how can we derive the expectation for a Poisson distribution?

Answer 1

The expectation is

∞
Σ [k λ^k e^(-λ)]/k!
k=0

Note that the k=0 term is zero, so we can drop it and write the expectation as

∞
Σ [k λ^k e^(-λ)]/k!
k=1

or after simplifying

∞
Σ [λ^k e^(-λ)]/(k-1)!
k=1

Now, replace k by j+1 in this sum:

∞
Σ [λ^(j+1) e^(-λ)]/(j+1-1)! or after simplifying,
j+1=1

∞
Σ [λ^(j+1) e^(-λ)]/ j!
j=0

Factor out a λ:

∞
Σ [λ^j e^(-λ)]/ j! * λ
j=0

Note that the term in the sum is just p_X(λ), so the sum equals 1. Therefore, the expectation of the Poisson distribution is its parameter λ.

Answer 2

That dosen't really make any sense, well at least, not to me. :\