Re: How to prove that the expectation and variance of a Poisson random variable = lam
Thank you for posting and letting me know about the protocol.
For question (a) I had this:
E(X)=
n
∑ x * p(x)
x=0
n
∑ x * λ^x * e^(-λ)/ x!
x=0
n
∑ x * λ^x * e^(-λ)/ (x-1)! * x
x=1
n
∑ λ^x * e^(-λ)/ (x-1)!
x=1
it was here that i had trouble and i got shown the next step, without explaination. I was wondering how the λ was factored across leaving λ^(x-1)?
n
λ ∑ λ^(x-1) * e^(-λ)/ (x-1)! then let y= x-1
x=1
n
λ ∑ λ^y * e^(-λ)/ y!
y=0
= λ
For Question (b) i was having more trouble. I started with:
Var(X)= E(X^2) - E(X)^2
where we know E(X^2 - X) = E(X^2) - E(X)
Therefore: Var(X)= E[X(X-1)] + E(X) - E(X)^2 so we first need to solve E[X(X-1)]
E[X(X-1)]=
n
∑ x * (x-1) * λ^x * e^(-λ)/ x! this is where i got stuck
x=0
Any help with explanations and help as to where to go with these, or if i have made a mistake will be very much appreciated