$Y$ 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.

  • 2
    $\begingroup$ 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 $Y$? $\endgroup$
    – jbowman
    Oct 23, 2021 at 3:35
  • $\begingroup$ such a intersting question $\endgroup$ Oct 23, 2021 at 3:37
  • $\begingroup$ do you guys know how a problem like this ($n$ as r.v) is called? $\endgroup$ Oct 23, 2021 at 3:54
  • $\begingroup$ I can't find any information about this composite distribution. $\endgroup$
    – Kenn Tie
    Oct 23, 2021 at 4:25
  • 1
    $\begingroup$ Have you considered applying the Law of Total variance? $\endgroup$
    – Glen_b
    Oct 23, 2021 at 9:03

1 Answer 1


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 $\lambda$.

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

  • $\begingroup$ 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. $\endgroup$
    – Kenn Tie
    Oct 25, 2021 at 1:17
  • $\begingroup$ KennTie the law of total variance is analogous to the solution with the viewpoint as the compound mixture distribution. $\endgroup$ Oct 25, 2021 at 5:02

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