Variance of Exponential Distribution/Proof 2

Theorem

Let $X$ be a continuous random variable with the exponential distribution with parameter $\beta$.

Then the variance of $X$ is:

$\var X = \beta^2$

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:

\(\ds \map {M_X} t\)

\(=\)

\(\ds \frac \d {\d t} \paren {\frac \beta {\paren {1 - \beta t}^2} }\)

\(\ds \)

\(=\)

\(\ds \frac {2 \beta^2} {\paren {1 - \beta t}^3}\)

Chain Rule for Derivatives, Derivative of Power

Setting $t = 0$ gives:

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

\(=\)

\(\ds \frac {2 \beta^2} {\paren {1 - 0 \beta}^3}\)

\(\ds \)

\(=\)

\(\ds 2 \beta^2\)

By Expectation of Exponential Distribution, we have:

$\expect X = \beta$

So:

\(\ds \var X\)

\(=\)

\(\ds 2 \beta^2 - \beta^2\)

\(\ds \)

\(=\)

\(\ds \beta^2\)

$\blacksquare$