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\textbf{Question}


\textbf{Answer}

The covariance matrix $\sigma_i$, must be positive definite as assets
cannot have arbitrary covariances. For example, it is evidently
impossible to have 3 risky assets where all the covariances are $-1$ as
3 assets cannot all move independently in opposite directions to each
other.

Now consider the portfolio $\Pi =\lambda_1S_1+ \lambda_2S_2+
\lambda_3S_3$ where the $\lambda_i$ sum to 1 (but need not be between
0 and 1).

We have
$$\overline{R}_{\Pi} = \frac{1}{100}(10\lambda_1+ 12\lambda_2+
15\lambda_3)$$
\begin{eqnarray*} \sigma_{\Pi}^2 & = & \sum_{i=1}^3 \! \sum_{j=1}^3
\sigma_{ij}\lambda_i\lambda_j\\
& = & \frac{1}{100^2} ( 9\lambda_1^2
+25\lambda_2^2 +100\lambda_3^2 +12\lambda_1\lambda_2
-12\lambda_1\lambda_3 +20\lambda_2\lambda_3)
\end{eqnarray*}

An investment consisting of $\Pi$ and cash along the straight line
joining $(0.06, 0)$ to $(\overline{R}_{\Pi}, \sigma_{\Pi})$ in the
Risk/Reward diagram. This line has slope
$$\theta = \frac{\overline{R}_{\Pi}-0.06}{\sigma_{\Pi}} $$
$$\theta = \frac{\overline{R}_{\Pi}-0.06}{\sigma_{\Pi}} = \frac{
(10\lambda_1 +12\lambda_2 +15\lambda_3) -6(\lambda_1 +\lambda_2
+\lambda_3)}{ (9\lambda_1^2 +25\lambda_2^2 +100\lambda_3^2
+12\lambda_1\lambda_2 -12\lambda_1\lambda_3
+20\lambda_2\lambda_3)^{\frac{1}{2}}}$$ 

Let $A=$ the bottom $\Rightarrow \ \theta=\frac{4\lambda_1 +6\lambda_2
+9\lambda_3}{A}$

\begin{eqnarray*}
\frac{\partial \theta}{\partial \lambda_1} & = & \frac{4}{A} -
\frac{1}{2} \left ( \frac{4\lambda_1 +6\lambda_2 + 9\lambda_3}{A^3}
\right ) (18\lambda_1 +12\lambda_2 -12\lambda_3) = 0\\
\frac{\partial \theta}{\partial \lambda_2} & = & \frac{6}{A} -
\frac{1}{2} \left ( \frac{4\lambda_1 +6\lambda_2 +9\lambda_3}{A^3}
\right ) (50\lambda_2 +12\lambda_2 +20\lambda_3) = 0\\ 
\frac{\partial \theta}{\partial \lambda_3} & = & \frac{9}{A} -
\frac{1}{2} \left ( \frac{4\lambda_1 +6\lambda_2 +9\lambda_3}{A^3}
\right ) (20\lambda_2 -12\lambda_2 +20\lambda_3) = 0
\end{eqnarray*}  

Now multiply through by $2A$, put $\alpha =(4\lambda_1 +6\lambda_2
+9\lambda_3)/A^2$ and we get (justified by the usual argument)
\begin{eqnarray*}
18\lambda_1+ 12\lambda_2 -12\lambda_3 & = & 8/\alpha\\
50\lambda_2+ 12\lambda_1 +20\lambda_3 & = & 12/\alpha\\
200\lambda_3 - 12\lambda_1 +20\lambda_2 & = & 18/\alpha
\end{eqnarray*}
now put $u=\alpha\lambda_1$, $ v=\alpha\lambda_2$, $w=\alpha\lambda_3$

$\begin{array}{rllrl}
18u+12v-12w & = 8 & \Rightarrow & 12u+8v-8w & = 16/3\\
12u+50v+20w & =12 & \Rightarrow & 70v+220w & =30\\
-12u+20v+200w & =18 \ \ & \Rightarrow & 28v+192w & = 70/3
\end{array}$

$\Rightarrow$ finally
$$u=\frac{251}{546}, \ \ \ v=\frac{47}{546}, \ \ \ w=\frac{17}{156}$$
$$u+v+w=\frac{55}{84} = \alpha(\lambda_1 +\lambda_2 +\lambda_3) =
\alpha$$
$\displaystyle \Rightarrow \ \lambda_1=\frac{502}{715},\ \ \
\lambda_2=\frac{94}{715}, \ \ \ \lambda_3=\frac{119}{715}$
($\Rightarrow$ no short selling.)

\begin{eqnarray*}
\Rightarrow \overline{R}_{\Pi} & = & 11.095 \%\\
\sigma_{\Pi} & = & 2.790\\
\theta & = &\frac{\overline{R}_{\Pi} -6/100}{\sigma_{\Pi}} \sim 1.8265
\end{eqnarray*}


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