Expectation of Poisson Distribution/Proof 1

Theorem

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

Then the expectation of $X$ is given by:

$\expect X = \lambda$

Proof

From the definition of expectation:

$\ds \expect X = \sum_{x \mathop \in \Img X} x \map \Pr {X = x}$

By definition of Poisson distribution:

$\ds \expect X = \sum_{k \mathop \ge 0} k \frac 1 {k!} \lambda^k e^{-\lambda}$

Then:

\(\ds \expect X\)

\(=\)

\(\ds \lambda e^{-\lambda} \sum_{k \mathop \ge 1} \frac 1 {\paren {k - 1}!} \lambda^{k-1}\)

as the $k = 0$ term vanishes

\(\ds \)

\(=\)

\(\ds \lambda e^{-\lambda} \sum_{j \mathop \ge 0} \frac {\lambda^j} {j!}\)

putting $j = k - 1$

\(\ds \)

\(=\)

\(\ds \lambda e^{-\lambda} e^{\lambda}\)

Taylor Series Expansion for Exponential Function

\(\ds \)

\(=\)

\(\ds \lambda\)

$\blacksquare$