# Poisson distribution

• Jul 25th 2013, 05:46 AM
alexander9408
Poisson distribution
If X~Po(λ) and E(X^2) = 2, find λ.
• Jul 25th 2013, 06:37 AM
HallsofIvy
Re: Poisson distribution
That's pretty straight forward, isn't it? The Poisson distribution is given by $P(k)= \frac{\lambda^k}{k! e^k}$ where k is a positive integer. So the Expected value of $X^2$ is $\sum_{k= 1}^\infty \frac{\lambda^k k^2}{k!e^k}= 2$.
What must $\lambda$ be in order that that equation be true?
• Jul 25th 2013, 07:56 AM
alexander9408
Re: Poisson distribution
Quote:

Originally Posted by HallsofIvy
$X^2$ is $\sum_{k= 1}^\infty \frac{\lambda^k k^2}{k!e^k}= 2$.

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.
• Jul 25th 2013, 08:56 PM
casio415
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