Variance of Exponential Distribution

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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

From Variance as Expectation of Square minus Square of Expectation:

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

From Expectation of Exponential Distribution:

$\expect X = \beta$

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:

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

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:

\(\ds \var X\) \(=\) \(\ds 2 \beta^2 - \beta^2\)
\(\ds \) \(=\) \(\ds \beta^2\)

$\blacksquare$