Proof that each random variable $Y_{j}$ has the same probability distribution

Question

Let $X_{1},X_{2},\cdots,X_{n}$ be nonnegative, independent and identically distributed random variables.Show that,if $k\leq n,$ then $$\mathbb{E}\left(\frac{X_1+\cdots+X_k}{X_1+\cdots+X_n}\right)=\frac{k}{n}.$$

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Set $$Y_j=\frac{X_j}{X_1+\cdots+X_n}\quad(j=1,2,\cdots,n),$$ we have that $\sum^{n}_{j=1}Y_j=1$ and each $\mathbb{E}(Y_j)$ dose exist (since $0<Y_j\leq 1$).

Also $$1=\mathbb{E}(\frac{X_1+\cdots+X_n}{X_1+\cdots+X_n})=\mathbb{E}(\sum^{n}_{j=1}Y_j).$$

If we can proof that each random variable $Y_{j}$ has the same probability distribution, then $$1=\mathbb{E}(\sum^{n}_{j=1}Y_j)=n\mathbb{E}(Y_j)\quad(j=1,2,\cdots,n),$$ and therefore $$\mathbb{E}\left(\frac{X_1+\cdots+X_k}{X_1+\cdots+X_n}\right)=\mathbb{E}(\sum^{k}_{j=1}Y_j)=k\mathbb{E}(Y_1)=\frac{k}{n}.$$

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I feel those $Y_j$ should be identically distributed,but how can we rigorously prove it instead of judging by intuition.

Answer 1

If the random variables $X_1,X_2,\ldots,X_n$ are independent and identically distributed, then they are also exchangeable. This means that the joint distribution of $(X_1,X_2,\ldots,X_n)$ is same as the joint distribution of $(X_{\sigma(1)},X_{\sigma(2)},\ldots,X_{\sigma(n)})$ for any permutation $\sigma$ on $\{1,2,\ldots,n\}$. This is the reason behind $\frac{X_j}{\sum_{i=1}^n X_i}$ having the same distribution for every $j=1,2,\ldots,n$.

So, for example if $n=3$, we have the following equality in distribution:

$$(X_1,X_2,X_3)\stackrel{d}= (X_2,X_3,X_1) \stackrel{d}=(X_3,X_1,X_2)$$

This implies

$$\frac{X_1}{X_1+X_2+X_3}\stackrel{d}=\frac{X_2}{X_2+X_3+X_1}\stackrel{d}=\frac{X_3}{X_3+X_1+X_2}$$