Variance property

Question

I know that $$Var(X + Y) = Var(X) + Var(Y) + 2Cov(X,Y)$$

Now if I applied that same rule to $Var(\beta Y + (1-\beta)C)$ would we have $$Var(\beta Y + (1-\beta)C) = \beta^2 Var(Y) + (1-\beta)^{2}Var(C) + 2(1-\beta)Cov(Y,C)$$

I am not sure if this is correct any suggestions are appreciated.

Answer 1

As @Ramil pointed out in the comment section, the correct answer would be:

$$ Var(\beta Y + (1-\beta)C) = \beta^2Var(Y) + (1-\beta)^2Var(C) + 2\beta(1-\beta)Cov(Y, C) $$

I just want to add a tip so that it's easier for you to expand the variance of a sum better in the future. Think of it as similar to the popular property:

$(a + b)^2 = a^2 + b^2 + 2ab$

and it should show you how to treat the constant in the variance case.

Answer 2

You need to check out the definition of variance and covariance; then it will be crystal clear.

However, note that $$ cov(\beta Y, (1-\beta)C) = E(\beta Y - E(\beta Y))((1-\beta)C - E((1-\beta)C)) = \beta(1-\beta)cov(Y,C). $$