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

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

$\label{eq:exp} X \sim \mathrm{Exp}(\lambda) \; .$

Then, the mean or expected value of $X$ is

$\label{eq:exp-mean} \mathrm{E}(X) = \frac{1}{\lambda} \; .$

Proof: The expected value is the probability-weighted average over all possible values:

$\label{eq:mean} \mathrm{E}(X) = \int_{\mathcal{X}} x \cdot f_\mathrm{X}(x) \, \mathrm{d}x \; .$

With the probability density function of the exponential distribution, this reads:

$\label{eq:exp-mean-s1} \begin{split} \mathrm{E}(X) &= \int_{0}^{+\infty} x \cdot \lambda \exp(-\lambda x) \, \mathrm{d}x \\ &= \lambda \int_{0}^{+\infty} x \cdot \exp(-\lambda x) \, \mathrm{d}x \; . \end{split}$

Using the following anti-deriative

$\label{eq:exp-mean-s2} \int x \cdot \exp(-\lambda x) \, \mathrm{d}x = \left( - \frac{1}{\lambda} x - \frac{1}{\lambda^2} \right) \exp(-\lambda x) \; ,$

the expected value becomes

$\label{eq:exp-mean-s3} \begin{split} \mathrm{E}(X) &= \lambda \left[ \left( - \frac{1}{\lambda} x - \frac{1}{\lambda^2} \right) \exp(-\lambda x) \right]_{0}^{+\infty} \\ &= \lambda \left[ \lim_{x \to \infty} \left( - \frac{1}{\lambda} x - \frac{1}{\lambda^2} \right) \exp(-\lambda x) - \left( - \frac{1}{\lambda} \cdot 0 - \frac{1}{\lambda^2} \right) \exp(-\lambda \cdot 0) \right] \\ &= \lambda \left[ 0 + \frac{1}{\lambda^2} \right] \\ &= \frac{1}{\lambda} \; . \end{split}$
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Metadata: ID: P47 | shortcut: exp-mean | author: JoramSoch | date: 2020-02-10, 21:57.