Uniform Distribution Statistic

Question

Let $X_1, X_2, X_3$ be i.i.d. $U(0,1)$ random variables. Then $E(X_1+X_2/X_1+X_2+X_3)=?$

Here $U(0,1)$ stands for uniform distribution which is independent and identically distributed.

Answer 1

I am not 100% sure, but I believe that $E({X_2\over X_1})$ does not converge and so the expectation you're looking for doesn't exist. I believe this can be demonstrated by finding the density of $X_2\over X_1$ and then using that density to find the expected value, which can be shown to be divergent:

$$ f_{X_2\over X_1} \left({x_2\over x_1} \right)=\left\{ \begin{matrix} {1\over 2} & 0 \le{x_2\over x_1} \le1 \\ {{x_1}^2\over {2{x_2}^2}} & {x_2\over x_1} >1 \\ 0 & \text{otherwise} \\ \end{matrix} \right.$$

Taking the expectation:

$$ \int_0^{\infty} {x_2\over x_1} f_{X_2\over X_1} \left({x_2\over x_1} \right)d{x_2\over x_1}=\text{integral is divergent}$$

So, just a disclaimer, I am pretty shaky on this material, but I don't see anyone else answering so I figured I'd share the little I know. If you search google for the expectation of $X_2\over X_1$, it seems that this does not converge, which is the result I get when evaluating the integral above.