# Variance of product of Gaussian random variables

Suppose I have $$r = [r_1, r_2, ..., r_n]$$, which are iid and follow normal distribution of $$N(\mu, \sigma^2)$$, then I have weight vector of $$h = [h_1, h_2, ...,h_n]$$, which iid followed $$N(0, \sigma_h^2)$$, how can I calculate the $$Var(\Sigma_i^nh_ir_i)$$? suppose $$h, r$$ independent.

How should I deal with the product of two random variables, what is the formula to expand it, I am a bit confused.

First just consider the individual components, which are gaussian r.v., call them $$r,h$$, $$r\sim N(\mu,\sigma^2),h\sim N(0,\sigma_h^2)$$ $$Var(rh)=\mathbb E(r^2h^2)-\mathbb E(rh)^2=\mathbb E(r^2)\mathbb E(h^2)-(\mathbb E r \mathbb Eh)^2 =\mathbb E(r^2)\mathbb E(h^2)$$ Under the given conditions, $$\mathbb E(h^2)=Var(h)=\sigma_h^2$$

$$\mathbb E(r^2)=\mathbb E[\sigma^2(z+\frac \mu\sigma)^2]\\ = \sigma^2\mathbb E(z+\frac \mu\sigma)^2\\ =\sigma^2\mathbb E[z^2+2\frac \mu\sigma z+\frac {\mu^2}{\sigma^2}]\\ =\sigma^2+\mu^2$$ $$z\sim N(0,1)$$ is standard gaussian random variables with unit standard deviation. Note the non-central Chi sq distribution is the sum $$k$$independent, normally distributed random variables with means $$\mu_i$$ and unit variances. Then $$r^2/\sigma^2$$ is such an RV.

Put it all together. $$Var(r^Th)=nVar(r_ih_i)=n \mathbb E(r_i^2)\mathbb E(h_i^2) = n(\sigma^2 +\mu^2)\sigma_h^2$$

If we are not too sure of the result, take a special case where $$n=1,\mu=0,\sigma=\sigma_h$$, then we know $$Var(rh)=\mathbb E(r^2h^2)=\mathbb E(r^2)\mathbb E(h^2) =Var(r)Var(h)=\sigma^4$$ which equals the result we obtained above.

I largely re-written the answer. The post that the original answer is based on is this.

Is the product of two Gaussian random variables also a Gaussian?

I found that the previous answer is wrong when $$\sigma\neq \sigma_h$$ since there will be a dependency between the rotated variables, which makes computation even harder. The answer above is simpler and correct.

• This approach feels slightly unnecessary under the assumptions set in the question. May 16 at 1:12
• @FD_bfa You are right! if variance is the only thing needed, I'm getting a bit too complicated May 16 at 1:18
• Thank you, that's the answer I derived, but I used the MGF to get $E(r^2)$, I am not quite familiar with Chi sq and will check out, but thanks!!! May 16 at 2:34

The first thing to say is that if we define a new random variable $$X_i$$=$$h_ir_i$$, then each possible $$X_i$$,$$X_j$$ where $$i\neq j$$, will be independent.

Therefore, we are able to say

$$Var(\sum_i^nX_i)=\sum_i^nVar(X_i)$$

Now, since the variance of each $$X_i$$ will be the same (as they are iid), we are able to say

$$\sum_i^nVar(X_i)=nVar(X_1)$$

So now let's pay attention to $$X_1$$. We know that $$h$$ and $$r$$ are independent which allows us to conclude that

$$Var(X_1)=Var(h_1r_1)=E(h^2_1r^2_1)-E(h_1r_1)^2=E(h^2_1)E(r^2_1)-E(h_1)^2E(r_1)^2$$

(by Fubini's Theorem).

We know that $$E(h_1)=0$$ and so we can immediately eliminate the second term to give us

$$Var(h_1r_1)=E(h^2_1)E(r^2_1)$$

And so substituting this back into our desired value gives us

$$\sum_i^nVar(X_i)=nE(h^2_1)E(r^2_1)$$

Using the fact that $$Var(A)=E(A^2)-E(A)^2$$ (and that the expected value of $$h_i$$ is $$0$$), we note that for $$h_1$$ it follows that

$$Var(h_1)=E(h^2_1)=\sigma^2_h$$

And using the same formula for $$r_1$$, we observe that

$$Var(r_1)=E(r^2_1)-\mu^2=\sigma^2$$

Rearranging and substituting into our desired expression, we find that

$$\sum_i^nVar(X_i)=n\sigma^2_h (\sigma^2+\mu^2)$$

Note: the other answer provides a broader approach, however, by independence of each $$r_i$$ with each other, and each $$h_i$$ with each other, and each $$r_i$$ with each $$h_i$$, the problem simplifies down quite a lot.

• $Var(h_1r_1)=E(h^2_1)E(r^2_1)=E(h_1)E(h_1)E(r_1)E(r_1)=0$ this line is incorrect... $r_i$ and itself is not independent so cannot be separated May 16 at 1:29
• Even from intuition, the final answer doesn't make sense ... $Var(h_iv_i)$ cannot be $0$ right? May 16 at 1:42
• Thanks for the answer, but as Wang points out, it seems to be broken at the $Var(h_1,r_1) = 0$, and the variance equals 0 which does not make sense. May 16 at 2:02
• @BinxuWang王彬旭 thanks for the answer, since $E(h_1^2)$ is just the variance of $h$, note that $Eh = 0$, I just need to calculate $E(r_1^2)$, is there a way to do it May 16 at 2:05
• I used the moment generating function of normal distribution and take derivative wrt t twice and set it to zero and got it. But thanks for the answer I will check it! May 16 at 2:22