# Find expectation of $\frac{X_1 + \cdots + X_m}{X_1 + \cdots + X_n}$ when $X_1,\ldots,X_n$ are i.i.d

Let $$X_1, \ldots, X_n$$ be i.i.d. random variables with expectation $$a$$ and variance $$\sigma^2$$, taking only positive values. Let $$m < n$$. Find the expectiation of $$\displaystyle\frac{X_1 + \cdots + X_m}{X_1 + \cdots + X_n}$$.

My attemps to solve this probles are rather straightforward. Denote $$X = X_1 + \cdots + X_m$$ and $$Y = X_{m+1} + \dots + X_n$$. So, $$X$$ has the expectation $$ma$$ and the variance $$m\sigma^2$$. And $$Y$$ has the expectation $$(n-m)a$$ and variance $$(n-m)\sigma^2$$. And also $$X$$ and $$Y$$ are independent. So we can compute the expectation by the definition $$\mathbb{E}\displaystyle\frac{X}{X+Y} = \int\limits_{\Omega^2}\frac{X(\omega_1)}{X(\omega_1) + Y(\omega_2)}\mathbb{P}(d\omega_1)\mathbb{P}(d\omega_2)$$. But we do not know the distribution, so we do not have chance to calculate it.

I would be glad to any help or ideas!

• Hint. You have $$\mathbb{E}\left[ \frac{X_1}{X_1+\cdots+X_n} \right] = \cdots = \mathbb{E}\left[ \frac{X_n}{X_1+\cdots+X_n} \right]$$ and $$\frac{X_1+\cdots+X_n}{X_1+\cdots+X_n} = 1.$$ – Sangchul Lee Sep 7 '17 at 10:29
• @SangchulLee how do you prove $\mathbb{E}\left[ \frac{X_1}{X_1+\cdots+X_n} \right] = \mathbb{E}\left[ \frac{X_n}{X_1+\cdots+X_n} \right]$ ? It doesn't look obvious to me. – Gabriel Romon Sep 7 '17 at 12:30
• @LeGrandDODOM, Since $X_1, \cdots, X_n$ are i.i.d., they are exchangable: for any permutation $\sigma$ on $\{1,\cdots,n\}$ we have the following equality in distribution: $$(X_1, \cdots, X_n) \stackrel{d}{=} (X_{\sigma(1)}, \cdots, X_{\sigma(n)} ).$$ And since the expectation depends only on the distribution, we have $$\mathbb{E}\left[\frac{X_1}{X_1+\cdots+X_n}\right] = \mathbb{E}\left[\frac{X_{\sigma(n)}}{X_{\sigma(1)}+\cdots+X_{\sigma(n)}}\right] = \mathbb{E}\left[\frac{X_{\sigma(n)}}{X_1+\cdots+X_n}\right].$$ – Sangchul Lee Sep 7 '17 at 12:33
• @SangchulLee if say $s_{i} = \frac{X_{i}}{\sum_{j=1}^{n}}X_{j}$, then while calculating expectation (in continuous case) we will have $s_{i}$ under the integrsl sign too,that is while calculating $E(s_{i})$ the term inside integral that is $s_{i}$ will be varing ,so how can we show that expectation of each $s_{i}$ are the same? or if $X_{i}$ are identically distributed then does that mean $s_{i}$ are identically distributed? I too get theintuition that they are same but how do i prove it? – BAYMAX Sep 16 '17 at 1:06
• @BAYMAX, If $X_i$ has common p.d.f. $f$, then $$\mathbb{E}\left[\frac{X_i}{X_1+\cdots+X_n}\right]=\int_{(0,\infty)^n}\frac{x_i}{x_1+\cdots+x_n}f(x_1)\cdots f(x_n)\,dx_1\cdots dx_n.$$ Now you can interchange the role of $x_1$ and $x_i$ to find that this expectation does not depend on $i$. This line of reasoning can be extended to arbitrary distribution on $(0,\infty)$. – Sangchul Lee Sep 16 '17 at 2:03

Suppose $$S_m=\sum\limits_{i=1}^{m} X_i$$ and $$S_n=\sum\limits_{i=1}^n X_i$$.

Now, $$\frac{X_1+X_2+\cdots+X_n}{S_n}=1\,, \text{ a.e. }$$

Therefore,

$$\mathbb E\left(\frac{X_1+X_2+\cdots+X_n}{S_n}\right)=1$$

Since $$X_1,\ldots,X_n$$ are i.i.d (see @SangchulLee's comments on main), we have for each $$i$$,

$$\mathbb E\left(\frac{X_i}{S_n}\right)=\frac{1}{n}$$

So for $$m\le n$$, $$\mathbb E\left(\frac{S_m}{S_n}\right)=\sum_{i=1}^m \mathbb E\left(\frac{X_i}{S_n}\right)=\frac{m}{n}$$

• How to prove that $\frac{X_i}{S_n}$ and $\frac{X_j}{S_n}$ are independent? – Alex Grey Sep 7 '17 at 11:10
• @AlexGrey That isn't really used anywhere. As the $X_i$'s are identically distributed, $E\left(\frac{X_1}{S_n}\right)=E\left(\frac{X_2}{S_n}\right)=...=E\left(\frac{X_n}{S_n}\right)$ and we simultaneously have $\sum_{i=1}^nE\left(\frac{X_i}{S_n}\right)=1$. – StubbornAtom Sep 7 '17 at 11:18
• @AlexGrey, And in general they are not independent. – Sangchul Lee Sep 7 '17 at 11:21
• @StubbornAtom How do you prove $E\left(\frac{X_1}{S_n}\right)=E\left(\frac{X_2}{S_n}\right)$ ? It doesn't look obvious to me. – Gabriel Romon Sep 7 '17 at 12:30
• @LeGrandDODOM I feel we don't need anything more after Sangchul's comment. – StubbornAtom Sep 7 '17 at 13:10