4

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and the amount they each deposit/withdraw is a normal distribution, what is the distribution for the amount deposited/withdrawn in a day (not just the EV)?

Steven Noble
  • 384
  • Are those Normal r.v.'s independent from each other (and also from $N$)? – Stat Nov 27 '12 at 05:27
  • 1
    Are you interested in conditional distribution of $X$ given $N$ or unconditional distribution? Try to write your question properly so we can help. – Stat Nov 27 '12 at 05:32
  • I'm interested in the distribution of the sum of n independent normal variables where n is determined by an independent binomial distribution. and I'm also interested in the case when n is determined by an independent Poisson distribution. I don't think a conditional distribution is relevant here. do you mean compound? – Steven Noble Nov 27 '12 at 15:32
  • 1
    When you wrote $Normal(N\mu,\sqrt(N).\sigma)$, then it means your $N$ is fixed (i.e. conditional distribution of $X$ given $N$). Otherwise, if this is unconditional distribution, then it is wrong to write $Normal(N\mu,\sqrt(N).\sigma)$, since $N$ is a random variable and not a cosntant. – Stat Nov 27 '12 at 16:37
  • Ok, the syntax you reference is a little loose (though I'm pretty certain I've seen this in actuarial texts like Klugman; I feel the natural interpretation is obvious. in terms of R the random generator would look like "n <- rpois(1, lambda)" "return(rnorm(1, nmu, sqrt(n)sigma))"). The 3 definitions I give are equivalent. Feel free to ignore this offending one and use the definition involving the sum or the story about the bank. I'd be grateful if you can tell me the name of the resulting distribution. – Steven Noble Nov 27 '12 at 17:31

2 Answers2

5

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 $E[X] = \mu E[N]$.

Community
  • 1
Dilip Sarwate
  • 46,658
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.

Scortchi - Reinstate Monica
  • 29,907
Stat
  • 7,486