Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ and $Y$ be independent exponential random variables with means 1 and 2 respectively. Let $Z = 2X + Y$. How can I find $E(X|Z)$?

share|improve this question
add comment

2 Answers

up vote 0 down vote accepted

Let $U = 2x + Y$ and $V = X$

$f_{u,v}(u,v) = f_{X,Y}(v, u-2v) \left| \begin{array}{cc} \frac{\partial v}{\partial u} & \frac{\partial u}{\partial v} \\ \frac{\partial (u-2v)}{\partial u} & \frac{\partial (u-2v)}{\partial v} \end{array} \right|$

= $\frac{1}{2} e^{-u/2}.1_{0, u/2}(v).1_{0,\infty}(u)$

Therefore, the conditional expectation is

$E[V|U] = \frac{\int_0^{u/2} v/2 \ exp[-U/2]dv}{\int_0^{U/2} 1/2 \ exp[-U/2]dv} = \frac{U}{4}$

share|improve this answer
add comment

Hint: Note that $Y=2T$ with $(X,T)$ i.i.d., hence $\mathbb E(X\mid Z)=\mathbb E(X\mid X+T)$. Furthermore, $X+T=\mathbb E(X+T\mid X+T)=\mathbb E(X\mid X+T)+\mathbb E(T\mid X+T)$ and, for every random variables $(\xi,\eta)$, $\mathbb E(\xi\mid\eta)$ depends only on the distribution of $(\xi,\eta)$ in the following sense:

If $(\xi',\eta')$ is distributed like $(\xi,\eta)$ and $\mathbb E(\xi\mid\eta)=a(\eta)$, then $\mathbb E(\xi'\mid\eta')=a(\eta')$.

Can you take it from here?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.