Expectation of Exponential Distribution

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Theorem

Let $X$ be a continuous random variable of the exponential distribution with parameter $\beta$ for some $\beta \in \R_{> 0}$


Then the expectation of $X$ is given by:

$\expect X = \beta$


Proof 1

The expectation of a continuous random variable $X$ with sample space $\Omega_X$ is given by:

$\ds \expect X := \int_{x \mathop \in \Omega_X} x \map {f_X} x \rd x$

where $f_X$ is the probability density function of $X$.


For the exponential distribution:

$\Omega_X = \hointr 0 \infty$

From Probability Density Function of Exponential Distribution:

$\ds \map {f_X} x = \frac 1 \beta \map \exp {- \frac x \beta}$

So:

$\ds \expect X = \int_0^\infty x \frac 1 \beta \map \exp {- \frac x \beta} \rd x$

where $\exp$ is the exponential function.

Substituting $u = \dfrac x \beta$, we have:

$\ds \expect X = \beta \int_0^\infty u \map \exp {-u} \rd u$

The integral evaluates to:

$\ds \expect X = \bigintlimits {-\beta \paren {u + 1} \map \exp {-u} } 0 \infty$

So:

$\ds \expect X = \beta - \beta \lim_{u \mathop \to \infty} \frac {u + 1} {\exp u}$


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By Limit at Infinity of Polynomial over Complex Exponential, it follows that this limit is zero, so that:

$\expect X = \beta$

$\blacksquare$


Proof 2

From Moment Generating Function of Exponential Distribution, the moment generating function $M_X$ of $X$, is given by:

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

By Moment in terms of Moment Generating Function:

$\expect X = \map {M_X'} 0$

We have:

\(\ds \map {M_X'} t\) \(=\) \(\ds \map {\frac \d {\d t} } {\frac 1 {1 - \beta t} }\)
\(\ds \) \(=\) \(\ds \frac {-\beta} {-1} \frac 1 {\paren {1 - \beta t}^2}\) Chain Rule for Derivatives, Derivative of Power
\(\ds \) \(=\) \(\ds \frac \beta {\paren {1 - \beta t}^2}\)

Setting $t = 0$ gives:

\(\ds \expect X\) \(=\) \(\ds \frac \beta {\paren {1 - 0 \beta}^2}\)
\(\ds \) \(=\) \(\ds \beta\)

$\blacksquare$


Also see


Sources

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