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How do I prove that entropy is maximal for $P(A_2) = \cdots = P(A_n) = (1-a) /(n-1)$ while $P(A_1) = a$ (a fixed number) and $A_1,…, A_n$ is a partition of the sample space?

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  • $\begingroup$ Try to use convexity/concavity. There are also other ways. $\endgroup$ – Yuval Filmus Oct 9 '18 at 21:03
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Suppose that $X$ is a random variable over the set $\{1,\ldots,n\}$ satisfying $\Pr[X=1] = a$. Let $Y$ be the indicator of the event $X=1$. Then $$ \begin{align*} H(X) &\stackrel{(1)} = H(Y) + H(X|Y) \\ &\stackrel{(2)}= h(a) + aH(X|Y=1) + (1-a)H(X|Y=0) \\ &\stackrel{(3)}\leq h(a) + (1-a) \log (n-1). \end{align*} $$ Here (1) is the chain rule (using $H(X) = H(X,Y)$, since $Y$ is determined by $X$), (2) follows from $\Pr[Y=1]=a$ and the definition of $H(X|Y)$, and (3) follows from $H(X|Y=1) = 0$ (since when $Y=1$ we have $X=1$) and $H(X|Y=0) \leq \log (n-1)$ (since when $Y=0$, the r.v. $X$ only attains the values $2,\ldots,n$).

Furthermore, assuming $a \neq 1$, equality holds only if $X|Y=0$ is uniform, that is, if $\Pr[X=i|Y=0] = 1/(n-1)$ for $2 \leq i \leq n$, or in other words, if $\Pr[X=i] = (1-a)/(n-1)$ for $2 \leq i \leq n$.

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  • $\begingroup$ Thank you for your answer, in (1) I think you mean H(X,Y)? $\endgroup$ – Sydney.Ka Oct 10 '18 at 10:23
  • $\begingroup$ They’re equal, since Y is determined by X. $\endgroup$ – Yuval Filmus Oct 10 '18 at 15:28

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