Variance of Exponential Distribution/Proof 1
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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
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \int_0^\infty x^2 \frac 1 \beta \map \exp {-\frac x \beta} \rd x\) | Probability Density Function 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:
\(\ds \var X\) | \(=\) | \(\ds \expect {X^2} - \paren {\expect X}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \beta^2 - \beta^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \beta^2\) |
$\blacksquare$