Variance of Exponential Distribution

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 1

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

The expectation of $X^2$ is:

 $\ds \expect {X^2}$ $=$ $\ds \int_{x \mathop \in \Omega_X} x^2 \, \map {f_X} x \rd x$ Definition of Expectation of Continuous Random Variable $\ds$ $=$ $\ds \int_0^\infty x^2 \frac 1 \beta \map \exp {-\frac x \beta} \rd x$ PDF of Exponential Distribution $\ds$ $=$ $\ds \intlimits {-x^2 \map \exp {-\frac x \beta} } 0 \infty + \int_0^\infty 2 x \map \exp {-\frac x \beta} \rd x$ Integration by Parts $\ds$ $=$ $\ds 0 + 2 \beta \int_0^\infty x \frac 1 \beta \, \map \exp {-\frac x \beta} \rd x$ algebraic manipulation $\ds$ $=$ $\ds 2 \beta \, \expect X$ Expectation of Exponential Distribution $\ds$ $=$ $\ds 2 \beta^2$

Thus the variance of $X$ is:

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

$\blacksquare$

Proof 2

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

$\map {M_X} t = \dfrac 1 {1 - \beta t}$
$\var X = \expect {X^2} - \paren {\expect X}^2$

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:

 $\ds \var X$ $=$ $\ds 2 \beta^2 - \beta^2$ $\ds$ $=$ $\ds \beta^2$

$\blacksquare$