# Entropy of Order Statistic

Consider $n$ independent and identically distributed random variables $\{X_i\}_{i=1,...n}$ with support on some interval $[a,b]$ and its $n$'th order statistic $\max_{i \in \{1,...n\}} X_i$ . The entropy of the maximum is

$$- \int_a^b F^n(x) \ln F^n(x) dx ,$$ where $F(x)= \Pr (X \le x)$. It seems natural that the entropy should be decreasing in $n$ (just think about $n$ very large). Is this a known result?

I did in fact prove that the entropy is monotone, but the proof turned out to be lengthy and messy. I would expect that there is a simple argument. Does anyone know?

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The integral quoted does not correspond to the Shannon entropy of the maximal order statistics, I am afraid. By definition the entropy $S_Z = - \int \ln(f_Z(z)) f_Z(z) \mathrm{d} z$, and for the $\max$, $f_{X_{n:n}}(x) = \left(F_X(x)^n\right)^\prime$. –  Sasha Jul 15 '12 at 19:36
If X has a density as Sasha said the cumulative distribution for the maximum is F$^n$(x) and the density f(x) =nF$^n$$^-$$^1(x)$F'(x). –  Michael Chernick Jul 15 '12 at 19:45
Now ln f(x)= ln n + ln F'(x) + (n-1)ln F(x). –  Michael Chernick Jul 15 '12 at 19:50
Sorry, I did use the wrong definition. -S –  Stephan Jul 15 '12 at 20:03

No, the entropy is not monotone. For example, consider $F_X(x) = x^{1/N}$ on $[0,1]$. Then $\max(X_1,\ldots,X_N)$ is uniform on $[0,1]$. The entropy of $\max(X_1,\ldots,X_n)$ increases as a function of $n$ for $1 \le n \le N$, reaching $0$ at $n=N$, then decreases after that.