Derive the Mean or Expected Value of Random Variable that has Poisson Distribution

Derive the Mean or Expected Value of Random Variable that has Poisson Distribution

or Prove that for Poisson Distribution, \( μ = E(X) = λ \)

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

We now derive the Expected Value(μ) of Poisson Distribution i.e. \(E(X)\)

By definition, we have \( E(X) = \displaystyle \sum_{k=0}^∞ k \cdot P(X) \)

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

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

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

Changing the lower index of summation to 1 as k=0 leads nothing

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

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

for the expression \( \displaystyle \sum_{k=1}^∞ \dfrac{{λ^{(k-1)}}}{{(k-1)}!} \), let k-1=r.

The expression now becomes \( \displaystyle \sum_{r=0}^∞ \dfrac{{λ^{(r)}}}{{(r)}!} \) which is nothing but \(e^{λ} \) only.

Hence \( E(X) = λ \cdot e^{-λ} \cdot\displaystyle \sum_{k=1}^∞ \dfrac{{λ^{(k-1)}}}{{(k-1)}!} \) = \( λ \cdot e^{-λ} \cdot e^{λ} \)

or \( E(X) = λ \)