# Exercise 5.1

We will use the following properties of the variance:

\begin{align} \newcommand{\Var}{\operatorname{Var}} \newcommand{\Cov}{\operatorname{Cov}} \Var(X + Y) & = \Var(X)+\Var(Y) + 2\Cov(X,Y)\\ \Var(\alpha X) & = \alpha^2 \Var(X) \\ \Cov(\alpha X, \beta Y) & = \alpha \beta \Cov(X,Y) \end{align}

With these properties, we get:

\begin{align} \Var(\alpha X + (1-\alpha) Y) & = \Var(\alpha X) + \Var((1-\alpha)Y) + 2\Cov(\alpha X,(1-\alpha) Y)\\ \ & = \alpha^2 \Var(X) + (1-\alpha)^2 \Var(Y) + 2\alpha\beta\Cov(X,Y) \end{align}

Now, we should find the minimum in the expression above. To do so, we differentiate the expression, equal it to 0 and solve for $\alpha$:

\begin{align} \ 0 & = 2 \alpha\Var(X) - 2(1-\alpha) \Var(Y) + 2(1-2\alpha)\Cov(X,Y) \Leftrightarrow \\ \ \Leftrightarrow 0 & = \alpha\Var(X) - (1-\alpha) \Var(Y) + (1-2\alpha)\Cov(X,Y) \Leftrightarrow \\ \ \Leftrightarrow \alpha\Var(X) + \alpha\Var(Y) - 2\alpha\Cov(X,Y) & = \Var(Y) -\Cov(X,Y) \Leftrightarrow \\ \ \Leftrightarrow \alpha & = \frac{\Var(Y) -\Cov(X,Y)}{\Var(X) + \Var(Y) - 2\Cov(X,Y)} = \frac{\sigma^2_Y - \sigma_{XY}}{\sigma^2_X + \sigma^2_Y - 2\sigma_X\sigma_Y}\\ \end{align}