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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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The right hand side of the first (in)equation is the same as: $$ \frac{\prod_{n=1}^N (1-\mathrm{rnd}_n)}{N^N}, $$ which is clearly different from the left hand side. –  Srivatsan Sep 14 '11 at 15:27
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Note that e.g. $$\frac{1}{N} = \frac{\prod_{n=1}^N 1}{N} \neq \prod_{n=1}^N \frac{1}{N} = \frac{1}{N^N}$$ –  TMM Sep 14 '11 at 15:29
    
thanks for the fast feedback. i will give my problem another thought. perhaps i can reformulate. –  tarrasch Sep 14 '11 at 15:58
    
why the downvote? –  tarrasch Sep 15 '11 at 6:57

2 Answers 2

up vote 3 down vote accepted

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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thanks for the answer, i will take a look at it tomorrow :) btw, what does iid mean? –  tarrasch Sep 14 '11 at 16:34
    
    
after giving it some more thought, i think you are right. The only thing i can do is giving an expected value and the maybe the confidence interval at 5% or sth. could you also include the variance of the distribution, then i will accept the answer. –  tarrasch Sep 15 '11 at 9:49
    
@tarrasch: here you are. –  Ilya Sep 15 '11 at 10:53

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 up –  tarrasch Sep 14 '11 at 15:59

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