Computation of the Second Moment of Poisson Distribution

or in other words,

Prove that for Poisson Distribution, the second moment is given by \( E(X^2) = λ(λ+1) \)

For Poisson Distribution, we have

\( P(X=k) = λ^{k} . \dfrac{e^{-λ}}{k!} \) for k = 0,1,2,…

We now derive the second moment of Poisson Distribution i.e. \(E(X^2)\)

By definition, we have \( E(X^2) = \displaystyle \sum_{k=0}^∞ k^2 . \dfrac{e^{-λ}}{k!} \)

or \( E(X^2) = e^{-λ} . \displaystyle \sum_{k=0}^∞ \dfrac { {k^2} . {λ^k} } {k!} \)
or \( E(X^2) = e^{-λ} . \displaystyle \sum_{k=0}^∞ \dfrac { {k} . {λ^k} } {(k-1)!} \)

or \( E(X^2) = e^{-λ} . \displaystyle \sum_{k=1}^∞ \dfrac { {k} . {λ^k} } {(k-1)!}  \)

Starting the summation from 1 as k=0 would lead to nothing

or \( E(X^2) = e^{-λ} . \displaystyle \sum_{k=1}^∞ \dfrac { {(k-1)} . {λ^k} } {(k-1)!} + e^{-λ} . \displaystyle \sum_{k=1}^∞ \dfrac { {1} . {λ^k} } {(k-1)!} \)

or \( E(X^2) = e^{-λ} . \displaystyle \sum_{k=2}^∞ \dfrac { {(k-1)} . {λ^k} } {(k-1)!} + e^{-λ} . \displaystyle \sum_{k=1}^∞ \dfrac { {1} . {λ^k} } {(k-1)!} \)

For the first summation, we changed the start index to 2 as k=1 would lead to 0 and is meaningless.

 

or \( E(X^2) = e^{-λ} . \displaystyle \sum_{k=2}^∞ \dfrac { {λ^k} } {(k-2)!} + e^{-λ} . \displaystyle \sum_{k=1}^∞ \dfrac { {λ^k} } {(k-1)!} \)

or \( E(X^2) = {λ^2} . e^{-λ} . \displaystyle \sum_{k=2}^∞ \dfrac { {λ^{k-2} } } {(k-2)!} + λ . e^{-λ} . \displaystyle \sum_{k=1}^∞ \dfrac { {λ^{k-1}} } {(k-1)!} \qquad\)    – – – – – (1)

Now, for \( \displaystyle \sum_{k=2}^∞ \dfrac { {λ^{k-2} } } {(k-2)!} \quad\) Putting k-2 = r , we get

\( \displaystyle \sum_{k=2}^∞ \dfrac { {λ^{k-2} } } {(k-2)!} = \displaystyle \sum_{r=0}^∞ \dfrac { {λ^{r} } } {(r)!} = e^λ \quad\)

Similarly,

\( \displaystyle \sum_{k=1}^∞ \dfrac { {λ^{k-1}} } {(k-1)!} = e^λ \quad\)

Thus, our expression in eqn(1) becomes

\( E(X^2) = {λ^2} . e^{-λ} . e^λ + λ . e^{-λ} . e^λ\)
or \( E(X^2) = {λ^2} + λ \)
or \( E(X^2) = λ(λ+1) \) Our Required Expression for Poisson Distribution Second Moment