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Let $b,c \in (0,1)$ be such that $b+c<1.$ Define the following function for $p \in (0,1) :$ $$ I(p;b,c):=(b+c)[p \log\frac{1}{p}+(1-p)\log\frac{1}{1-p}]-cH(p;b,c)-bH(p;c,b)$$ where $H(x;\epsilon,\delta):= p \log\frac{\epsilon (1-p)+ \delta p}{\delta p}+ (1-p) \log\frac{\delta (1-p)+\epsilon p}{\delta (1-p)} .$

Ploting this function for various values of $b,c$ we see that the maximum is obtained at $p=\frac{1}{2}.$ Can we prove rigorously this fact for any $b,c$ described before ?

(Since $I(p;b,c)=I(1-p;b,c)$ it suffices to show for example that $H'(p;b,c)>0$ for $0<p<\frac{1}{2}.$ However this seems more difficult than the original question.)

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1 Answer 1

Note that the function $H(x;\epsilon,\delta)$ satisfies this nice property that $$ H(x; \epsilon, \delta) = H(x; k\epsilon, k\delta)$$ for any $k \neq 0$.

So suppose we scale $b$ and $c$ by some $k$ such that $kb +kc = 1$. We notice that $$ I(p; kb,kc) = kI(p; b,c) $$ So maximising $I(p;b,c)$ is the same as maximising $I(p; kb, kc)$. Without loss of generality, we can assume $b+c = 1$.

Now the problem reduces just simple expansion and cancellation and arriving at the result. $$H(p) - (1-b)H(x;b,1-b) - bH(x;1-b,b) = - H(b) + H(b(1-p)+(1-b)p)$$ Since $H(b)$ does not depend on $p$, we need to maximise $H(b(1-p)+(1-b)p)$, which clearly happens when $p = \frac{1}{2}$.

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