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i have some trouble with averages. Here are two questions rolled in one:

why is : $\frac{\prod _{n=1}^N \left(1-\text{rnd}_n\right)}{N} \neq \prod _{n=1}^N \frac{1-\text{rnd}_n}{N} \mbox{where $rnd_n$ is a random gaussian real} $

And how can i get $\frac{\prod _{n=1}^N \left(1-\text{rnd}_n\right)}{N}$ using only the mean and the variance of rnd, not the actual values ? So i only know how rnd is shaped, but not the values, that are supposed to average out anyway.

What rule about averaging am i violating?

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    why the downvote?2011-09-15

2 Answers 2

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As Ross has mentioned, you cannot know the actual value of the expressions you wrote based only on the characteristics of random variables such as mean or a variance. You can only ask for the distribution of these expressions.

E.g. in the case $\xi_n$ (rnd$_n$) are iid random variables, you can use the fact that $ \mathsf E[(1-\xi_i)(1-\xi_j)]=\mathsf E(1-\xi_i)\mathsf E(1-\xi_j) = (\mathsf E(1-\xi_i))^2$ which leads to the fact that $ \mathsf E \pi_N = \frac1N[\mathsf E(1-\xi_1)]^N = \frac{(1-a)^N}{N} $ where $a = \mathsf E\xi_1$. Here I denoted $ \pi_N = \frac{\prod\limits_{n=1}^N(1-\xi_n)}{N}. $

This holds regardless of the distribution of $\xi$, just integrability is needed. In the same way you can also easily calculate the variance of this expression based only on the variance and expectation of $\xi$ (if you want, I can also provide it).

Finally, there is a small hope that for the Gaussian r.v. $\xi$ the distribution of this expression will be nice since it includes the products of normal random variables.

On your request: variance.

Recall that for any r.v. $\eta$ holds $V \eta = \mathsf E \eta^2 - (\mathsf E\eta)^2$ hence $\mathsf E\eta^2 = V\eta+(\mathsf E\eta)^2$. As I told, you don't need to know the distribution of $\xi$, just its expectation $a$ and variance $\sigma^2$. Since we already calculated $\mathsf E\pi_N$, we just need to calculate $\mathsf E\pi^2_N$: $ \mathsf E\pi_N^2 = \frac1{N^2}\mathsf E\prod\limits_{n=1}^N(1-\xi_n)^2 = \frac{1}{N^2}\prod\limits_{n=1}^N\mathsf E(1-\xi_n)^2 = \frac{1}{N^2}\left(\mathsf E(1-\xi_1)^2\right)^N. $ Now, $ \mathsf E(1-\xi_1)^2 = \mathsf E(1-2\xi_1+\xi^2_1) = 1-2a+\mathsf E\xi_1^2 = 1-2a+a^2+\sigma^2 = (1-a)^2+\sigma^2 $ and $ \mathsf E\pi_N^2 = \frac{1}{N^2}\left((1-a)^2+\sigma^2\right)^N. $

As a consequence, $ V\pi_N = \frac{1}{N^2}\left[\left((1-a)^2+\sigma^2\right)^N - (1-a)^{2N}\right]. $

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    @tarrasch: here you are.2011-09-15
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For the first question, on the left the denominator is $N$, on the right it is $N^N$. For the second, you can't calculate $\frac{\prod _{n=1}^N \left(1-\text{rnd}_n\right)}{N}$ without the actual values. Imagine the random numbers are on $(0,1)$ and in one case you get a very small one. One (not me) could calculate the expectation or distribution of $\frac{\prod _{n=1}^N \left(1-\text{rnd}_n\right)}{N}$ from the distribution of $\text{rnd}_n$

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    thanks for the suggestion. i will upvote, but leave it open for a while still, to see what comes up2011-09-14