Expectation of Poisson Distribution

Theorem
Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.

Then the expectation of $X$ is given by:
 * $E \left({X}\right) = \lambda$

Proof 1
From the definition of expectation:
 * $\displaystyle E \left({X}\right) = \sum_{x \in \operatorname{Im} \left({X}\right)} x \Pr \left({X = x}\right)$

By definition of Poisson distribution:
 * $\displaystyle E \left({X}\right) = \sum_{k \ge 0} k \frac 1 {k!} \lambda^k e^{-\lambda}$

Then:

Proof 2
From the Probability Generating Function of Poisson Distribution, we have:
 * $\Pi_X \left({s}\right) = e^{-\lambda \left({1-s}\right)}$

From Expectation of Discrete Random Variable from P.G.F., we have:
 * $E \left({X}\right) = \Pi'_X \left({1}\right)$

We have:

Plugging in $s = 1$:
 * $\Pi'_X \left({1}\right) = \lambda e^{- \lambda \left({1-1}\right)} = \lambda e^0$

Hence the result from Exponential of Zero and One: $e^0 = 1$.