# Differential Entropy of Exponential Distribution

## Theorem

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

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

$\map h X = 1 + \map \ln \beta$

## Proof

From the definition of the exponential distribution, $X$ has probability density function:

$\map {f_X} x = \dfrac 1 \beta e^{-\frac x \beta}$

From the definition of differential entropy:

$\ds \map h X = -\int_0^\infty \map {f_X} x \map \ln {\map {f_X} x} \rd x$

So:

 $\ds \map h X$ $=$ $\ds -\frac 1 \beta \int_0^\infty e^{-\frac x \beta} \map \ln {\frac 1 \beta e^{-\frac x \beta} } \rd x$ $\ds$ $=$ $\ds \frac 1 \beta \int_0^\infty e^{-\frac x \beta} \map \ln {\beta e^{\frac x \beta} } \rd x$ Reciprocal of Logarithm $\ds$ $=$ $\ds \frac {\map \ln \beta} \beta \int_0^\infty e^{-\frac x \beta} \rd x + \frac 1 {\beta^2} \int_0^\infty x e^{-\frac x \beta} \rd x$ Logarithm of Product, Definition of Natural Logarithm $\ds$ $=$ $\ds \frac {\map \ln \beta} \beta \sqbrk {-\beta e^{-\frac x \beta} }_0^\infty + \frac 1 {\beta^2} \paren {\sqbrk {-\beta x e^{-\frac x \beta} }_0^\infty + \beta \int_0^\infty e^{-\frac x \beta} \rd x}$ Primitive of Exponential Function, Integration by Parts $\ds$ $=$ $\ds \frac {\beta \map \ln \beta} \beta + \frac 1 \beta \int_0^\infty e^{-\frac x \beta} \rd x$ Exponential Tends to Zero and Infinity, Limit at Infinity of Polynomial over Real Exponential $\ds$ $=$ $\ds \map \ln \beta + \frac 1 \beta \sqbrk {-\beta e^{-\frac x \beta} }_0^\infty$ Primitive of Exponential Function $\ds$ $=$ $\ds \map \ln \beta + \frac \beta \beta$ Exponential Tends to Zero and Infinity, Exponential of Zero $\ds$ $=$ $\ds 1 + \map \ln \beta$

$\blacksquare$