How to compute the mean of this ratio? Let $X_i$ be i.i.d. exponential random variables with mean $1$. How to compute the mean of this ratio?
$$ Y = \dfrac{X_i}{\sum\limits_{j=1,j\neq i}^{n}X_j}$$
Is it ?
\begin{align}\mathbb{E}[Y] &= \dfrac{\mathbb{E}[X_i]}{\sum\limits_{j=1,j\neq i}^{n}\mathbb{E}[X_j]}= \dfrac{1}{n-1}\end{align}
 A: I will call the denominator $S$. Then recognize that $S\sim\text{Gamma}(n-1, 1)$ and so $\frac{1}{S}$ follows an inverse gamma distribution with expectation $\frac{1}{(n-1)-1} = \frac{1}{n-2}$. If you didn't know that, then you can check with some math
\begin{align*}
E\left[\frac{1}{S}\right] &= \int_0^\infty \frac{1}{s}\cdot f_S(s)\,ds\\
&=\int_0^\infty \frac{1}{s}\cdot \frac{1}{\Gamma(n-1)}1^{n-1}s^{(n-1)-1}e^{-s}\,ds\\
&=\frac{\Gamma(n-2)}{\Gamma(n-1)}\int_0^\infty \frac{1}{\Gamma(n-2)}1^{n-2}s^{(n-2)-1}e^{-s}\,ds\tag 1\\
&=\frac{1}{n-2},
\end{align*}
where the integral in $(1)$ is a gamma density.
Finally, since $S$ and $X_i$ are independent, then
$$E\left[\frac{X_i}{S}\right] = E[X_i]E\left[\frac{1}{S}\right] = 1\cdot \frac{1}{n-2} = \frac{1}{n-2}.$$
A: Yes, it is. The sum (call it $S$) in the denominator of $Y$ has the gamma distribution with scale parameter $1$ and shape parameter $n-1$. That is, $S$ has density $f_S(s) = s^{n-2}e^{-s}/(n-2)!$ for $s>0$. By an easy integration, $\Bbb E[1/S]=1/(n-2)$, so $\Bbb E[X_i/S]=1\cdot 1/(n-2)=1/(n-2)$ by independence.
