Variance of the difference of two iid sample means Let $X_{1}, ..., X_{n}$ be random variables independent of $Y_{1}, ..., Y_{n}$, where both groups are iid with associated population means $\mu_{1}$ and $\mu_{2}$ and population variances $\sigma_{1}^{2}$ and $\sigma_{2}^{2}$, respectively.
Let $\bar{X}$ and $\bar{Y}$ be their sample means.
What is the variance of $\bar{X} - \bar{Y}$?
Given what I understand about the properties of variances and means, I have arrived at $(1/n^{2})Var(nX-nY)$, but I am not sure how to proceed from this point.
I am aware of the variance property $Var(X + Y) = Var(X) + Var(Y) + 2Cov(X,Y)$, but I am not sure how to relate a subtraction of two random variables to this situation. Is it as simple as affixing a negative sign to the entire equation?

As you note, $Var(X + Y) = Var(X) + Var(Y) + 2 Cov(X, Y)$. The variance of the difference of two means just involves a negative covariance term, namely $Var(X - Y) = Var(X) + Var(Y) - 2 Cov(X, Y)$.
