Differential Entropy 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 $\beta \in \R_{> 0}$.

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
\(\ds \) \(=\) \(\ds 1 + \map \ln \beta\)

$\blacksquare$