$\newcommand{\Var}{\operatorname{Var}}$I'm currently trying to solve/complete a statistics project. So far, I have finished the first part of the project. However, one of the tasks in the second part is causing me trouble.
I'm given a random variable, let's call it $C_1$, which describes a portfolio consisting of $1$ and $2$.
$C_1$ is defined as as the following: $C_1 = \alpha X_1 + (1 – \alpha) X_2$.
I then had to define an expression for the variance of $C_1$, i.e. $\Var(C_1)$. Considering the definition of $C1$, that means I had to define an expression for $\Var(α X_1 + (1 – α) X_2)$.
So according to my book, I ended up with the following:
$$\Var(C1) = \Var(α X_1 + (1 – α) X_2) = α^2 \Var(X_1) + (1 – α)^2 \Var(X_2).$$
I don't know if my expression is correct, but that is how I think it should be according to the book.
Now, my task is to express $\Var(C_1$) as a function of $α$, more specifically $V(α)$. I'm pretty sure $V$ is just an abbreviation for variance. So I set up the following:
$$V(α) = α^2 \Var(X_1) + (1 – α)^2 \Var(X_2).$$
Now, according to the project, I have to insert the values for $\Var(X_1), \Var(X_2)$ and $\operatorname{Cov}(X_1, X_2)$ into the equation. I have calculated these values in the first part of the project and it was pretty easy to insert the values for $\Var(X_1)$ and $\Var(X_2)$ into the equation. However, I can't figure out what is meant by inserting the covariance of $X_1$ and $X_2$ into it. I have calculated the covariance of $X_1$ and $X_2$, but don't know where to insert it into the equation. I have been looking at this for days now and re-read a lot of the pages in my book, but I'm stuck.
I know you haven't seen the project tasks, but maybe someone could point out what's going wrong.
Thank you!
\begin{align} & \operatorname{var}(\alpha X_1 + (1-\alpha)X_2) \\[10pt] = {} & \operatorname{var}(\alpha X_1) + \operatorname{var}((1-\alpha)X_2) + 2\operatorname{cov}(\alpha X_1, (1-\alpha)X_2) \\[10pt] = {} & \alpha^2 \operatorname{var}(X_1) + (1-\alpha)^2 \operatorname{var}(X_2) + 2\alpha(1-\alpha)\operatorname{cov}(X_1,X_2). \end{align}
You don't need the last term if the covariance is $0$, and that happens if (but not only if) the two random variables being added are independent.