Raw Moment of Exponential Distribution

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $n$ be a strictly positive integer.


Then the $n$th raw moment of $X$ is given by:

$\expect {X^n} = n! \beta^n$


Proof

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

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

By Moment in terms of Moment Generating Function:

$\expect {X^n} = \map {M^{\paren n}_X} 0$

We have:

\(\ds \map {M^{\paren n}_X} t\) \(=\) \(\ds \frac {\d^n} {\d t^n} \paren {\frac 1 {1 - \beta t} }\)
\(\ds \) \(=\) \(\ds -\frac 1 \beta \frac {\d^n} {\d t^n} \paren {\frac 1 {t - \frac 1 \beta} }\)
\(\ds \) \(=\) \(\ds -\frac {\paren {-1}^n n!} {\beta \paren {t - \frac 1 \beta}^{n + 1} }\) Nth Derivative of Reciprocal of Mth Power: Corollary, Chain Rule for Derivatives
\(\ds \) \(=\) \(\ds \frac {\paren {-1}^{n + 1} n!} {\frac \beta {\beta^{n + 1} } \paren {\beta t - 1}^{n + 1} }\)
\(\ds \) \(=\) \(\ds \frac {\paren {-1}^{n + 1} n! \beta^n} {\paren {\beta t - 1}^{n + 1} }\)

Setting $t = 0$ gives:

\(\ds \expect {X^n}\) \(=\) \(\ds \frac {\paren {-1}^{n + 1} n! \beta^n} {\paren {-1}^{n + 1} }\)
\(\ds \) \(=\) \(\ds n! \beta^n\)

$\blacksquare$