Study of the function -p log p

Question

Show, by studying the function, that $h(p)=-p\log_2 p$ is increasing on $[0,\,1/e]$ and decreasing on $[1/e,\,1]$ (with maximum $\log_2 e\cdot e^{-1}\approx 0{,}531$).

Solution

Write $h(p) = -p\dfrac{\ln p}{\ln 2}$. We differentiate (product rule): $h'(p) = -\frac{1}{\ln 2}\left(\ln p + p\cdot\frac{1}{p}\right) = -\frac{1}{\ln 2}\bigl(\ln p + 1\bigr).$ Since $\dfrac{1}{\ln 2}>0$, the sign of $h'(p)$ is opposite to that of $\ln p + 1$: $h'(p) = 0 \iff \ln p = -1 \iff p = \frac1e.$ For $p<1/e$ we have $\ln p + 1 < 0$, hence $h'(p)>0$ (increasing); for $p>1/e$ we have $h'(p)<0$ (decreasing). The maximum is therefore at $p=1/e$: $h\!\left(\frac1e\right) = -\frac1e\log_2\frac1e = \frac1e\log_2 e = \frac{\log_2 e}{e} \approx \frac{1{,}4427}{2{,}718} \approx \boxed{0{,}531}. \qquad \blacksquare$

Links

Topics: Probability distributions
Concepts: Entropy
Skills: Proving · Studying a function
Exercise type: Function study