# Thread: Poisson distribution

1. ## Poisson distribution

If X~Po(λ) and E(X^2) = 2, find λ.

2. ## Re: Poisson distribution

That's pretty straight forward, isn't it? The Poisson distribution is given by $\displaystyle P(k)= \frac{\lambda^k}{k! e^k}$ where k is a positive integer. So the Expected value of $\displaystyle X^2$ is $\displaystyle \sum_{k= 1}^\infty \frac{\lambda^k k^2}{k!e^k}= 2$.
What must $\displaystyle \lambda$ be in order that that equation be true?

3. ## Re: Poisson distribution

Originally Posted by HallsofIvy
$\displaystyle X^2$ is $\displaystyle \sum_{k= 1}^\infty \frac{\lambda^k k^2}{k!e^k}= 2$.
That's exactly where I get stuck, I can't solve that, I don't know what to do after I expanded it in the form of (λ^1)(1^2)/1!(e^1) + (λ^2)(2^2)/2!(e^2) + (λ^3)(3^2)/3!(e^3) +..... = 2.
Please give some hint or guide, thank you.

4. ## Re: Poisson distribution

Can also try something like this.

Since X~Po(λ), we know two things:

1) E(X) = λ
2) Var(X) = λ = E(X^2) - E(X)^2

combining these two gives:
λ = E(X^2) - λ^2
E(X^2) = λ + λ^2

If this is equal to 2 then its the same as solving the quadratic λ + λ^2 = 2