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I was trying a past paper from http://www.abacus.utwente.nl/tentamens/M%20-%20Stochastic%20Processes/1a%20-%20Stochastic%20Processes%20Februari%202007.pdf

Hint: Use the fact that $p \log (1 + x) + q \log (1 − x)$ is maximized when $x = p − q$.

Secondly it was written: Let $f : (-1,1) \rightarrow \mathbb{R} : x \mapsto p\log (1+x) + q \log (1-x).$ Then using the hint above, $f$ is maximised at $p-q$ and $f(p-q) = \log 2 + p\log p + q\log q.$


My first question is, how (possibly using analysis/calculus?) would you deduce/get that $p \log (1 + x) + q \log (1 − x)$ is maximized when $x = p − q?$

The other thing, I have trouble understanding what exactly does $f : (-1,1)$ mean? And how would I intepret/read the following line? Its a bit new to me as I have not seen functions written like the following before. $f : (-1,1) \rightarrow \mathbb{R} : x \mapsto p\log (1+x) + q \log (1-x).$

Lastly, it says that $f$ is maximised at $p-q$ and $f(p-q) = \log 2 + p\log p + q\log q.$ How do you get that $f(p-q) = \log 2 + p\log p + q\log q?$ I tried substituting it into $f(x)=p\log (1+x) + q \log (1-x)$, getting $f(p-q) = p\log(1+p-q) + q\log(1-p+q). $ How would I go about from there?

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    Dear Heijden, we do not delete questions which have been answered (and which have already had one of their answers accepted!) except in extraordinary situations. In this particular case, two people have been kind enough to spend their time writing down very nice answers, and deleting the question would also delete *their* work.2012-04-03

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For your first question, you can see that $p \log (1 + x) + q \log (1 − x)$ is maximized when $x=p-q$ by taking the derivative at setting it equal to $0$. We have $0=\frac{d}{dx}(p \log (1 + x) + q \log (1 − x))=-\frac{p}{1+x}+\frac{q}{1-x}=\frac{-p(1-x)+q(1+x)}{(x+1)(x-1)}$ and so $(q-p)+(p+q)x=0$ hence $x=\frac{p-q}{p+q}$, and since $q=1-p$ this simplifies to $x=p-q$. This tells us that $p \log (1 + x) + q \log (1 − x)$ has an extremum at $x=p-q$, and this extremum must be the maximum as making $x$ near $1$ or $-1$ makes $\log(1-x)$ or $\log(1+x)$ very negative, respectively.

For your second question, "$f: (-1,1)$" doesn't mean anything. However, the full statement "$f:(-1,1)\to \mathbb R$" means "$f$ is a function from $(-1,1)$ to $\mathbb R$ (the real numbers)". In your case, the statement $f : (-1,1) \rightarrow \mathbb{R} : x \mapsto p\log (1+x) + q \log (1-x)$ means "$f$ is a function from $(-1,1)$ to $\mathbb R$ such that $f(x)=p\log (1+x) + q \log (1-x)$ for any $x\in (-1,1)$ (any $x$ between $-1$ and $1$)".

Edit: To get that $f(p-q)=\log 2+p\log p+q\log q$, use the fact that $p=1-q$ and $q=1-p$ so $\begin{eqnarray} f(p-q)&=&p\log(1+p-q) + q\log(1-p+q)\\ &=&p\log(1+p-(1-p)) + q\log(1-(1-q)+q)\\ &=&p\log(2p) + q\log(2q)\\ &=&p(\log 2+\log p) + q(\log 2+\log q)\\ &=&(p+q)\log 2+p\log p + q\log q\\ &=&\log 2+p\log p+q\log q. \end{eqnarray}$

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    Why the downvote?2012-04-02
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$f: (-1,1) \to \mathbb R$ just means that $f$ is a function defined on the interval $(-1,1)$ with values in the real line.

To find critical points of $f(x) = p \log(1+x) + q \log(1-x)$, solve f'(x)=0 for $x$. You should find exactly one critical point, at $x=p-q$ (if you remember that $p+q=1$). Note that this is a local maximum (e.g. by using the second derivative test, or noting that $\log(1+x)$ and $\log(1-x)$ are concave functions). If a differentiable function has only one critical point in an interval and it is a local maximum, then it is a global maximum on that interval.