Expectation of Exponential Distribution/Proof 2

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

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:

\(\ds \expect X\)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \beta\)

$\blacksquare$