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Let X,Y be i.i.d. exponentially distributed with parameter $\lambda$. Show that for $Z:=X+Y$ and a measurable, non-negative function $h$ we have:
$\mathbb{E}(h(X)|Z)=\frac{1}{Z}\int_0^Zh(u)\mathrm{du}$.

I know that $Z$ is gamma-distributed with shape $2$ and rate $\lambda$. Then using the joint distribution of $X$ and $Y$ I calculated the joint density of $X$ and $Z$ to be $f_{X,Z}(x,z)=\chi_{z\ge x}\lambda^2\exp(-\lambda z)$. Using this, and the formula for conditional expectation, I got $\mathbb{E}(h(X)|Z=z)=\int_0^\infty h(u)\cdot f_{X,Z}(x,z)/f_Z(z) \mathrm{du}=\int_0^Zh(u)\mathrm{du}$. I can't see why I didn't get the correct result. Also, do we need the assumption that $h$ is non-negative?

Edit: Of course, right after posting this, I find that a accidentally lost the $1/z$ along the way in the last equality. The second part of the question is still not clear to me though.

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    $\begingroup$ Because you forgot that $f_Z(z)$ is not $\lambda^2\exp(-\lambda z)$ but $\lambda^2z\exp(-\lambda z)$. "Also, do we need the assumpation that h is non-negativ(e)?" No, why should we? $\endgroup$ – Did Apr 17 '15 at 11:32
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We have that,

$$f_{X|Z}(x|z)=\frac{f_{Z|X}(z|x)\,f_X(x)}{f_Z(z)}=\frac{f_{Y}(z-x)\,f_X(x)}{f_Z(z)}=\frac{\exp\{-\lambda(z-x)\}\exp\{(-\lambda x)\}}{z\exp\{(-\lambda z)\}}$$

leading to $$f_{X|Z}(x|z)=\frac{1}{z}\mathbb{1}{\{x\leq z\}}.$$

It is easy to see that $X|Z=z \sim U(0,z)$ and therefore $$\mathbb{E}_{X|Z} [h(X)|z]=\frac{1}{z}\int_0^z h(X) dx.$$ Q.E.D. Note that, in particular, $\mathbb{E}_{X|Z} [X|z] = \frac{z}{2}$ in light of the expectation of an Uniform distribution.

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