# Variance with Poisson distribution

• MrsTesla
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.

#### 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?

Thank you for your time.

Last edited by a moderator:
You have ##n!=n(n-1)!##.
Use this to simplify ##n/(n-1)!##.

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.##

Yeah, I got it.

Thank you!

@MrsTesla

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

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.

Yes, I solved the problem.

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.