Expectation of Poisson Distribution/Proof 2

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
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 PGF, 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: $e^0 = 1$.