Variance with Poisson distribution

In summary: Thank you both for offering your help. In summary, the problem involved calculating the probability of a given number of events in a counting experiment using the Poisson distribution. The mean and standard deviation were also calculated, with the help of the series expansion of the exponential function.
  • #1
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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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  • #2
You have ##n!=n(n-1)!##.
Use this to simplify ##n/(n-1)!##.
 
  • #3
MrsTesla said:
<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
Yeah, I got it.

Thank you!
 
  • #5
@MrsTesla

Have you solved the problem? if not I can help.
 
  • #6
gleem said:
@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
Yes, I solved the problem.
 
  • Like
Likes gleem

What is the Poisson distribution?

The Poisson distribution is a probability distribution that is often used in situations where the outcome of interest is a count or number of occurrences within a specific time or space interval. It is named after the French mathematician Siméon Denis Poisson.

How is the Poisson distribution different from other probability distributions?

The Poisson distribution is unique in that it describes the probability of a specific number of events occurring in a fixed interval, given the average rate of occurrence. Most other probability distributions, such as the normal or binomial distributions, describe the probability of a range of values occurring.

What is the variance of a Poisson distribution?

The variance of a Poisson distribution is equal to its mean, which is also known as its lambda parameter. This means that the spread or variability of the data in a Poisson distribution is determined by the average rate of occurrence.

How is variance calculated for a Poisson distribution?

The formula for calculating the variance of a Poisson distribution is Var(X) = λ, where λ is the mean or expected value of the distribution. This means that for a Poisson distribution with a mean of 5, the variance would also be 5.

What is the importance of understanding variance in a Poisson distribution?

Understanding the variance of a Poisson distribution is important because it allows us to make predictions about the likelihood of certain numbers of events occurring within a specific time or space interval. It also helps us to identify any unusual or unexpected patterns in the data, which could be indicative of underlying factors or causes.

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