Variance of Poisson Distribution/Proof 3

Theorem

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

Then the variance of $X$ is given by:

$\var X = \lambda$

Proof

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

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

From Variance as Expectation of Square minus Square of Expectation, we have:

$\var X = \expect {X^2} - \paren {\expect X}^2$

From Moment in terms of Moment Generating Function:

$\expect {X^2} = \map {M_X''} 0$

In Expectation of Poisson Distribution, it is shown that:

$\map {M_X'} t = \lambda e^t e^{\lambda \paren {e^t - 1} }$

Then:

\(\ds \map {M''_X} t\)

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

Chain Rule for Derivatives

\(\ds \)

\(=\)

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

Derivative of Power, Derivative of Exponential Function

Setting $t = 0$:

\(\ds \expect {X^2}\)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \lambda \paren {\lambda + 1}\)

Exponential of Zero

\(\ds \)

\(=\)

\(\ds \lambda^2 + \lambda\)

From Expectation of Poisson Distribution:

$\expect X = \lambda$

So:

$\var X = \lambda^2 + \lambda - \lambda^2 = \lambda$

$\blacksquare$