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

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 \( E(X) = λ  \)