Differential Entropy of Continuous Uniform Distribution

Theorem

Let $X \sim \ContinuousUniform a b$ for some $a, b \in \R$, $a \ne b$, where $\operatorname U$ is the continuous uniform distribution.

Then the differential entropy of $X$, $\map h X$, is given by:

$\map h X = \map \ln {b - a}$

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 differential entropy:

$\ds \map h X = - \int_{-\infty}^\infty \map {f_X} x \map \ln {\map {f_X} x} \rd x$

So:

\(\ds \map h X\)

\(=\)

\(\ds -\int_a^b \frac 1 {b - a} \map \ln {\frac 1 {b - a} } \rd x\)

\(\ds \)

\(=\)

\(\ds \frac {\map \ln {b - a} } {b - a} \int_a^b \rd x\)

Logarithm of Reciprocal

\(\ds \)

\(=\)

\(\ds \frac {\map \ln {b - a} } {b - a} \bigintlimits x a b\)

Primitive of Constant, Fundamental Theorem of Calculus

\(\ds \)

\(=\)

\(\ds \frac {\paren {b - a} \map \ln {b - a} } {b - a}\)

\(\ds \)

\(=\)

\(\ds \map \ln {b - a}\)

$\blacksquare$