# Variance with Poisson distribution

MrsTesla
<Moderator's note: Moved from a technical forum and thus no template.>

So, I have this problem and I am stuck on a sum. The problem I was given is the following:

The probability of a given number n of events (0 ≤ n < ∞) in a counting experiment per time (e.g. radioactive decay events per second) follows the function (Poisson distribution) P(n) = µn/n! *e , where µ is a real number. Recalling the series expansion of the exponential function ex = ∑n=0 xn/n!

a) show that ∞ ∑ n=0 P(n) = 1, i.e. that P(n) is normalised
b) compute the mean <n> = ∑n=0nP(n)
c) compute the standard deviation

I already did a) and b), where for b) I got <n>=μ. For c) I know that the variance is given by σ2=<n2> - <n>2. I know that <n>22 but for <n2>, which I know is given by ∑n=0 n2*P(n), I evolve until I get stuck with μen=0(n*μn-1)/(n-1)!

I have no idea how to go from there in order to find <n2>, do you have any idea of how to go from there?

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eys_physics
You have ##n!=n(n-1)!##.
Use this to simplify ##n/(n-1)!##.

Homework Helper
Dearly Missed
<Moderator's note: Moved from a technical forum and thus no template.>

So, I have this problem and I am stuck on a sum. The problem I was given is the following:

The probability of a given number n of events (0 ≤ n < ∞) in a counting experiment per time (e.g. radioactive decay events per second) follows the function (Poisson distribution) P(n) = µn/n! *e , where µ is a real number. Recalling the series expansion of the exponential function ex = ∑n=0 xn/n!

a) show that ∞ ∑ n=0 P(n) = 1, i.e. that P(n) is normalised
b) compute the mean <n> = ∑n=0nP(n)
c) compute the standard deviation

I already did a) and b), where for b) I got <n>=μ. For c) I know that the variance is given by σ2=<n2> - <n>2. I know that <n>22 but for <n2>, which I know is given by ∑n=0 n2*P(n), I evolve until I get stuck with μen=0(n*μn-1)/(n-1)!

I have no idea how to go from there in order to find <n2>, do you have any idea of how to go from there?

Perhaps the easiest way is to compute ##\langle n \rangle## and ##\langle n(n-1)\rangle = \langle n^2 - n \rangle.## Then you can easily get ##\langle n^2 \rangle.##

MrsTesla
Yeah, I got it.

Thank you!

@MrsTesla

Have you solved the problem? if not I can help.

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