Variance of Poisson Distribution/Proof 3

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


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

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


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

From Moment in terms of Moment Generating Function:


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

In Expectation of Poisson Distribution, it is shown that:


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

Then:

Setting $t = 0$:

From Expectation of Poisson Distribution:


 * $\displaystyle \expect X = \lambda$

So:


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