Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate continuous distributions ▷ Exponential distribution ▷ Variance

Theorem: Let $X$ be a random variable following an exponential distribution:

$X \sim \mathrm{Exp}(\lambda) .$

Then, the variance of $X$ is

$\label{eq:exp-var} \mathrm{Var}(X) = \frac{1}{\lambda^2} \; .$

Proof: The variance of a random variable is defined as

$\label{eq:var} \mathrm{Var}(X) = \mathrm{E}\left[ (X-\mathrm{E}(X))^2 \right]$

which, partitioned into expected values, reads:

$\label{eq:var2} \mathrm{Var}(X) = \mathrm{E}\left[ X^2 \right] - \mathrm{E}\left[ X \right]^2 \; .$ $\label{eq:exp-mean-ref} \mathrm{E}[X] = \frac{1}{\lambda}$

The second moment $\mathrm{E}[X^2]$ can be derived as follows:

$\label{eq:second-moment-s1} \begin{split} \mathrm{E} [X^2] &= \int_{- \infty}^{+\infty} x^2 \cdot f_\mathrm{X}(x) \, \mathrm{d}x \\ &= \int_{0}^{+\infty} x^2 \cdot \lambda \exp(-\lambda x) \, \mathrm{d}x \\ &= \lambda \int_{0}^{+\infty} x^2 \cdot \exp(-\lambda x) \, \mathrm{d}x \\ \end{split}$

Using the following anti-derivative

$\label{eq:second-moment-s2} \begin{split} \int x^2 \cdot \exp(-\lambda x) \, \mathrm{d}x &= \left[ - \frac{1}{\lambda} x^2 \cdot \mathrm{exp}(-\lambda x) \right]_{0}^{+\infty} - \int 2x \left( - \frac{1}{\lambda} x \cdot \mathrm{exp}(-\lambda x) \right) \mathrm{d}x \\ &= \left[ - \frac{1}{\lambda} x^2 \cdot \mathrm{exp}(-\lambda x) \right]_{0}^{+\infty} - \left( \left[ \frac{1}{\lambda^2} 2x \cdot \mathrm{exp}(-\lambda x) \right]_{0}^{+\infty} - \int 2 \left( \frac{1}{\lambda^2} \cdot \mathrm{exp}(-\lambda x) \right) \mathrm{d}x \right) \\ &= \left[ - \frac{x^2}{\lambda} \cdot \mathrm{exp}(-\lambda x) \right]_{0}^{+\infty} - \left( \left[ \frac{2x}{\lambda^2} \cdot \mathrm{exp}(-\lambda x) \right]_{0}^{+\infty} - \left[ - \frac{2}{\lambda^3} \cdot \mathrm{exp}(-\lambda x) \right]_{0}^{+\infty} \right) \\ &= \left[ \left( - \frac{x^2}{\lambda} - \frac{2x}{\lambda^2} - \frac{2}{\lambda^3} \right) \exp(-\lambda x) \right]_{0}^{+\infty} \; , \end{split}$

the second moment becomes

$\label{eq:second-moment-s3} \begin{split} \mathrm{E} [X^2] &= \lambda \left[ \left( - \frac{x^2}{\lambda} - \frac{2x}{\lambda^2} - \frac{2}{\lambda^3} \right) \exp(-\lambda x) \right]_{0}^{+\infty} \\ &= \lambda \left[ \lim_{x \to \infty} \left( - \frac{x^2}{\lambda} - \frac{2x}{\lambda^2} - \frac{2}{\lambda^3} \right) \exp(-\lambda x) - \left( 0 - 0 - \frac{2}{\lambda^3} \right) \exp(-\lambda \cdot 0) \right] \\ &= \lambda \left[ 0 + \frac{2}{\lambda^3} \right] \\ &= \frac{2}{\lambda^2} \; . \end{split}$

Plugging \eqref{eq:second-moment-s3} and \eqref{eq:exp-mean-ref} into \eqref{eq:var2}, we have:

$\label{eq:exp-var-2} \begin{split} \mathrm{Var}(X) &= \mathrm{E}\left[ X^2 \right] - \mathrm{E}\left[ X \right]^2 \\ &= \frac{2}{\lambda^2} - \left ( \frac{1}{\lambda} \right)^2 \\ &= \frac{2}{\lambda^2} - \frac{1}{\lambda^2} \\ &= \frac{1}{\lambda^2} \; . \end{split}$
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Metadata: ID: P401 | shortcut: exp-var | author: majapavlo | date: 2023-01-23, 09:02.