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\textbf{Question}

Consider a portfolio $\Pi$ which is composed of a proportion $\lambda
&lt; 1$ of a risk free asset $S_0$ with associated return $R_0$ and a
proportion $1-\lambda$ of a risky portfolio $S_1$ with associated
return $R_1$ and variance $\sigma_1^2$. Show that as $\lambda$ varies,
$\Pi$ lies along a straight line in the risk/reward diagram, the line
having slope $\theta = (R_1 - R_0)/\sigma_1$. Explain briefly why this
implies that the problem of finding the capital market line reduces to
that of maximizing $\theta$ over all risky portfolios.

Now consider a scenario where there are three risky assets $S_1$,
$S_2$ and $S_3$ with respective expected returns
$$R_1=0.08, \ \ \ R_2=0.10, \ \ \ R_3=0.12.$$

The variances and covariances between the assets are given by 
\begin{eqnarray*}
\sigma_1^2 &amp; = &amp; 0.008\\
\sigma_{12} &amp; = &amp; 0.004\\
\sigma_{13} &amp; = &amp; 0\\
\sigma_2^2 &amp; = &amp; 0.006\\
\sigma_{23} &amp; = &amp; 0.002\\
\sigma_3^2 &amp; = &amp; 0.008
\end{eqnarray*}
and the risk free rate is $0.05$. Short selling and borrowing are
allowed. Show that the optimal portfolio of risky assets consists of
investing proportions $1/11$, $4/11$ and $6/11$ of one's total wealth
in $S_1$, $S_2$ and $S_3$ respectively. Show that the associated risk
and return are $\sqrt{520/121} \sim 2.07 \% $ and $120/11 \sim 10.91
\% $ respectively, and the market price of risk is
$$\theta = \frac{\sqrt{130}}{4} \sim 2.85$$

\newpage
\textbf{Answer}

We have $\Pi=\lambda S_0+ (1-\lambda)S_1$ and thus
$$R_{\Pi}=\lambda R_0+ (1-\lambda)R_1.$$

For the portfolio variance $\sigma_{\Pi}^2$ we have
$$\sigma_{\Pi}^2 = \lambda^2\sigma_0^2+ 2\lambda(1-
\lambda)\sigma_{01} +(1-\lambda)^2 \sigma_1^2.$$

But $S_0$ is riskless, so by definition $\sigma_0=0$, $\sigma_{01}=1$.

Thus
\begin{eqnarray*}
\sigma_{\Pi}^2 &amp; = &amp; (1-\lambda)^2\sigma_1^2\\
\Rightarrow \sigma_{\Pi} &amp; = &amp; (1-\lambda)\sigma_1
\end{eqnarray*}

Now
\begin{eqnarray*}
\lambda&lt;1 &amp; \Rightarrow &amp; 1 -\lambda &gt;0\\
&amp; \Rightarrow &amp; \sigma_{\Pi}=(1-\lambda)\sigma_1\\
&amp; \Rightarrow &amp; \lambda = 1 - \frac{\sigma_{\Pi}}{\sigma_1}
\end{eqnarray*}

Thus
$$R_{\Pi}=\left ( 1 -\frac{\sigma_{\Pi}}{\sigma_1} \right ) R_0
+\frac{\sigma_{\Pi}}{ \sigma_1}R_1 = R_0+\sigma_{\Pi}\left ( \frac{
R_1- R_0}{\sigma_1} \right )$$
A straight line of slope $(R_1-R_0)/\sigma_1$ as required.

Now the CML is just the straight line through the risk free return
$R_0$ at $\sigma_{\Pi}$ which is tangent to the boundary of the
(convex) opportunity slope; obviously therefore it is the line that
passes through $(R_0,0)$ and some risky portfolio that has positive
slope.

Now we have to maximize $\theta =(R_{\Pi}-R_0)/\sigma_{\Pi}$ over all
possible risky portfolios where
\begin{eqnarray*}
\Pi &amp; = &amp; X_1S_1+X_2S_2+X_3S_3\\
with &amp; &amp; X_1+X_2+X_3-1.
\end{eqnarray*}

Now
$$R_{\Pi}=X_1R_1+X_2R_2+X_3R_3=\frac{1}{100}(8X_1+10X_2+12X_3)$$
and
\begin{eqnarray*}
\sigma_{\Pi} &amp; = &amp; (X_1^2\sigma_1^2+2X_1X_2\sigma_{12}+
X_2^2\sigma_2^2+2X_1X_3\sigma_{13} +X_3^2\sigma_3^2\\
&amp; &amp; +2X_2X_3\sigma_{23} )^{\frac{1}{2}}\\
&amp; = &amp; (X_1^2+8X_1X_2+6X_2^2+8X_3^2\\
&amp; &amp; +4X_2X_3)^{\frac{1}{2}}\left (
\frac{10^{\frac{1}{2}}}{100} \right )\\ 
&amp; = &amp; 10^{\frac{1}{2}}\frac{\alpha^{\frac{1}{2}}}{100}\ \rm{say}.
\end{eqnarray*}

Now we have to maximize $\displaystyle \theta =\frac{(8-5)X_1+
(10-5)X_2 +(12-5)X_3}{\alpha^{1/2}}$.
$$\rm{i.e.}\ \ \ \theta=\frac{3X_1+5X_2+7X_3}{\alpha^{1/2}}.$$

Let $3X-1+ 5X_2+ 7X_3 =\beta$. Then
\begin{eqnarray*}
\frac{\partial \theta}{\partial X_1} &amp; = &amp; \frac{3\alpha^{1/2}-
\beta\frac{1}{2} \alpha^{-1/2} (16X_1+8X_2)}{\alpha} =0\\
\frac{\partial \theta}{\partial X_2} &amp; = &amp; \frac{5\alpha^{1/2}-
\beta\frac{1}{2} \alpha^{-1/2}(8X_1+12X_2+4X_3)}{\alpha} =0\\
\frac{\partial \theta}{\partial X_3} &amp; = &amp; \frac{7\alpha^{1/2}-
\beta\frac{1}{2} \alpha^{-1/2}(16X_3+4X_2)}{\alpha}=0
\end{eqnarray*}
(n.b. factors of $10$ not important)

Let $\beta X_i/\alpha^2 =z_i$ (using the usual trick) then

$\begin{array}{rlr}
3 &amp; = 8z_1+ 4z_2 &amp; \Rightarrow z_1=3/8-z_2/2\\
5 &amp; = 4z_1+ 6z_2 +2z_3 &amp;\\
7 &amp; = 2z_2+ 8z_3 &amp; \Rightarrow z_3=7/8 -z_2/4
\end{array}$

$\Rightarrow 5-\frac{3}{2} = -2z_2+6z_2+\frac{7}{4}-z_2/2$ $\ \ \
z+2=\frac{1}{2}$

Thence $z_1=\frac{1}{8},\ \ \ z_3=\frac{3}{4}$.

Now $\sum X_i =1$
$$\Rightarrow \frac{\beta}{\alpha^2}=\sum z_i =\frac{11}{8}$$

Thus
\begin{eqnarray*}
X_1 &amp; = &amp; \frac{1}{11}\\
X_2 &amp; = &amp; \frac{4}{11}\\
X_3 &amp; = &amp; \frac{6}{11}
\end{eqnarray*}

Thence $\displaystyle R_{\Pi}=\frac{1}{100} \left ( \frac{8}{11}+
\frac{40}{11}+ \frac{77}{11} \right ) = \frac{120}{11}\% \sim 10.91\%$

\begin{eqnarray*}
\frac{100^2}{10}\sigma_{\Pi}^2 &amp; = &amp; \frac{8}{121}+ \frac{32}{121}+
\frac{96}{121}+ \frac{8.36}{121}\\
&amp; = &amp; \frac{520}{121}\\
&amp; \Rightarrow &amp; \sigma_{\Pi} =\left ( \sqrt{\frac{5200}{121}} \right )
\frac{1}{100}\\ &amp; &amp; \sigma_{\Pi} \sim 6.56 \%\\
\theta &amp; = &amp; \frac{\frac{120}{11}-
\frac{500}{100}}{\frac{100}{100}\sqrt{\frac{5200}{ 121}}}\\
&amp; = &amp; \sqrt{\frac{\sqrt{13}}{4}} \sim 0.90138
\end{eqnarray*}

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