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Given a sequence $X_1, X_2, ..., X_n$ of i.i.d. random values, the problem is to find the expectation of $E[\frac{X_1 + X_2 + ... + X_k}{X_1 + X_2 + ... + X_n}]$, where $k < n$.

There is no infromation about which distribution it is. We know that for i.i.d random values $E(\sum\limits_{i=0}^nX_i) = \sum\limits_{i=0}^nE(X_i)$. But what about the denominator?

I only had an idea that maybe we could present it as: $E[X_1 + X_2 + ... + X_k] = E[\frac{X_1 + X_2 + ... + X_k}{X_1 + X_2 + ... + X_n}*(X_1 + X_2 + ... + X_n)] = E[\frac{X_1 + X_2 + ... + X_k}{X_1 + X_2 + ... + X_n}]*E[X_1 + X_2 + ... + X_n] ?$

But then we have to prove that $\frac{X_1 + X_2 + ... + X_k}{X_1 + X_2 + ... + X_n}$ and $(X_1+X_2+...+X_n)$ are independent. And i got stuck.

That was only and idea, so i don't know what the right solution looks like. But the answer is $\frac{k}{n}$.

Nice guy
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1 Answers1

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$$E[\frac{X_1 + X_2 + ... + X_k}{X_1 + X_2 + ... + X_n}]=E(\frac{X_1}{X_1 + X_2 + ... + X_n})+...E(\frac{X_k}{X_1 + X_2 + ... + X_n}) \\=kE(\frac{X_1}{X_1 + X_2 + ... + X_n}) \quad \text{(symmetry)}\\=\frac{k}{n} \quad\text{(symmetry)}$$

The first "symmetry" should be clear. The second follows from the fact $$1=E(\frac{X_1+...+X_n}{X_1+...+X_n})=nE(\frac{X_1}{X_1+...+X_n})$$ using the first symmetry appeal again

user947124
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  • Thanks. But the thing that is not clear for me is that:we know for iid random values $E(\sum\limits_{i=0}^nX_i) = \sum\limits_{i=0}^nE(X_i)$. But do we have the independence here $E[\frac{X_1 + X_2 + ... + X_k}{X_1 + X_2 + ... + X_n}]=E(\frac{X_1}{X_1 + X_2 + ... + X_n})+...E(\frac{X_k}{X_1 + X_2 + ... + X_n})$ ?? If we had $E[\frac{X}{C}]$ where C is const, than yeah, $E[\frac{(\sum\limits_{i=0}^n X_i)}{C}] = \sum\limits_{i=0}^nE[\frac{X_i}{C}]$. But if we have $\sum\limits_{i=0}^nX_i$ instead of C, then is the indendence preserved so that we can exctract sum from expectation? – Nice guy Dec 16 '21 at 21:31
  • @Niceguy linearity of expectation (i.e. $E(X+Y) = E(X) + E(Y)$) holds without assuming independence. – Jose Avilez Dec 16 '21 at 21:34
  • @JoseAvilez oh, thanks. I was stuck at this moment. Easy, but not obvious – Nice guy Dec 16 '21 at 22:32
  • Linearity of expectation can be confusing (at the end comes from associativity and commutativity of addition) – Thomas Dec 17 '21 at 00:11