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Chapter 7: Q.51 (page 356)

The joint density of Xand Yis given by

f(x,y)=e-yy,0<x<y,0<y<∞

Compute EX3∣Y=y.

Short Answer

Expert verified

EX3∣Y=y=y34

Step by step solution

01

Given information

Given in the question that, The joint density of Xand Yis given by

f(x,y)=e-yy,0<x<y,0<y<∞.

02

Explanation

The joint density of Xand Yis given by f(x,y)=e-yyon the district 0<x<yand 0<y<∞. For simpler arrangement we'll introduce a diagram on the subsequent page. The goal is to work out the normal worth ofX3given Y=y. Initially note that:

f(x∣y)=f(x,y)fY(y)=e-yy·1e-y=1y

Now we simply calculate:

EX3∣Y=y=∫0yx3·1ydx=1y·x440y=y34

03

Graph

We integrate on the following region:

04

Final answer

EX3∣Y=y=y34

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Most popular questions from this chapter

Ten hunters are waiting for ducks to fly by. When a flock of ducks flies overhead, the hunters fire at the same time, but each chooses his target at random, independently of the others. If each hunter independently hits his target with probability .6, compute the expected number of ducks that are hit. Assume that the number of ducks in a flock is a Poisson random variable with mean 6.

7.4. If X and Y have joint density function fX,Y(x,y)={1/y,if0<y<1,0<x<y0,otherwisefind

(a) E[X Y]

(b) E[X]

(c) E[Y]

The best quadratic predictor ofYwith respect to Xis a + bX+cX2, where a, b, and c are chosen to minimize E[(Y−(a+bX+cX2))2]. Determine a, b, and c.

In the text, we noted that

E∑i=1∞Xi=∑i=1∞EXi

when the Xiare all nonnegative random variables. Since

an integral is a limit of sums, one might expect that E∫0∞X(t)dt=∫0∞E[X(t)]dt

whenever X(t),0≤t<∞,are all nonnegative random

variables; this result is indeed true. Use it to give another proof of the result that for a nonnegative random variable X,

E[X)=∫0∞P(X>t}dt

Hint: Define, for each nonnegative t, the random variable

X(t)by

role="math" localid="1647348183162" X(t)=1ift<X\\0ift≥X

Now relate4q

∫0∞X(t)dttoX

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