Suppose $X$ and $Y$ are two random variables. I would like to see if the solution to $ \min_w \quad \mathrm{Var}(wX+(1-w)Y) $ can be negative.
I know that \begin{align*} &\mathrm{Var}(wX+(1-w)Y) \\ &= w^2 \mathrm{Var} X + 2w(1-w)\mathrm{Cov}(X,Y) + (1-w)^2 \mathrm{Var}Y \\&= w^2 (\mathrm{Var} X - 2\mathrm{Cov}(X,Y) + \mathrm{Var}Y) + 2w(\mathrm{Cov}(X,Y) - \mathrm{Var}Y) + \mathrm{Var}Y \end{align*} Since $\mathrm{Var} X - 2\mathrm{Cov}(X,Y) + \mathrm{Var}Y \geq \mathrm{Var} X - 2\sqrt{\mathrm{Var} X \, \mathrm{Var} Y}+ \mathrm{Var}Y \geq 0, $ the minimizer is $ w^*=-\frac{\mathrm{Cov}(X,Y) - \mathrm{Var}Y}{\mathrm{Var} X - 2\mathrm{Cov}(X,Y) + \mathrm{Var}Y} $
So if I am correct so far, the problem of whether $w^*$ can be negative becomes whether it can be true that $ \mathrm{Cov}(X,Y) - \mathrm{Var}Y > 0? $
Thanks!