Pmf of Poisson Distribution is as follows:

\[f(X=k;\lambda)=\frac{\lambda^k e^{-\lambda}}{k!}\]

Our aim is to derive the the expectation of \(E(X)\) and the variance \(Var(X)\). Given that the formula of expectation: \[ E(X)=\sum_{k=0}^{\infty} k \frac{\lambda^k e^{-\lambda }}{k!} \]

Notice that when \(k=0\), the formula is equal to 0, that is:

\[\sum_{k=0}^{\infty} k \frac{\lambda^ke^{-\lambda}}{k!}\Large|_{k=0}=0\]

Then, the formula become as followed:

\[E(X)=\sum_{k=1}^{\infty} k \frac{\lambda^ke^{-\lambda}}{k!}\]

\[\begin{aligned}E(X)&=\sum_{k=0}^{\infty} k \frac{\lambda^ke^{-\lambda}}{k!}=\sum_{k=0}^{\infty} \frac{\lambda^ke^{-\lambda}}{(k-1)!}\\&=\sum_{k=0}^{\infty} \frac{\lambda^{k-1}\lambda e^{-\lambda}}{(k-1)!}\\&=\lambda e^{-\lambda}\sum_{k=1}^{\infty}\frac{\lambda^{k-1}}{(k-1)!}\end{aligned}\]

Now we need take advantage of Taylor Expansion, recall that:

\[e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots+\frac{x^{k-1}}{(k-1)!}=\sum_{k=1}^{\infty}\frac{x^{k-1}}{(k-1)!}\]

Compare \(E(X)\), we can get:

\[E(X)=\lambda e^{-\lambda}e^\lambda=\lambda\]

As known that \(Var(X)=E(X^2)-(E(x))^2\), we just get \(E(X^2)\). Given that:

\[E(X)=\sum_{k=1}^{\infty} k \frac{\lambda^ke^{-\lambda}}{k!}=\lambda\]

we can use this formula to derive the \(E(X^2)\),

\[\begin{aligned}E(X)=&\sum_{k=1}^{\infty} k \frac{\lambda^ke^{-\lambda}}{k!}=\lambda\\\Leftrightarrow&\sum_{k=1}^{\infty} k \frac{\lambda^k}{k!}=\lambda e^{\lambda}\\\Leftrightarrow&\frac{\partial\sum_{k=1}^{\infty} k \frac{\lambda^k}{k!}}{\partial \lambda}=\frac{\partial \lambda e^{\lambda}}{\partial \lambda}\\\Leftrightarrow&\sum_{k=1}^{\infty}k^2\frac{\lambda^{k-1}}{k!}=e^\lambda+\lambda e^\lambda\\\Leftrightarrow&\sum_{k=1}^{\infty}k^2\frac{\lambda^{k-1}e^{-\lambda}}{k!}=1+\lambda \\\Leftrightarrow&\sum_{k=1}^{\infty}k^2\frac{\lambda^{k}e^{-\lambda}}{k!}=\lambda+\lambda^2=E(X^2)\end{aligned}\]

then,

\[Var(X)=E(X^2)-(E(X))^2=\lambda+\lambda^2-(\lambda)^2=\lambda\]

Thus, we have proved that the Expectation and the Variance of Poisson Distribution are both \(\lambda\)