Questions about averaging 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?
 A: 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 $\text{rnd}_n$
A: 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,
$$
V\pi_N = \frac{1}{N^2}\left[\left((1-a)^2+\sigma^2\right)^N - (1-a)^{2N}\right].
$$
