Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.8 AIME, harder Prove it United States

Problem:

Yan and Jacob play the following game. Yan shows Jacob a weighted 4-sided die labelled $1$ , $2$ , $3$ , $4$ , with weights $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{1}{7}$ , $\frac{1}{42}$ , respectively. Then, Jacob specifies $4$ positive real numbers $x_{1}, x_{2}, x_{3}, x_{4}$ such that $x_{1}+x_{2}+x_{3}+x_{4}=1$ . Finally, Yan rolls the die, and Jacob earns $10+\log \left(x_{k}\right)$ dollars if the die shows $k$ (note this may be negative). Which $x_{i}$ should Jacob pick to maximize his expected payoff?

(Here log is the natural logarithm, which has base $e \approx 2.718$ .)

Solution

Solution:

Jacob should pick $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\left(\frac{1}{2}, \frac{1}{3}, \frac{1}{7}, \frac{1}{42}\right)$ . More generally, suppose the weights are $p_{1}, \ldots, p_{4}$ . Then Jacob's expected payoff is

$10+\sum_{i=1}^{4} p_{i} \log \left(x_{i}\right)=10+\sum_{i=1}^{4} p_{i} \log p_{i}+\sum_{i=1}^{4} p_{i} \log \left(\frac{x_{i}}{p_{i}}\right)$

Now, by JENSEN's INEQUALITY on the concave function $\log x$ , we obtain

$\sum_{i=1}^{4} p_{i} \log \left(\frac{x_{i}}{p_{i}}\right) \leq \log \left(\sum_{i=1}^{4} p_{i} \cdot \frac{x_{i}}{p_{i}}\right)=\log 1=0$

and equality occurs exactly when $\frac{x_{1}}{p_{1}}=\frac{x_{2}}{p_{2}}=\frac{x_{3}}{p_{3}}=\frac{x_{4}}{p_{4}}$ ; that is, when $x_{i}=p_{i}$ for every $i$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.