What's the resulting distribution when you take the N times self convolution of normal(mu, sigma) where N~Poisson or binomial

Question

So if you have $N \sim \mathrm{Poisson}(\lambda)$ or $N \sim \mathrm{Binomial}(n,p)$ I'm curious about the convolution $$X = \sum_{i=1}^N X_i$$ (where $X_i \sim \mathrm{Normal}(\mu, \sigma)$ and $X=0$ if $N=0$) I guess alternatively this is the compound distribution $$\mathrm{Normal}(N\cdot\mu, \sqrt{N}\cdot\sigma)$$ but that's not much more help.

Another way to think about this, if you know the number of customers entering a bank in a day is determined by a binomial distribution, and the amount they each deposit/withdraw is a normal distribution, what is the distribution for the amount deposited/withdrawn in a day (not just the EV)?

Answer 1

This is just a law of total probability analysis. Given $N = n > 0$, the conditional distribution of $\sum_{i=1}^n X_i$ is a normal distribution with mean $n\mu$ and variance $n\sigma^2$ as the OP correctly says. For $n = 0$, $P\{X \leq x \mid N = 0\}$ has value $0$ if $x < 0$ and value $1$ if $x \geq 0$. Thus, $$\begin{align*} P\{X \leq x\} &= \sum_{n=0}^{\infty} P\{X \leq x \mid N = n\}P\{N = n\}\\ &= P\{N = 0\}\mathbf 1_{[0,\infty)}(x) + \sum_{n=1}^{\infty} P\{X \leq x \mid N = n\}P\{N = n\}\\ &= P\{N = 0\}\mathbf 1_{[0,\infty)}(x) + \sum_{n=1}^{\infty} \Phi\left(\frac{x-n\mu}{\sqrt{n} \sigma}\right)P\{N = n\}\\ &= p_N(0)\mathbf 1_{[0,\infty)}(x) + \sum_{n=1}^{\infty} p_N(n)\Phi\left(\frac{x-n\mu}{\sqrt{n} \sigma}\right) \end{align*}$$ where $\Phi(\cdot)$ is the cumulative probability distribution function of the standard normal random variable. Note that $X$ is not a continuous random variable but rather a mixed random variable that has both a (degenerate) discrete component as well as a continuous component. Those who believe in impulses (also called Dirac delta) can write the density of $X$ as $$f_X(x) = p_N(0)\delta(x) + \sum_{n=1}^\infty p_N(n)\frac{1}{\sigma\sqrt{2\pi n}} \exp\left(-\frac{(x-n\mu)^2}{2n\sigma^2}\right)$$ which is a mixture density of normal random variables except for the discrete component. Of course, we can choose to regard the discrete component as a (degenerate) normal random variable with variance $0$ (and mean $0$ too) and simply say that the density of $X$ is a mixture of normal densities. However, some statisticians are uncomfortable with this extension of the notion of normal random variables (cf. the extensive discussion following this answer). Note incidentally that $E[X] = \mu E[N]$.

Answer 2

Let's do it first for the simple case. Assume $N∼\mathrm{Ber}(p)$ i.e. $P(N=1)=p=1-P(N=0)$ and let $q=1-p$. I am going to find distribution function of $X$, i.e. $F_X(x)=P(X\le x)$.

$F_X(x)=P(X\le x)=P(X\le x, N=0)+P(X\le x,N=1)=P(N=0).P(X\le x| N=0)+P(N=1).P(X\le x| N=1)$.

Now $P(X\le x| N=0)=0$ if $x<0$ and $P(X\le x| N=0)=1$ if $x\geq 0$.

So $F_X(x)=p.P(X\le x| N=1)$ if $x<0$ and $F_X(x)=q+p.P(X\le x| N=1)$ if $x\geq 0$.

Now $X|N=1$ has a normal distribution with mean $\mu$ and sd $\sigma$.

So $P(X\le x| N=1)=P\left(\left(\frac{X_1-\mu}{\sigma}\right)\le \left(\frac{x-\mu}{\sigma}\right)\right)=F_Z\left(\frac{x-\mu}{\sigma}\right)$, where $Z$ is a standard normal r.v. with mean zero and variance one.

Therefore $F_X(x)=p.F_Z\left(\frac{x-\mu}{\sigma}\right)$ if $x<0$ and $F_X(x)=q+p.F_Z\left(\frac{x-\mu}{\sigma}\right)$ if $x\geq 0$. You can simply generalize this method to the binomial case. The only difference is that the $p$ and $q$ will be replace by the corresponding probabilities. For the poisson case, it will be based on the infinite sumations, so there won't be any closed formula like binomial distribution.