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\usepackage{epsfig}
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\begin{document}
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\textbf{Question}

Let $\cal{O}$ be the opportunity set, in risk/return space, for a set
of risky assets (none of which are perfectly negatively
correlated). Assume that short selling is allowed and that there is
also a riskless investment with return $R_F$ available. Suppose that
an investor choses to invest $X \ge 0$ in a purely risky portfolio
$\cal{P}$, with variance $\sigma_{\cal{P}}^2$ and expected return
$R_{\cal{P}}$, and $1-X$ in the riskless investment. Show that as $X$
varies the investor's total portfolio lies along a line of slope
$(R_{\cal{P}}-R_F)/\sigma_{\cal{P}}$ and intercept $R_F$. Hence or
otherwise, deduce that the problem of finding the capital market line
reduces to maximizing the slope of the line above all possible
portfolios in $\cal{O}$.

Consider a situation where there are three risky assets $S_1$, $S_2$
and $S_3$ with respective expected returns
$$R_1=0.1, \ \ \ R_2=0.12, \ \ \ R_3=0.18,$$
variances and covariances given by
$$\sigma_1^2=0.0016, \ \ \ \sigma_{12}=0.0016, \ \ \ \sigma_{13}=0,$$
$$\sigma_2^2=0.01, \ \ \ \sigma_{23}=0.0012, \ \ \
\sigma_3^2=0.0144.$$
Further, assume that the risk free rate is $0.06$, short selling and
borrowing are allowed. Show that under these circumstances, the market
price of risk is
$$\theta = \frac{167}{\sqrt{13861}} \sim 1.418$$
and that the optimal portfolio of risky assets consists of the
following proportions of total wealth invested in $S_1$, $S_2$ and
$S_3$, respectively,
\begin{eqnarray*}
\frac{237}{331} &amp; \sim &amp; 0.716\\
\frac{12}{331} &amp; \sim &amp; 0.036\\
\frac{82}{331} &amp; \sim &amp; 0.247
\end{eqnarray*}

\newpage
\textbf{Answer}

Let $\Pi=XP+(1-X)R_f$, $R_f=$ risk free investment.

Then \begin{eqnarray*} R_{\Pi}=\rm{expected\ return\ on\ }\Pi &amp; = &amp;
E(XP+(1-X)R_f)\\ &amp; = &amp; XE(P)+(1-X)E(R_f)\\ &amp; = &amp; XR_P+(1-X)R_F
\end{eqnarray*}
\begin{eqnarray*} \sigma_{\Pi}^2=\rm{variance\ of\ }\Pi &amp; = &amp;
E((XP+(1-X)R_f)^2)\\
&amp; &amp; - E(XP+(1-X)R_f)^2\\ &amp; = &amp;
E(X^2P^2+2X(1-X)PR_f+(1-X)^2R_f^2)\\ &amp; &amp;
-X^2R_P^2-2X(1-X)R_PR_F-(1-X)^2R_F^2\\ &amp; = &amp; X^2(E(P^2))-R_P^2)
+2X(1-X)[R_FE(P)\\
&amp; &amp; -R_FR_P]+(1-X)^2(E(R_f^2)-R_F^2)\\ &amp; = &amp;
X^2\rm{Var}(P) = X^2\sigma_P^2 \ \ \ (\rm{note\ }X \ge 0)
\end{eqnarray*}

$\Rightarrow \sigma_\Pi$, $R_{\Pi}=XR_P+(1-X)R_F$, so

\begin{center}
$\begin{array}{c}
\epsfig{file=448-1998-1.eps, width=40mm}
\end{array}
\ \ \ 
\begin{array}{rcl}
\displaystyle \Rightarrow R_{\Pi} &amp; = &amp; \frac{\sigma_{\Pi}}{\sigma_P}+\left ( 1-
\frac{\sigma_{\Pi}}{\sigma_P} \right ) R_F\\
&amp; = &amp; \displaystyle R_F+ \left ( \frac{R_P -R_F}{\sigma_P} \right ) \sigma_{\Pi}
\end{array}$
\end{center}

Capital market line is just the straight line through the risk free
rate $R_F$ at $\sigma=0$ which is tangent to opportunity sets
boundary; (can assume opp set convex!)

\begin{center}
$\begin{array}{c}
\epsfig{file=448-1998-2.eps, width=40mm}
\end{array}
\begin{array}{c}
\rm{Evidently\ it\ is\ the\ line\ passing\ }\\
\rm{through\ }(R_F, 0)\ \rm{and\ some\ risky\ portfolio}\\
\rm{which\ has\ greatest\ possible\ slope.}
\end{array}$
\end{center}

Aim is to maximize $\displaystyle \frac{R_P-R_F}{\sigma_P}$ over all
possible risky assets.

$\begin{array}{rcl} Let P &amp; = &amp; X_1S_1+X_2S_2+X_3S_3 \ \ \ \rm{with} \
X_1+X_2+X_3=1\\
 R+P &amp; = &amp; X_1R_1+X_2R_2+X_3R_3\\
\sigma_{P} &amp; = &amp; (X_1^2\sigma_1^2+2X_1X_2\sigma_{12}+
X_2^2\sigma_2^2+ 2X_2X_3\sigma_{23} +X_3^2\sigma_3^2\\
&amp; &amp; + 2X_1X_3\sigma_1\sigma_3)^{\frac{1}{2}}
\end{array}$

$\begin{array}{rl} \rm{Thus\ \ } R_P-R_F &amp; = (4X_1+ 6X_2+ 12X_3)/100\\
\sigma+P &amp; = (16X_1^2+ 32X_1X_2+ 100X_2^2+ 24X_2X_3\\
&amp; +144X_3^2)^{\frac{1}{2}}/100
\end{array}$

So we want to maximize $$(*) \ \ \displaystyle \theta= \frac{4X_1+ 6X_2+
12X_3}{(16X_1^2+ 32X_1X_2+ 100X_2^2+ 24X_2X_3+
144X_3^2)^{\frac{1}{2}}}$$
$$ = \frac{4X_1+ 6X_2+ 12X_3}{\alpha}$$

$\begin{array}{rl} \displaystyle \frac{\partial\theta}{\partial X_1} &amp;
\displaystyle \frac{4}{\alpha}-\frac{1}{2}\frac{(4X_1+ 6X_2+
12X_3)}{\alpha^3}(32X_1+ 32X_2) = 0\\
&amp; \displaystyle \Rightarrow
\frac{4X_1+6X_2+12X_3}{\alpha^2}(32X_1+32X_2)=8\\
\displaystyle \frac{\partial\theta}{\partial X_2} &amp; =0\\
&amp; \displaystyle \Rightarrow \frac{4X_1+ 6X_2+ 12X_3}{\alpha^2}(32X_1+
200X_2+ 24X_3)=12\\
\displaystyle \frac{\partial\theta}{\partial X_3} &amp; =0\\
&amp; \displaystyle \Rightarrow \frac{4X_1+ 6X_2+ 12X_3}{\alpha^2}(24X_2+
288X_3)=24
\end{array}$

Put $z_i= \displaystyle \left ( \frac{4X_1+ 6X_2+ 12X-3}{\alpha^2}
\right ) X_i$ so we get

$\begin{array}{llll}
&amp; \left (
\begin{array}{ccc} 32 &amp; 32 &amp; 0\\ 32 &amp; 200 &amp; 24\\ 0 &amp; 24 &amp; 288
\end{array} \right ) \left ( \begin{array}{c} z_1\\ z_2\\ z_3
\end{array} \right ) &amp; = \left ( \begin{array}{c} 8\\ 12\\ 24
\end{array} \right ) &amp; \\
\Rightarrow &amp; \left (
\begin{array}{ccc} 4 &amp; 4 &amp; 0\\ 16 &amp; 100 &amp; 12\\ 0 &amp; 1 &amp; 12 \end{array}
\right ) \left ( \begin{array}{c} z_1\\ z_2\\ z_2 \end{array} \right )
&amp; = \left ( \begin{array}{c} 1\\ 6 \\ 1 \end{array} \right )
 \end{array}$

$$\Rightarrow \left ( \left. \begin{array}{ccc} 4 &amp; 4 &amp; 0\\ 16 &amp; 100 &amp;
12\\ 0 &amp; 1 &amp; 12 \end{array} \right | \begin{array}{c} 1\\ 6\\ 1
\end{array} \right ) \sim \left ( \left. \begin{array}{ccc} 4 &amp; 4 &amp; 0\\ 0 &amp; 84 &amp;
12\\ 0 &amp; 1 &amp; 12 \end{array} \right | \begin{array}{c} 1\\ 2\\ 1
\end{array} \right ) \sim \left ( \left. \begin{array}{ccc} 4 &amp; 4 &amp; 0\\ 0 &amp; 83 &amp;
0\\ 0 &amp; 1 &amp; 12 \end{array} \right | \begin{array}{c} 1\\ 1\\ 1
\end{array} \right ) $$

So

$\begin{array}{rl} z_2 &amp; = 1/83\\ 4z_1 &amp; =1 - 4/83 = 79/83\\ 12z_3
&amp; =1 - 1/83 = 82/83 \end{array}$

$\Rightarrow \ \ z_1=79/(4\times83)$, $z_2=1/83$, $82/(12\times83)$

Now note that
\begin{eqnarray*} z_1+z_2+z_3 &amp; = &amp; \displaystyle \left
( \frac{4X_1+ 6x_2+ 12X_3}{\alpha^2} \right )(X_1+X_2+X_3)\\ &amp; = &amp;
\displaystyle \left ( \frac{4X_1+ 6X_2+ 12X_3}{\alpha^2} \right )
\end{eqnarray*}

hence $X_i = z_i/(z_1+z_2+z_3)$; $z_1+z_2+z_3=
\frac{1}{12\times83}(82+3\times79+12) =\frac{1}{12\times83}(331)$

$\begin{array}{lrll} i.e. &amp; X_1 = &amp;
\frac{79}{4\times83}\times\frac{12\times83}{331} &amp;= \displaystyle
\frac{237}{331}\\ &amp; X_2 = &amp; \frac{1}{83}\times\frac{12\times83}{331}
&amp;= \displaystyle \frac{12}{331}\\ &amp; X_3 = &amp;
\frac{82}{12\times83}\times\frac{12\times83}{331} &amp;= \displaystyle
\frac{82}{331} \end{array}$

Finally put these numbers back into $(*)$ to give $$\displaystyle
\theta = \frac{167}{\sqrt{13861}}$$
\begin{eqnarray*} z_1 &amp; = &amp; \displaystyle \frac{79}{332} =
\frac{237}{996}\\ \displaystyle z_2 &amp; = &amp; \frac{1}{83}=\frac{12}{996}\\
\displaystyle z_3 &amp; = &amp; \frac{82}{996}
\end{eqnarray*}

\end{document}



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