Expectation of Poisson Distribution/Proof 3

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 Moment Generating Function of Poisson Distribution, the moment generating function of $X$, $M_X$, is given by:

$\map {M_X} t = e^{\lambda \paren {e^t - 1} }$

By Moment in terms of Moment Generating Function:

$\expect X = \map {M_X'} 0$

We have:

\(\ds \map {M_X'} t\)

\(=\)

\(\ds \map {\frac \d {\d t} } {e^{\lambda \paren {e^t - 1} } }\)

\(\ds \)

\(=\)

\(\ds \map {\frac \d {\d t} } {\lambda \paren {e^t - 1} } \frac \d {\map \d {\lambda \paren {e^t - 1} } } \paren {e^{\lambda \paren {e^t - 1} } }\)

Chain Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \lambda e^t e^{\lambda \paren {e^t - 1} }\)

Derivative of Exponential Function

Setting $t = 0$ gives:

\(\ds \expect X\)

\(=\)

\(\ds \lambda e^0 e^{\lambda \paren {e^0 - 1} }\)

\(\ds \)

\(=\)

\(\ds \lambda\)

Exponential of Zero

$\blacksquare$