I have the following homework assignment:
Let $X,Y$ be two independent $\operatorname{Exp}(\lambda)$ r.v. and $Z := X + Y$. Show that for any non-negative measurable $h$ we have $\mathbb{E}(h(X)|Z) = \frac{1}{Z} \int_0^Z h(t) dt$.
So far I've done this:
We know that $Z \sim \Gamma(2, \lambda)$, so $$ f_Z(z) = \frac{\lambda^2}{\Gamma(2)} z^{2-1} e^{-\lambda z} = \frac{\lambda^2}{1!} z^1 e^{-\lambda z} = z \cdot \lambda^2 \cdot e^{-\lambda z}. $$ Furthermore \begin{align*} F_{X,Z}(x,z) &= \mathbb{P}(X \leq x, Z \leq z) \\ &= \mathbb{P}(X \leq x, X + Y \leq z) \\ &= \mathbb{P}(0 \leq x, Y \leq z) \\ &= 1 \cdot \mathbb{P}(Y \leq z), \quad \text{if } x \geq 0 \\ &= 1 - e^{-\lambda z}, \quad \text{if } x \geq 0, \end{align*} derivating, the p.d.f. is $f_{X,Z}(x,z) = \lambda e^{-\lambda z}$, $x \geq 0$.
Thus the expected value is $$ \mathbb{E}(h(X)|Z) = \int\limits_\Omega h(x) \frac{f(x,z)}{f_Z(z)} dx = \int\limits_0^Z h(x) \frac{\lambda \cdot e^{-\lambda Z}}{Z \cdot \lambda^2 \cdot e^{-\lambda Z}} dx = \frac{1}{Z} \int\limits_0^Z h(x) \frac{1}{\lambda} dx. $$ As you can see, this is almost correct, with the exception of $\frac{1}{\lambda}$ in the integral. My question would be, how do I get rid of that?
As noted in the comments, the joint density of $(X,Z)$ is \begin{align} f_{X,Z}(x,z) &=f_{X,Y}(x,z-x)\\ &= \left(\lambda e^{-\lambda x}\mathsf 1_{(0,\infty)}(x)\right) \left(\lambda e^{-\lambda (z-x)}\mathsf 1_{(0,\infty)}(z-x)\right)\\ &= \lambda^2 e^{-\lambda z}\mathsf 1_{(0,z)}(x). \end{align} Hence the density of $Z$ is $$f_Z(z)=\int_\mathbb R f_{X,Z}(x,z)\,\mathsf dx=\lambda^2 e^{-\lambda z}\mathsf 1_{(0,\infty)}(z)\int_0^z\,\mathsf dx=\lambda^2 ze^{-\lambda z}\mathsf 1_{(0,\infty)}(z).$$ For $z>0$ we compute \begin{align} f_{X\mid Z=z}(x) &= \frac{f_{X,Z}(x,z)}{f_Z(z)}\\ &= \frac{\lambda^2 e^{-\lambda z}}{\lambda^2 z e^{-\lambda z}}\mathsf 1_{(0,z)}(x)\\ &= \frac1z \mathsf 1_{(0,z)}(x). \end{align}
It follows that for every nonnegative measurable $h$, \begin{align} \mathbb E[h(X)\mid Z] &= \int_0^\infty h(x)f_{X\mid Z}(x\mid Z)\,\mathsf dx\\ &= \int_0^\infty h(x)\frac1Z\mathsf 1_{(0,Z)}(x)\,\mathsf dx\\ &= \frac1Z\int_0^Z h(x)\,\mathsf dx. \end{align}