Let $X_1, \dots, X_n$ be iid random variables with positive mean. Let $W_i=X_i/(X_1+\cdots+X_n)$, find $E(W_i)$.
The proof goes like this: since $E(W_1+\cdots+W_n)=E(1)$, by symmetry and linearity of expectation, the answer is $1/n$.
My question is, how does symmetry work here? I don't get why for example $E(X_1/(X_1+X_2))=E(X_2/(X_1+X_2))$. And more generally when can we use “by symmetry”? Thank you in advance.
Best Answer
The random variables $W_1,\dots,W_n$ are all equal in distribution since $X_1,\dots,X_n$ are identically distributed. Thus $E(W_1) = \dots = E(W_n)$. This is the "by symmetry" part of the argument.
Let $\mu$ denote the law of $X_1$ (which is the same as the law of any of $X_2,\dots,X_n$ by assumption). The intuition for why $W_1,\dots,W_n$ are equal in expectation is that we form $W_1,\dots,W_n$ by taking $n$ random samples from $\mu$, and then picking any one of the samples as the numerator, and using the sum of the samples as the denominator. Since the numerator samples are all coming from the same distribution, we expect $E(W_i)$ to be a quantity that does not depend on the particular index $i$.
You can prove for instance that $E(W_1) = E(W_2)$ as follows. By independence of $X_1,\dots,X_n$, the pushforward $(X_1,\dots,X_n)_*P = \mu\otimes\dots\otimes \mu$ ($n$ times), so by Fubini's theorem and the change of variable $$ (x_1,x_2,\dots,x_n)\mapsto (x_2,x_1,\dots,x_n), $$ we have: \begin{align*} E(W_1) &= E\bigg(\frac{X_1}{X_1+X_2+\dots+X_n}\bigg) \\ &= \int_{\mathbf R^{n}}\frac{x_1}{x_1+x_2+\dots+x_n}\,\mu(dx_1)\mu(dx_2)\dots\mu(dx_n)\\ &= \int_{\mathbf R^n}\frac{x_2}{x_1+x_2\dots+x_n}\,\mu(dx_2)\mu(dx_1)\dots\mu(dx_n)\\ &= E(W_2). \end{align*}