Given two independent and identically distributed random variables with exponential distribution $X,Y$, whose parameter is $\lambda$, compute $E[X^2\mid X+Y]$
My thoughts: let parameter be $\lambda$, then $P(X=x) = \lambda e^{-\lambda x}$
Let $Z = X+Y$, Then $P(Z=z) = \lambda^2 z e^{-\lambda z} $. But how do I go about computing $P(X^2\mid Z)$?
If you must: $$\begin{align}f_{\small X^2\mid Z}(s\mid z) &= \dfrac{f_{\small X^2,Z}(s,z)}{f_{\small Z}(z)} \\[1ex] &=\dfrac{f_{\small X^2}(s)\cdot f_{\small Y}(z-\surd s)}{f_{\small Z}(z)}\\[1ex] &=\dfrac{f_{\small X}(\surd s)\cdot f_{\small Y}(z-\surd s)}{2\sqrt{ s~}\cdot f_{\small Z}(z)\hspace{3ex}}\end{align}$$
However, you don't need to do that. You may use the Law of the Unconscious Statistician: $$\mathsf E(X^2\mid Z)=\int_\Bbb R x^2~f_{X\mid Z}(x\mid Z)\,\mathrm d x$$
Let $Z=X+Y$, then the conditional pdf for $X$ given $Z$ becomes:
$$f_{\small X\mid Z}(x\mid z)~=~\dfrac{f_{\small X}(x)\,f_{\small Y}(z-x)}{f_{\small Z}(z)}$$
Now since $f_{\small X}(x)=f_{\small Y}(x)=\lambda\mathrm e^{-\lambda x}\mathbf 1_{0\leqslant x}$ and $f_{\small Z}(z)=\lambda^2 z\mathrm e^{-\lambda z}\mathbf 1_{0\leqslant z}$ , then you can do the rest.