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Let be $t\in \mathbb{R}$, $n=1,2,\cdots$, $p\in \left[0,1\right]$ and $a\in \left(p,1\right]$.

Show that
$\underset{t}{sup}\left(ta-\mathrm{log}\left(p{e}^{t}+\left(1-p\right)\right)=a\mathrm{log}\left(\frac{a}{p}\right)+\left(1-a\right)\mathrm{log}\left(\frac{1-a}{1-p}\right)$
So, I've try to derivate, but I did'nt get sucess, since my result is different. I've get $\mathrm{log}\left(\frac{a\left(1-p\right)}{p\left(1-a\right)}\right)$. Any ideas?
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Carolyn Farmer
You are given a function $f$ of the variable $t$, with two parameters $p$ and $a$. You are asked to verify that the supremum of the function has a certain specific form in terms of the two parameters.

The correct procedure to find the supremum (or maximum) is to differentiate the function $f\left(t\right)$ with respect to the variable $t$. This give you the first derivative. The next step is to set the first derivative equal to zero, and solve the equation in terms of $t$. From your post I see that you have done so, and found that $t=log\left(a\right)-log\left(1-a\right)+log\left(1-p\right)-log\left(p\right)$. This answer is correct !

Now all you have to do is substitute this particular value fot $t$ into the expression for $f\left(t\right)$. This step will yield the supremum. If you perform this calculation (try it!), you will indeed find the result that was specified in the exercise.