Minimizing the variance of weighted sum of two random variables with respect to the weights 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!
 A: Yes, of course.  For example, try $X = t Y$ where $t > 1$.
A: $$\mathrm{Var}(wX+(1-w)Y) = w^2 (\mathrm{Var} X - 2\mathrm{Cov}(X,Y) + \mathrm{Var}Y) + 2w(\mathrm{Cov}(X,Y) - \mathrm{Var}Y) + \mathrm{Var}Y$$ is a quadratic function of $w$ whose minimum value occurs, as you noted,  at 
$$-\frac{\mathrm{Cov}(X,Y) - \mathrm{Var}Y}{\mathrm{Var} X - 2\mathrm{Cov}(X,Y) + \mathrm{Var}Y}$$
which is negative if $\mathrm{Cov}(X,Y) > \mathrm{Var}Y.$
Note that $\mathrm{Var}(wX+(1-w)Y)$ has value $\mathrm{Var}Y$ 
at $w = 0$ and value 
$\mathrm{Var} X$ at $w = 1$.
Suppose $\mathrm{Var} X > \mathrm{Var} Y$.  Then
$w^*$ must be smaller than $1$.  In fact, 
$$w^* < 0 ~~\mathrm{if}~~ \frac{\mathrm d}{\mathrm dw}\mathrm{Var}(wX+(1-w)Y)\biggr|_{w=0} > 0,
~~\mathrm{i.e.\ if}~\mathrm{Cov}(X,Y) - \mathrm{Var}Y > 0.$$ 
A similar analysis can be done to figure out the conditions
under which $w^* > 1$.
So, when $\mathrm{Var} X > \mathrm{Var} Y$, what does it mean to
have $\mathrm{Cov}(X,Y) > \mathrm{Var}Y$?  Equivalent ways of
saying the same thing are that $(X-Y)$ and $Y$ are positively correlated
or that 
$$\rho_{X,Y} > \sqrt{\frac{\mathrm{Var} Y}{\mathrm{Var} X}}.$$
The minimum value of $\mathrm{Var}(wX+(1-w)Y)$ is positive
unless $Y = aX+b$ for some real numbers $a$ and $b$ in which case
the minimum value of variance is $0$.
