Solution to Exercise 5.22.

  We have reproduced the table of the joint p.m.f. but summed the rows and columns to get the marginal p.m.f.'s.

 

 

y=0 y=5 y=10 y=15 p_X

x=0 0.02 0.06 0.02 0.10 0.20

x=5 0.04 0.15 0.20 0.10 0.49

x=10 0.01 0.15 0.14 0.01 0.31

p_Y 0.07 0.36 0.36 0.21 1.00

(a) Using the marginal p.m.f.'s we get

$$E[X] \; = \; 0*0.20 + 5*0.49 + 10*0.31 \; = \; 5.55 .$$

$$E[Y] \; = \; 0*0.07 + 5*0.36 + 10*0.36 + 15*0.21 \; = \; 8.55 .$$

(b) Let $Z = \max{X,Y}$. Getting the p.m.f. of Z:

$$p_Z (z) \; = \; \left\{ \begin{array} {rcl} .02 & & \mbox{if... ... .10 + .10 + .01 = .21 & & \mbox{if } z = 15\end{array} \right.$$

For example, looking at z=5, the event [Z=10] can be decomposed into the mutually exclusive events $[X = 0 \, \& \, Y=5]$ or $[X = 5 \, \& \, Y=5]$ or $[X = 5 \, \& \, Y=0]$. Note the pattern of the values that are ``traced out'' in the table in this event. Checking that the p.m.f. values sum to 1:

.02 + .25 + .52 + .21 = 1 .

Thus,

$$E[Z] \; = \; 0*.02 + 5*.25 + 10*.52 + 15*.21 \; = \; 9.6 .$$