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

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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?

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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

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

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