If X~Po(λ) and E(X^2) = 2, find λ.

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- Jul 25th 2013, 04:46 AMalexander9408Poisson distribution
If X~Po(λ) and E(X^2) = 2, find λ.

- Jul 25th 2013, 05:37 AMHallsofIvyRe: 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? - Jul 25th 2013, 06:56 AMalexander9408Re: Poisson distribution
- Jul 25th 2013, 07:56 PMcasio415Re: 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