# Distribution of sum of product of normal random variables.

The distribution of a product $$Z=XY$$ of two normally distributed random variables is given by the product distribution https://mathworld.wolfram.com/NormalProductDistribution.html.

What is the distribution of $$Q=\sum_{i=1}^n Z_i$$, where $$Z_i= X_i Y_i$$, and $$X_i,Y_i\sim \mathcal N(0,\sigma_i)$$?

A nice answer was given in: Distribution of sum of product-normal distributions. by @wolfies, for the case where $$\sigma_i=1$$ (that is, all the $$X_i,Y_i$$ are identically distributed standard normal). The distribution can be expressed in terms of the modified Bessel function of the second kind. And as $$n\to\infty$$, the distribution approaches a normal distribution.

But I am interested in the case where the $$\sigma_i$$ are different. If no-closed form solution can be found, I am curious about when $$p(Q)$$ will approach a normal distribution. Is there an "effective" $$n$$ in terms of the $$\sigma_i$$? My first guess was the participation ratio $$n_\text{eff} = (\sum_{i=1}^n \sigma_i^2)^2/\sum_{i=1}^n \sigma_i^4$$.

Here is how I've approached it so far. The characteristic function of each $$Z_i$$ is given, I believe, by

$$\varphi_{Z_i}(t)=\frac{1}{\sqrt{t^2\sigma_i^2 + 1}}$$

So by independence the characteristic function of $$Q$$ is given by

$$\varphi_Q(t) = \prod_{i=1}^n\varphi_{Z_i}(t) = \prod_{i=1}^n \frac{1}{\sqrt{t^2\sigma_i^2 + 1}}$$

And $$p(Q)$$ should be given by the inverse Fourier transform of $$\varphi_Q(t)$$. I am stuck trying to perform the inverse Fourier transform, and any help would be greatly appreciated!

edit:

@Henry gave a very nice answer regarding the asymptotic behavior of $$p(Q)$$ as $$n\to\infty$$. but I am still curious about the behavior of $$p(Q)$$ for $$n$$ small. Can $$p(Q)$$ be computed exactly? If not, how large must $$n$$ be before $$p(Q)$$ is approximately normal, as a function of the $$\{\sigma_i\}$$?

If $$X_i$$ and $$Y_i$$ are independent and have zero mean then the variance of $$X_iY_i$$ will be $$\sigma_i^2$$ (or $$\sigma_{X_i}\sigma_{Y_i}$$ if these differ) so $$\sum\limits_{i=1}^n X_iY_i$$ will have zero mean and variance $$s_n^2=\sum\limits_{i=1}^n \sigma_i^2$$

The Central Limit Theorem will apply in the sense that $$\frac1{s_n} \sum\limits_{i=1}^n X_iY_i$$ will converge in distribution to $$\mathcal N(0,1)$$ as $$n$$ increases, providing that the $$\sigma_i^2$$ do not become too extreme, for example if they are bounded above and below by positive finite numbers, or if say the Lyapunov or Lindeberg conditions are met.

• Thanks! Those conditions are helpful. Is it possible to say what $p(Q)$ will look like for intermediate $n$? Or if not, given a set $\{\sigma_i\}$, how large must $n$ be before the distribution is approximately normal?
– Bean
Feb 3, 2021 at 23:55
• Do you suppose there is a closed form solution for $p(Q)$, or not?
– Bean
Feb 3, 2021 at 23:57
• I would be surprised if there was a closed form. If the $σ^2_i$ are not too different, I would expect something similar to the constant $σ^2$ case: a spike becoming bell shaped and then closer to a normal distribution, perhaps a little more slowly Feb 4, 2021 at 0:08
• That's right - but I am curious about how slowly, i.e. is there some function of the $\sigma_i$ that governs how large $n$ must be before the distribution looks normal?
– Bean
Feb 4, 2021 at 0:27