M < W 7 or W < M 7} When a current I (in amperes flows through a resistance R (in ohms, the power generated is given by W I 2 R (in watts. Suppose that I and R are independent random variables with densities f I (x 6x( x x, f R (x 2x x.

2 Then find the density of W. Use this to compute EW. Solution: We have { 2xy( x, if x, y, f I,R (x, y, otherwise. Thus, for all a, F W (a P {W a} P { I 2 R a } f I,R (x, y dx dy, R where R is the region inside the square ( x, y that is under the curve y a/x 2. This region is a union of a rectangle and one bounded by a hyperbola. Therefore, for all a, F W (a a 2xy( x dx dy + a 2 + 6a( a 8a 3/2 ( a /2. a a/y 2xy( x dx dy Because W, F W (a if a <. Also, F W (a if a >. Therefore, { f W (a F W 6a 2 a + 6, if a, (a, otherwise. Consequently, EW ( 6a 2 2a 3/2 + 6a da Suppose that 3 balls are chosen without replacement from an urn consisting of 5 white and 8 red balls. Let X i equal if the ith ball selected is white, and let X i equal otherwise. Compute the conditional probability mass function of X given that X 2. Use this to find E[X X 2 ]. Solution: Let W i denote the event that we drew white on the ith draw. Then, P (X X 2 P (W W 2 P (W 2 W P (W P (W 2 P (W 2 W 3. Also, P (X X 2 2/3. Thus, E [X X 2 ] ( ( P (X X 2 + P (X X The density of (X, Y is f(x, y { xe x(y+, if x > and y >,, otherwise.

3 (a Find the conditional density of X, given Y y. Use this to compute E[X Y y]. Solution: First of all the density of Y at y > is f Y (y xe x(y+ dx If y then f Y (y. As a result, we have and f X Y (x y for x. Thus, ( + y 2. f X Y (x y x(y + 2 e x(y+, x >, E[X Y y] x 2 (y + 2 e x(y+ dx 2 + y. (b Find the conditional density of Y, given X x. Use this conditional density to compute E[Y X x]. Solution: First of all the density of X at x > is f X (x If x then f X (x. As a result, we have and f Y X (y x for y. Thus, xe x(y+ dy e x. f Y X (y x xe xy, y >, E[Y X x] xye xy dy x. (c Find the density of Z XY. Use this to compute E[Z] and Var(Z. Solution: F Z (a if a <. But if a then Therefore, F Z (a P {XY a} f Z (a df Z(a da a/y xe x(y+ dx dy. ( d a/y xe x(y+ dx dy. da By the fundamental theorem of calculus, f Z (a a y 2 e a(y+/y dy ay 2 e a(+y dy.

4 Set z : y to find that dz y 2 dy, and so f Z (a ae a(+z dz e a, a >. And f Z (a if a. Therefore, Z is exponential with mean one. So EZ and Var(Z. Theoretical Problems:. Suppose X is exponentially distributed with paramater λ >. Find P {[X] n, X [X] x}, where [a] denotes the largest integer a. Are X and X [X] independent? Solution: If x is a positive number and n is a positive integer, then [x] n is synonymous to n x n +. Therefore, P {[X] n, X [X] x} P {n X n +, X n + x}. There are three cases to consider: (i If x then P {[X] n, X [X] x} P {n X n + } (ii If x, then e n e n. P {[X] n, X [X] x} P {n X n + x} e n e n x. n+ n n+x n e z dz e z dz (iii If x <, then the probability in question is zero. No, X and X [X] are not independent, as can be seen from the above discussion. 2. Suppose X and Y are independent, standard normal random variables. Prove that Z : X/Y has the Cauchy density. That is, prove that the density of Z is f Z (a π( + a 2, < a <. Solution: We start, as before, and compute F Z first, and then differentiate. You will need to draw the region of integration in order to follow this discussion. We begin with the observation that { } X F Z (a P Y a P {X ay, Y } + P {X ay, Y < }. (eq.

5 Now, P {X ay, Y } ( ay e x2 /2 e y2 /2 dx dy. 2π 2π By the fundamental theorem of calculus, (d/da ( ay e y2 /2 dy ye a2 y 2 /2. Therefore, d ye y2 a 2 /2 e y2 /2 P {X ay, Y } dy ye (+a2 y 2 /2 dy da 2π 2π 2π 2π( + a 2 e z [ dz z : y 2 ( + a 2 /2 ] 2π( + a 2. (eq. Very similar computations show that d da P {X ay, Y < } 2π( + a 2. (eq.2 Combine (eq., (eq., and (eq.2 to deduce the result. 3. Suppose X is a standard normal random variable. Compute the density of Y X 2. Solution: Once more, we start with the distribution function: If a < then F Y (a because Y. Else if a then F Y (a P { a X a } Φ ( a Φ ( a. But if α then Φ( α Φ(α, by symmetry. Hence, Differentiate to find that if a >. Else, f Y (a. F Y (a 2Φ ( a. f Y (a F Y (a 2Φ ( a 2 a e a/2 2πa,

STAT/MA 416 Answers Homework 6 November 15, 2007 Solutions by Mark Daniel Ward PROBLEMS

STAT/MA 416 Answers Homework 6 November 15, 2007 Solutions by Mark Daniel Ward PROBLEMS STAT/MA 4 Answers Homework November 5, 27 Solutions by Mark Daniel Ward PROBLEMS Chapter Problems 2a. The mass p, corresponds to neither of the first two balls being white, so p, 8 7 4/39. The mass p,