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:
 * $$E \left({X}\right) = \sum_{x \in \operatorname{Im} \left({X}\right)} x \Pr \left({X = x}\right)$$

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

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 Basic Properties of Exponential Function: $$e^0 = 1$$.