Compute $E(\sqrt{X+Y})$ given that $X,Y$ are iid.
Assume $X,Y$ are iid both having an Exp($\lambda=1)$ distribution.
Although this is the only information provided in the question, I know that since they are independent, then the joint distribution must be $$f_{X,Y}(X,Y)=e^{-x} e^{-y}$$
Then we must integrate:
$$\int_{0}^{\infty}\int_{0}^{\infty}e^{-x}e^{-y}(x+y)^{1/2}dxdy=\int_{0}^{\infty}e^{-y}\int_{0}^{\infty}e^{-x}(x+y)^{1/2}dxdy$$
This integral seems complicated to evaluate. I tried it out and it seems like I'll have to do parts atleast twice. Wolfram says the answer is $1.32934$ and I have $1.33$ as one of the options to this question.
I want to know if there's an easier way to evaluate this. Do I even need the integral? Or perhaps is there some trick to get the answer with this integral?
One way of getting the integral slightly faster is to see that everything is in terms of $x+y$ only.
So let $z=x+y$, which ranges from $0$ to $\infty$, and then given $z$ we have $x$ ranges from $0$ to $z$. The Jacobian for this is just $1$.
Then the double integral becomes $$ \int_0^{\infty}\int_0^ze^{-z}z^{\frac{1}{2}}\,\mathrm{d}x\,\mathrm{d}z=\int_0^{\infty}e^{-z}z^{\frac{3}{2}}\,\mathrm{d}z=\Gamma\left(\frac{5}{2}\right)=\frac{3\sqrt{\pi}}{4}$$