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^-μ , where µ is a real number. Recalling the series expansion of the exponential function e^x = ∑_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=0^∞nP(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>²=μ² but for <n²>, which I know is given by ∑_n=0^∞ n²*P(n), I evolve until I get stuck with μe^-μ∑_n=0^∞(n*μ^n-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?

Thank you for your time.

 

[eys_physics]

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

 

[Ray Vickson]

<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^-μ , where µ is a real number. Recalling the series expansion of the exponential function e^x = ∑_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=0^∞nP(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>²=μ² but for <n²>, which I know is given by ∑_n=0^∞ n²*P(n), I evolve until I get stuck with μe^-μ∑_n=0^∞(n*μ^n-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?

Thank you for your time.

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!

 

[gleem]

@MrsTesla

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

 

[Ray Vickson]

@MrsTesla

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

Her post #4 suggests that she solved the problem, although she does not say so explicitly.

 

[MrsTesla]

Yes, I solved the problem.