STAT/MATH 395 A - PROBABILITY II UW Winter Quarter 7 Néhémy Lim HW4 : Bivariate Distributions () Solutions Problem. The joint probability mass function of X and Y is given by the following table : X Y 8 8 4 (a) Give the marginal probability mass functions of X and Y. Answer. The marginal probability mass function of X is given by : p X () p XY (, ) + p XY (, ) 8 + 4 3 8 p X () p XY (, ) + p XY (, ) 8 + 5 8 The marginal probability mass function of Y is given by : p Y () p XY (, ) + p XY (, ) 8 + 8 4 p Y () p XY (, ) + p XY (, ) 4 + 3 4 (b) Are X and Y independent? Answer. No since p X () p Y () 3/8 /4 3/3 /8 p XY (, ). (c) Compute P(XY 3). Answer. The set of values that satisfy xy 3 is : B {(, ), (, ), (, )}. Thus,
P(XY 3) P ((X, Y ) B ) p XY (, ) + p XY (, ) + p XY (, ) 8 + 4 + 8 (d) Compute P(X/Y > ). Answer. The only couple (x, y) that satisfies x/y > is (,). Thus, P(X/Y > ) p XY (, ) 8 Problem. Assume that X and Y have joint probability density function : with S {(x, y) < y < x < } f XY (x, y) x S(x, y) (a) Verify that f XY is a valid joint probability density function. Answer. It is clear that f XY (x, y) for all x, y R. Second, we have : { x } f XY (x, y) dx dy dy x dx Therefore, f XY dx is a valid joint probability density function. (b) Give the marginal probability density function of X. Answer. The marginal probability density function of X, for x (, ), is given by : f X (x) and f X (x) otherwise. Thus : x f XY (x, y) dy x dy f X (x) (,) (x)
(c) Give the marginal probability density function of Y. Answer. The marginal probability density function of Y, for y (, ), is given by : f Y (y) and f Y (y) otherwise. Thus : (d) Find the expected value of X. y x dx ln(x) y ln(y) f XY (x, y) dx f Y (y) ln(y) (,) (y) Answer. Using the marginal probability density function of X, the expected value of X is : E[X] xf X (x) dx x dx (e) Find the expected value of Y. Answer. Using the marginal probability density function of Y, the expected value of Y is : E[Y ] yf Y (y) dy y ln(y) dy 3
Integrating by parts u ln(y) dv y dy du dy/y v y / gives : [ y ln(y) E[Y ] [ ] y + 4 4 ] y + dy Problem 3.. The joint probability density function of X and Y is given by : with S {(x, y) x >, y > } f XY (x, y) xe (x+y) S (x, y) (a) Give the marginal probability density function of X. Answer. The marginal probability density function of X for x > is given by : f X (x) and f X (x) otherwise. Thus, f XY (x, y) dy xe x e y dy xe x [ e y] xe x f X (x) xe x (, ) (x) (b) Give the marginal probability density function of Y. Answer. The marginal probability density function of Y for y > is given by : f Y (y) f XY (x, y) dx e y xe x dx e y e y 4
The third equality comes from the fact that we recognize the expected value of an exponential distribution with parameter. Hence, f Y (y) e y (, ) (y) (c) Are X and Y independent? Answer. Yes since for all x, y R, f X (x)f Y (y) xe x (, ) (x) e y (, ) (y) xe (x+y) (, ) (x, y) f XY (x, y). The joint probability density function of X and Y is given by : f XY (x, y) S (x, y) with S {(x, y) < x < y, < y < } (a) Give the marginal probability density function of X. Answer. The marginal probability density function of X for x (, ) is given by : f X (x) and f X (x) otherwise. Thus, x f XY (x, y) dy dy ( x) f X (x) ( x) (,) (x) (b) Give the marginal probability density function of Y. Answer. The marginal probability density function of Y for y (, ) is given by : f Y (y) y y and f Y (y) otherwise. Thus, (c) Are X and Y independent? Answer. No since for instance, f XY (x, y) dx dx f Y (y) y (,) (y) f X (.5)f Y (.5) (.5).5 f XY (.5,.5) 5
Problem 4. Let X and Y be two discrete random variables with joint probability mass function given by : p XY (x, y) 6 min(x,y) S(x, y) where S {(x, y) Z x, y, x y }. Reminder : Z is the set of all integers. (a) Plot the points in S in the xy-plane for x 3 and y 3. Answer. There are points in S such that x 3 and y 3. 3 y.5.5.5 3 x (b) Compute p X () the marginal probability mass function of X for x. Answer. For x, p XY (x, y) is nonzero for y and y. Therefore, the marginal probability mass function of X for x is p X () p XY (, ) + p XY (, ) 6 + 6 3. (c) Compute p X (x) the marginal probability mass function of X for x,, 3. Answer. The marginal probability mass function of X for x,, 3 is given by p X () p XY (, ) + p XY (, ) + p XY (, ) 6 + 6 + 6 3 p X () p XY (, ) + p XY (, ) + p XY (, 3) 6 + 6 + 6 6 p X (3) p XY (3, ) + p XY (3, 3) + p XY (3, 4) 6 + 6 3 + 6 3. 6
(d) Derive from (c) a general formula for p X (x) the marginal probability mass function of X for x N, x. Answer. In general, we have, for x N, x, p X (x) p XY (x, x ) + p XY (x, x) + p XY (x, x + ) 6 x + 6 x + 6 x 3 x. (e) Using a simple argument, give p Y (y) the marginal probability mass function of Y. Answer. Noting the symmetry of the joint probability mass function, we have 3 y p Y (y) 3 y N, y y otherwise (f) Compute P(X Y, X < ). Answer. The set of values in S that satisfy both x y and x < is : B {(, ), (, )}. Hence, P(X Y, X < ) P ((X, Y ) B) p XY (, ) + p XY (, ) 6 + 6 4 7