What would be the standard deviation for $A+B$, $AB$ and $\frac{A}{B}$ for $A$ and $B$ Poisson distributed?
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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.) $\endgroup$– Stephan KolassaCommented Jul 25, 2020 at 12:29
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$\begingroup$ @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? $\endgroup$– Abhinna SundarCommented Jul 25, 2020 at 14:38
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$\begingroup$ 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)? $\endgroup$– Stephan KolassaCommented Jul 25, 2020 at 14:46
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$\begingroup$ yes ! I'll edit the question, I think that was the confusion. $\endgroup$– Abhinna SundarCommented Jul 25, 2020 at 14:48
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1$\begingroup$ Thank you. Please do edit the question, so later users can find it easily. $\endgroup$– Stephan KolassaCommented Jul 25, 2020 at 15:11
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1 Answer
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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 expectedyields
> 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 20.0as 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.
answered Jul 25, 2020 at 15:10