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Let $p_1,\ldots,p_n$ be numbers such that $0 < p_i < 1$ and $\sum p_i = 1$.

Let also $x_1,\ldots,x_n$ be numbers such that $x_i \ge -1$.

Define the function (for $0<y<1$)

$$f(y) = \sum_{i=1}^n \log\left(1+x_iy\right)p_i.$$

Now let $p_{n+1}$ be a new number such that $0 < p_{n+1} < 1$, and let $x_{n+1}$ be a new number such that $x_{n+1} > x_i$ for all $i=1,\ldots,n$. Define the new function (for $0<y<1$)

$$f^*(y) = \sum_{i=1}^n \log\left(1+x_iy\right)\left(p_i - \frac{p_{n+1}}{n}\right) + \log\left(1+x_{n+1}y\right)p_{n+1}.$$

I believe I have already managed to show, with just basic algebra and some calculus, that $f^*(y) > f(y)$ for all $y$.

Problem: However, can someone help me show also that the maximum value of $f^*(y)$ lies to the right of the maximum value of $f(y)$? (In other words, that the $y$ that maximizes $f^*$ is greater than the $y$ that maximizes $f$.)

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What does $\log(1+x_iy)p_i$ mean? Is it $\log((1+x_i)p_i)$ or is it $p_i\log(1+x_i y)$? –  André Nicolas Jun 9 '12 at 6:08

1 Answer 1

Note that $$ f^\prime(y)=\sum_{i=1}^n\frac{p_ix_i}{1+x_iy}\tag{1} $$ and $$ \begin{align} f^{\ast\prime}(y) &=\sum_{i=1}^n\frac{\left(p_i-\frac{p_{n+1}}{n}\right)x_i}{1+x_iy}\;+\;\frac{p_{n+1}x_{n+1}}{1+x_{n+1}y}\\ &=f^\prime(y)+\frac{p_{n+1}}{n}\sum_{i=1}^n\left(\frac{x_{n+1}}{1+x_{n+1}y}-\frac{x_i}{1+x_iy}\right)\\ &\ge f^\prime(y)\tag{2} \end{align} $$ Since $y\ge0$, $\frac{x}{1+xy}$ is an increasing function of $x$, so $\frac{x_{n+1}}{1+x_{n+1}y}\ge\frac{x_i}{1+x_iy}$.

Both $f^{\prime}$ and $f^{\ast\prime}$ are obviously decreasing functions.

When $f^\prime(y)=0$, we have $f^{\ast\prime}(y)\ge0$. Since $f^{\ast\prime}$ is a decreasing function, if $f^{\ast\prime}(y^\ast)=0$, we must have $y^\ast\ge y$.

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