# Standard deviation/variance for the sum, product and quotient of two Poisson distributions

What would be the standard deviation for $$A+B$$, $$AB$$ and $$\frac{A}{B}$$ for $$A$$ and $$B$$ Poisson distributed?

• What do you mean by "error"? Are you looking for the variance, or the standard deviation? Also, what is your first distribution you are interested in? (The second is $AB$, the third is $\frac{A}{B}$ - note that the third one is not defined, because $B$ has a nonzero probability of being zero.) – Stephan Kolassa Jul 25 '20 at 12:29
• @StephanKolassa I am looking for std dev. I am working on Gamma Ray data, so that corresponds to the Poisson distribution, so the error or the std dev is the sqrt(data value) , so what would be these qts asked above? – Abhinna Sundar Jul 25 '20 at 14:38
• OK, so by "error" you mean the standard deviation, good. What do you mean by "2 data A and B"? Are you looking for the standard deviation of a random variable $A+B$ where $A$ and $B$ are both Poisson (and presumably independent)? – Stephan Kolassa Jul 25 '20 at 14:46
• yes ! I'll edit the question, I think that was the confusion. – Abhinna Sundar Jul 25 '20 at 14:48
• Thank you. Please do edit the question, so later users can find it easily. – Stephan Kolassa Jul 25 '20 at 15:11

We will assume that $$A\sim\text{Pois}(\lambda)$$ and $$B\sim\text{Pois}(\mu)$$ are independent Poisson variables. Then

$$EA=\text{var} A=\lambda\quad\text{and}\quad EB=\text{var} B=\mu.$$

1. Let's consider $$A+B$$. It's a standard property of the Poisson distribution that the sum of independent Poisson variables is again Poisson, with a parameter that is just the sum of the separate Poisson parameters:

$$A+B\sim\text{Pois}(\lambda+\mu).$$

Thus,

$$\text{sd}(A+B)=\sqrt{\text{var}(A+B)} = \sqrt{\lambda+\mu}.$$

1. Let's consider the product $$AB$$. This distribution does not have a specific name, but we can derive its variance from the general formula for the variance of a product distribution: $$\text{var}(AB) =\text{var}A\,\text{var}B+\text{var}A(EB)^2+(EA)^2\text{var}B = \lambda\mu+\lambda\mu^2+\lambda^2\mu,$$ so the standard deviation is just $$\text{sd}(AB) = \sqrt{\text{var}(AB)} = \sqrt{\lambda\mu+\lambda\mu^2+\lambda^2\mu}.$$ I like verifying things like this by a quick simulation. In R:

 n_sims <- 1e6
lambda <- mu <- c(0.5,1,2)
observed <- expected <- matrix(nrow=length(lambda),ncol=length(mu),dimnames=list(lambda,mu))
for ( ii in seq_along(lambda) ) {
for ( jj in seq_along(mu) ) {
set.seed(1) # for reproducibility
observed[ii,jj] <- var(rpois(n_sims,lambda[ii])*rpois(n_sims,mu[jj]))
expected[ii,jj] <- lambda[ii]*mu[jj]+lambda[ii]*mu[jj]^2+lambda[ii]^2*mu[jj]
}
}
observed
expected


yields

 > observed
0.5        1         2
0.5 0.4945624 1.232162  3.470369
1   1.2362181 2.959398  7.946589
2   3.4516277 7.870455 19.795708
> expected
0.5    1    2
0.5 0.50 1.25  3.5
1   1.25 3.00  8.0
2   3.50 8.00 20.0


as expected.

2. Finally, how about $$\frac{A}{B}$$? This is not a well-defined distribution, since $$P(A\neq 0, B=0)\neq 0$$, and we would divide by zero.