# Error for operations for Two Poisson distributions

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 ?

• $A/B$ is undefined. In the other cases, please explain what you mean by "error:" we need some context to understand your question.
– whuber
Commented Aug 5, 2020 at 13:36
• Question editied Commented Aug 6, 2020 at 10:01
• As said by @whuber X/Y is only defined for Y distributed as a truncated Poisson. Commented Aug 8, 2020 at 8:39

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

• I have been thinking about products of Poissons, but plan to post nothing here unless you say what application and type of analysis you have in mind. Why do you want to know standard error? No use writing up stuff of unpredictable value. Commented Aug 9, 2020 at 13:29
• I was working on GRB data and was trying to fit an exponential model to it. So, before that most of the data was just random noise, to I was trying to remove it by fitting a constant line to it, and removing all the data which has chi sq <=2. I was plotting a residual plot, which is chi-sq vs time since trigger(sec). So, for residual plots I want to fit with error bars, for which I need this y= u/v since chi-sq = (model-data)^2/err^2, where err= sqrt(data) corresponding to each point. Commented Aug 9, 2020 at 21:00