91Ó°ÊÓ

If the person wins the bet, then the fortune is \(A + b\)

If the person loses the bet, then the fortune is\(A - b\)

The probability of a person winning is p.

The probability of a person losing is\(1 - p\).

The utility function is,

\(U\left( x \right) = \log x\)

For any given value of b,

\(E\left( {U\left( X \right)} \right) = p\log \left( {A + b} \right) + \left( {1 - p} \right)\log \left( {A - b} \right).\)

Therefore,

\(\begin{align}\frac{{\partial E\left( {U\left( X \right)} \right)}}{{\partial b}} &= \frac{\partial }{{\partial b}}\left( {p\log \left( {A + b} \right) + \left( {1 - p} \right)\log \left( {A - b} \right)} \right)\\ &= \frac{p}{{\left( {A + b} \right)}} - \frac{{\left( {1 - p} \right)}}{{\left( {A - b} \right)}}\end{align}\)

Equating the derivative to 0,

\(b = \left( {2p - 1} \right)A.\)

Since, \(\frac{{{\partial ^2}E\left( {U\left( X \right)} \right)}}{{\partial {b^2}}} < 0,\) this value of b does yield a maximum value of \(E\left( {U\left( X \right)} \right)\).If \(p \ge {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.\ } \!\lower0.7ex\hbox{$2$}}\), this value of b lies between 0 and A as required.

However, if \(p < {\raise0.7ex\hbox{$1$} \!\mathord{\left/ {\vphantom {1 2}}\right.\ } \!\lower0.7ex\hbox{$2$}}\), this value of b is negative and not permissible.

In this case, it can be shown that the maximum value of\(E\left( {U\left( X \right)} \right)\)for\(0 \le b \le A\)occurs when person bet an amount\(b = \left( {2p - 1} \right)A\).