Variance of Exponential Distribution/Proof 2
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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
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:
\(\ds \var X\) | \(=\) | \(\ds 2 \beta^2 - \beta^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \beta^2\) |
$\blacksquare$