Value of $E\left[ \frac{X_1 +\cdots + X_k}{X_1 + \cdots + X_n} \right]$ if $X_1, \ldots , X_n$ are i.i.d. random variables

Question

Let $X_1, \ldots , X_n$ be independent and identically distributed random variables. Calculate: $$E\left[ \frac{x_1 + \cdots + x_k}{x_1 + \cdots + x_n} \right]$$

I was wondering if I went about this the right way. $$E \left[ \frac{x_1 + \cdots + x_k}{x_1 + \cdots + x_n} \right]$$ $$= \frac{E[x_1 + \cdots + x_k]}{E[x_1 + \cdots + x_n]}, \text{ since each is i.i.d } \implies $$ $$\frac{\mu_1 + \cdots + \mu_k}{\mu_1 + \cdots + \mu_n}$$ $$= \frac{k\mu}{n\mu} = \frac{k}{n}$$ $$= 0 \text{ if $k = n$ and }\frac{k}{n} \text{ if n > k}$$

Answer 1

It is unfortunately not true that $E[X/Y]=E[X]/E[Y]$ even if $X$ and $Y$ are independent.

However, we can show that $$\tag{*} \mathbb E\left[\frac{X_j}{X_1+\dots+X_n}\right]=\mathbb E\left[\frac{X_1}{X_1+\dots+X_n}\right]=a. $$ (this is due to the fact that $(X_1,\dots,X_n)$ and $(X_{\sigma(1)},\dots,X_{\sigma(n)})$, where $\sigma$ is a permutation, have the same law and $\sum_{i=1}^nX_{\sigma(i)}= \sum_{i=1}^nX_i$.

Also, note that $na=1$, by summing over $j$ in (*).