# $Y$ is equal to the sum of $n$ independent identically distributed Gaussian distribution variables, where $n$ is Poisson distribution

$$Y$$ is equal to the sum of $$n$$ independent identically distributed Gaussian distribution variables, where $$n$$ is Poisson distribution. If $$Y$$ is approximated to Gaussian distribution, what is its variance?$$Y = \sum\limits_{i = 0}^n {{x_i}},$$ where $${x_i} \sim N\left( {0,{\sigma ^2}} \right)$$, $$n \sim {\rm{Pois}}\left( \lambda \right)$$.

If we directly calculate the variance of $$Y$$ by summation, the variance is $$n{\sigma ^2}$$. When $$\lambda$$ is large, $$n{\sigma ^2}$$ is a Gaussian distribution variable. That is, $$Y$$ can be regarded as a Gaussian distribution with $${\sigma ^2_Y}$$ as Gaussian r.v. But I can't find any information about this composite distribution.

• Please show us what you have done and where you are stuck, and we will try to guide you to an answer; just asking for a solution is grounds for closing your question! Having said that, can you calculate the mean and variance of $Y$? Commented Oct 23, 2021 at 3:35
• such a intersting question Commented Oct 23, 2021 at 3:37
• do you guys know how a problem like this ($n$ as r.v) is called? Commented Oct 23, 2021 at 3:54
• Have you considered applying the Law of Total variance? Commented Oct 23, 2021 at 9:03

You can compute first the distribution of $$Y$$ conditional on the value of $$n$$. It is a normal distribution

$$Y_n \sim N(0,n\sigma^2)$$

### Compound mixture distribution

And the distribution of $$Y$$ unconditional on $$n$$ is like a compound distribution or mixture distribution.

$$Y \sim N(0, n\sigma^2) \qquad \text{where n \sim Pois(\lambda)}$$

To compute the raw moments you can take the sum of the individual components

$$E(Y) = \sum p(n) E(Y_n) = 0$$ $$E(Y^2) = \sum p(n) E(Y_n^2) = \sum p(n) n \sigma^2 = \bar{n} \sigma^2 = \lambda \sigma^2$$

And for the variance

$$Var(Y) = E(Y^2) - E(Y)^2 = \lambda \sigma^2$$

### Product distribution

You can view this also as a product distribution

$$Y = \sqrt{N}X \qquad \text{where N \sim Pois(\lambda) and X\sim N(0,\sigma^2)}$$

And then use this rule for the variance of a product of variables

$$Var(XY) = (\sigma_X^2 + \mu_X^2) (\sigma_Y^2 + \mu_Y^2) - \mu_X^2\mu_Y^2$$

We have $$\mu_X = 0$$ and

$$Var(XY) = \sigma_X^2 (\sigma_Y^2 + \mu_Y^2)$$

note that $$(\sigma_Y^2 + \mu_Y^2)$$ is equal to the raw moment $$E(Y^2)$$, and in our case $$Y$$ is the square root of a Poisson distributed variable so $$Y^2$$ is distributed as a Poisson distributed variable and it's expectation is $$\lambda$$.

$$Var(X\sqrt{N}) = \lambda\sigma^2$$

• Thank you for your generous help, the law of total variance can solve this problem well, and the answer is the same as you gave. Commented Oct 25, 2021 at 1:17
• KennTie the law of total variance is analogous to the solution with the viewpoint as the compound mixture distribution. Commented Oct 25, 2021 at 5:02