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?
 A: We will assume that $A\sim\text{Pois}(\lambda)$ and $B\sim\text{Pois}(\mu)$ are independent Poisson variables. Then
$$ EA=\text{var} A=\lambda\quad\text{and}\quad EB=\text{var} B=\mu. $$

*

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


*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.


*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.
