Error for operations for Two Poisson distributions

Question

Can error (std dev) in A +B or A-B be $\sqrt{A}$+$\sqrt{B}$ if A and B are Poissonian? If yes, what would be similar expressions for AB and A/B ?

Answer 1

If $X \sim \mathsf{Pois}(\lambda)$ and you have random observations $X_1, \dots, X_n,$ then $T = X_1 + \cdots + X_n \sim \mathsf{Pois}(n\lambda).$ If your goal is to make confidence intervals, then the following may be relevant for making CIs for Poisson means.

The logic is somewhat similar that of the Agresti-Coull binomial 95% CI based on $\hat p = (x+2)/(n+4)$ with standard error $\sqrt{\hat p(1-\hat p)/(n+4)},$ which is superior to the Wald CI based on $\hat p = x/n$ with standard error $\sqrt{\hat p(1-\hat p)/n}.$

If $\lambda$ is not too small, a reasonable 95% CI for $n\lambda$ is of the form $T+2 \pm 1.96\sqrt{T+1}.$ Divide endpoints by $n$ to get a CI for $\lambda.$ For the case $n=10, \lambda=5$ with data simulated below in R, the 95% CI for $\lambda$ is $(2.86, 5.34).$

set.seed(2020)
lam = 5; n = 10; pm=c(-1,1)
t = sum(rpois(n, lam));  t
t
[1] 39
CI = t+2 + pm*1.96*sqrt(t+1)
CI.lam = CI/n;  CI.lam
[1] 2.860387 5.339613

I suppose that a CI for the difference in two Poisson means could be arise from inverting the test of $H_0: \lambda_1=\lambda_2$ against $H_a: \lambda_1 \ne \lambda_2.$

If you can give us some idea what you intend to do with the product of two Poisson random variables, maybe one of us can give a useful response. If you're interested in a CI, maybe the appropriate standard error could come from bootstrapping.

Note: With $n=10,\lambda=4,$ the style of CI suggested above for $\lambda$ has about 95% coverage probability based on a million such 95% CIs.

set.seed(808)
t = rpois(10^6, 10*4)
ucl = (t+2 + 1.96*sqrt(t+1))/10
lcl = (t+2 - 1.96*sqrt(t+1))/10
mean(lcl <= 4 & ucl >= 4)
[1] 0.952753