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I have a random variable denoted $N$. Then I have random variables denoted $X_1, X_2, \dots, X_N$ distributed according to a uniform distribution. I have also random variables denoted $Y_1, Y_2, \dots, Y_N$ distributed according to this same probability distribution.

If we consider the variable $S_X$ such that $S_X=\sum_{i=1}^NX_i$ and $S_Y$ such that $S_Y=\sum_{i=1}^NY_i$.

Are $S_X$ and $S_Y$ mutually independent ?

If we extend this problem to the case of any number $A$ of sums, for instance $S_X$, $S_Y$ and $S_Z$ when $A=3$. Are these $A$ variables (these 3 sums) mutually independent ?

Thank you.

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Ok, thank you very much. In fact you could write this partial answer. I think this intuition is sufficient. – Dingo13 Jul 10 '13 at 17:05
rewrote it as you suggested. – gt6989b Jul 10 '13 at 17:06
Suppose the $X_i,Y_i$ are uniform on $[0,1]$ and we know that $S_X$ has taken on value $2013$. That makes it quite unlikely that $S_Y \lt \frac{1}{10}$! – André Nicolas Jul 10 '13 at 17:09

Assume $X_i,Y_i$ are already pairwise independent and let $\mathbb{E}[N] = \nu$.

Assuming $N$ is independent of the $X_i,Y_i$, Wald's indentity forces $$ \mathbb{E}[S_X] = \mathbb{E}[X_1] \mathbb{E}[N] = \nu/2 = \mathbb{E}[Y_1] \mathbb{E}[N] = \mathbb{E}[S_Y]. $$ There are similar results for other moments, so there is an explicit dependence on the distribution of $N$ for both $S_X$ and $S_Y$.

You can see that knowing the value of $S_X$ makes certain values of $N$ more likely than others, which will have an influnence on the distribution of $S_Y$.

On the other hand, fix $N$ to be a constant (not a random variable) and the dependence disappears.

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Thank you for the prompt answer. – Dingo13 Jul 10 '13 at 17:07

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