Stein–Chen identity for the Poisson

Question

1-) Solve the following:

a-) If X ∼ Pois(λ) and g is any function, show that :

E [Xg(X)] = λ E [g(X + 1)]

(This is called the Stein–Chen identity for the Poisson.)

b-) Find the third moment E [X3 ] for X ∼ Pois(λ) by using the identity from (a) and the fact that X has mean λ and variance λ. Simplify your answer.

Answer 1

If X ~ P(λ), f(x) = e^-λ λ^x/x!

For the Poisson distribution, E(X) = λ and E(X²) = λ² + λ

Then, E(Xg(X)) = Σ_x=0^∞ xg(x)e^-λ λ^x/x! = Σ_x=1^∞ xg(x)e^-λ λ^x/x! = Σ_x=1^∞ xg(x)e^-λ λ * λ^x-1/x! =

λ Σ_x=1^∞ g(x)e^-λ λ^x-1/x-1! = letting y = x-1 λΣ_y=0^∞ g(y+1)e^-λ λ^y/y! = letting x = y λΣ_x=0^∞ g(x+1)e^-λ λ^x/x! =

λ E(g(X+1))

b) E(X³) = E(X*X²) = λ E((X+1)²) = λ E(X²+2X+1) = λ(λ² + λ+ 2λ + 1) = λ³ + 3λ² + λ