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!

  • 7
    $\begingroup$ 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 $$ \frac{X_1+\cdots+X_n}{X_1+\cdots+X_n} = 1. $$ $\endgroup$ Sep 7, 2017 at 10:29
  • $\begingroup$ @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. $\endgroup$ Sep 7, 2017 at 12:30
  • 6
    $\begingroup$ @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 $$ \mathbb{E}\left[\frac{X_1}{X_1+\cdots+X_n}\right] = \mathbb{E}\left[\frac{X_{\sigma(n)}}{X_{\sigma(1)}+\cdots+X_{\sigma(n)}}\right] = \mathbb{E}\left[\frac{X_{\sigma(n)}}{X_1+\cdots+X_n}\right]. $$ $\endgroup$ Sep 7, 2017 at 12:33
  • $\begingroup$ @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? $\endgroup$
    – BAYMAX
    Sep 16, 2017 at 1:06
  • $\begingroup$ @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 $(0,\infty)$. $\endgroup$ Sep 16, 2017 at 2:03

1 Answer 1


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

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


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

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

  • $\begingroup$ How to prove that $\frac{X_i}{S_n}$ and $\frac{X_j}{S_n}$ are independent? $\endgroup$
    – Alex Grey
    Sep 7, 2017 at 11:10
  • 1
    $\begingroup$ @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 $\sum_{i=1}^nE\left(\frac{X_i}{S_n}\right)=1$. $\endgroup$ Sep 7, 2017 at 11:18
  • 6
    $\begingroup$ @AlexGrey, And in general they are not independent. $\endgroup$ Sep 7, 2017 at 11:21
  • $\begingroup$ @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. $\endgroup$ Sep 7, 2017 at 12:30
  • 1
    $\begingroup$ @LeGrandDODOM I feel we don't need anything more after Sangchul's comment. $\endgroup$ Sep 7, 2017 at 13:10

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