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

  • 1
    $\begingroup$ 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.) $\endgroup$ Jul 25, 2020 at 12:29
  • $\begingroup$ @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? $\endgroup$ Jul 25, 2020 at 14:38
  • $\begingroup$ 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)? $\endgroup$ Jul 25, 2020 at 14:46
  • $\begingroup$ yes ! I'll edit the question, I think that was the confusion. $\endgroup$ Jul 25, 2020 at 14:48
  • 1
    $\begingroup$ Thank you. Please do edit the question, so later users can find it easily. $\endgroup$ Jul 25, 2020 at 15:11

1 Answer 1


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:

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


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

  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
               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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.