91Ó°ÊÓ

\(\Pr \left( {X = a} \right) = p\,\,\,and\,\,\,\Pr \left( {X = 1 - a} \right) = 1 - p\)

The utility function is given by,

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

For any given value of a,

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

The maximum of this expected utility can be found by elementary difference, then we have,

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

When this derivative is set to equal to 0, we find the value of a, i.e.,

\(\begin{align}\frac{{\partial \left( {E\left( {U\left( X \right)} \right)} \right)}}{{\partial a}} &= 0\\\frac{p}{a} - \frac{{1 - p}}{{1 - a}} &= 0\\\frac{{\left( {p\left( {1 - a} \right)} \right) - \left( {a\left( {1 - p} \right)} \right)}}{{a\left( {1 - a} \right)}} &= 0\\\frac{{p - ap - a + ap}}{{a\left( {1 - a} \right)}} &= 0\end{align}\)

\(\begin{align}p - a &= 0\\a &= p\end{align}\)

Then the second-order derivative is minimum to 0, then\(a = p\)

So,

\(\begin{align}\frac{{{\partial ^2}E\left( {U\left( X \right)} \right)}}{{\partial {a^2}}} &= \frac{\partial }{{\partial a}}\left( {\frac{p}{a} - \frac{{1 - p}}{{1 - a}}} \right)\\ &= - \frac{p}{{{a^2}}} - \frac{{1 - p}}{{{{\left( {1 - a} \right)}^2}}} < 0\end{align}\)

Therefore, \(E\left( {U\left( X \right)} \right)\)it is maximum when \(a = p\)