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:

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

By definition of Poisson distribution:

$\displaystyle E \left({X}\right) = \sum_{k \mathop \ge 0} k \frac 1 {k!} \lambda^k e^{-\lambda}$

Then:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle E \left({X}\right)$$ $$=$$ $$\displaystyle$$ $$\displaystyle \lambda e^{-\lambda} \sum_{k \mathop \ge 1} \frac 1 {\left({k-1}\right)!} \lambda^{k-1}$$ $$\displaystyle$$ $$\displaystyle$$ the $k = 0$ term vanishes $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \lambda e^{-\lambda} \sum_{j \mathop \ge 0} \frac {\lambda^j} {j!}$$ $$\displaystyle$$ $$\displaystyle$$ putting $j = k - 1$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \lambda e^{-\lambda} e^{\lambda}$$ $$\displaystyle$$ $$\displaystyle$$ Taylor Series Expansion for Exponential Function $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \lambda$$ $$\displaystyle$$ $$\displaystyle$$

$\blacksquare$

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 PGF, we have:

$E \left({X}\right) = \Pi'_X \left({1}\right)$

We have:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle \Pi'_X \left({s}\right)$$ $$=$$ $$\displaystyle$$ $$\displaystyle \frac {\mathrm d} {\mathrm ds} e^{-\lambda \left({1-s}\right)}$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$=$$ $$\displaystyle$$ $$\displaystyle \lambda e^{- \lambda \left({1-s}\right)}$$ $$\displaystyle$$ $$\displaystyle$$ Derivatives of PGF of Poisson Distribution

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:

$e^0 = 1$

$\blacksquare$