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Suppose that $X_1 ,X_2,...X_N$ are independent identically distributed positive random variables. How to find the variance of $\frac{X_i}{\sum\limits_{j=1}^n X_j}$, $i \in 1, 2, \ldots, n$?

AdamO
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bvbvbvvbv
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    Is this an assignment for a class? If so, please add the [self-study] tag & read its wiki. – gung - Reinstate Monica May 10 '14 at 16:21
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    It seems to me that the variance of $\frac{X_i}{\sum\limits_{j=1}^n X_j}$ must equal that of $\frac{X_k}{\sum\limits_{j=1}^n X_j}$ for any $k$ and this observation should help you make a good start in answering this question. – whuber May 10 '14 at 17:44

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I don't think this question can be fully answered without more information. But we can find a few quantities. Let $n=N$ and define $Y_i={{X_i} \over {\sum_{j=1}^N X_j}}.$ Let the variance of $Y_i$ be $\sigma^2$ and the covariance be $\sigma_{Y_iY_j}=\sigma_{12}.$ We can do this since the $Y_i$ are identically distributed. Calling their sum $Z,$ we know $Z=1$ so $E[Y_i]= {{1 \over N}}.$ The variance of $Z$ is zero, so we also know $$\sigma^2_Z=\sum^N \sigma^2 + 2 \sum_{i < j} \sigma_{Y_iY_j}=0.$$ This yields $$N \sigma^2 + N(N-1)\sigma_{12}=0$$ so we have $$\sigma_{12}={{-\sigma^2} \over {N-1}}$$ The correlation $\rho$ between $Y_i$ and $Y_j$ is then $$\rho ={{\sigma_{12}} \over {\sigma^2}}={{-1} \over {N-1}}.$$ This is far as I've been able to get. I've not been able to find any expression for $\sigma^2$ - perhaps another can provide further insight.

soakley
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