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Variance with Poisson distribution

  • #1
11
2
<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?

Thank you for your time.
 
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Answers and Replies

  • #2
264
70
You have ##n!=n(n-1)!##.
Use this to simplify ##n/(n-1)!##.
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
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<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?

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.##
 
  • #4
11
2
Yeah, I got it.

Thank you!
 
  • #5
gleem
Science Advisor
Education Advisor
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@MrsTesla

Have you solved the problem? if not I can help.
 
  • #6
Ray Vickson
Science Advisor
Homework Helper
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@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.
 
  • #7
11
2
Yes, I solved the problem.
 

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