Variance of Exponential Distribution/Proof 2

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


 * $\map {M_X} t = \dfrac 1 {1 - \beta t}$

From Variance as Expectation of Square minus Square of Expectation:


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

From Moment in terms of Moment Generating Function, we also have:


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

In Expectation of Exponential Distribution: Proof 2, it is shown that:


 * $\map {M_X'} t = \dfrac \beta {\paren {1 - \beta t}^2}$

We have:

Setting $t = 0$ gives:

By Expectation of Exponential Distribution, we have:


 * $\expect X = \beta$

So: