Moment Generating Function of Continuous Uniform Distribution
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Theorem
Let $X \sim \ContinuousUniform a b$ for some $a, b \in \R$ denote the continuous uniform distribution on the interval $\closedint a b$.
Then the moment generating function of $X$ is given by:
- $\map {M_X} t = \begin {cases} \dfrac {e^{t b} - e^{t a} } {t \paren {b - a} } & t \ne 0 \\ 1 & t = 0 \end{cases}$
Proof
From the definition of the continuous uniform distribution, $X$ has probability density function:
- $\map {f_X} x = \begin{cases} \dfrac 1 {b - a} & a \le x \le b \\ 0 & \text{otherwise} \end{cases}$
From the definition of a moment generating function:
- $\ds \map {M_X} t = \expect {e^{t X} } = \int_{-\infty}^\infty e^{t x} \map {f_X} x \rd x$
where $\expect \cdot$ denotes expectation.
First, consider the case $t \ne 0$.
Then:
\(\ds \map {M_X} t\) | \(=\) | \(\ds \int_{-\infty}^a 0 e^{t x} \rd x + \int_a^b \frac{ e^{t x} } {b - a} \rd x + \int_b^\infty 0 e^{t x} \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \intlimits {\frac {e^{t x} } {t \paren {b - a} } } a b\) | Primitive of $e^{a x}$, Fundamental Theorem of Calculus | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {e^{t b} - e^{t a} } {t \paren {b - a} }\) |
In the case $t = 0$, we have $\expect {X^0} = \expect 1 = 1$.
$\blacksquare$
Sources
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Appendix $13$: Probability distributions
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): Appendix $15$: Probability distributions