Show that $\min_{a \in \mathbb{R}} E \left[ \max \left( (1-a) V, a Z \right) \right]$ is minimized by $a$ such that $0Consider two independent random variables  $V$ and $Z$. Assume that $Z$ is standard normal. We only assume that $E[|V|]<\infty$ and $E[V]=0$.
Now consider the following optimization problem:
\begin{align}
\min_{a \in \mathbb{R}} E \left[ \max \left( (1-a) V, a Z \right) \right],
\end{align}
Can we show that the minimizer must be such that $0<a<1$? Or is there a counter-example?
Here is what it tried. Assume that $a<0$ and $|a|<1$ then
\begin{align}
E \left[ \max \left( (1-a) V, a Z \right) \right]&=E \left[ \max \left( (1+|a|) V, -|a| Z \right) \right]\\
&= E \left[ \max \left( (1+|a|) V, |a| Z \right) \right] \text{ using } Z \stackrel{d}{=}-Z\\
& \ge  E \left[ \max \left( (1-|a|) V, |a| Z \right) \right]
\end{align}
where in the last step we used that
\begin{align}
(1+|a|) V \ge  (1-|a|) V
\end{align}
provided that $V>0$. So this clearly is not correct, but I just wanted to show the idea I had.
Edit:  I implicitly assumed that $V$ is not equal to a constant (i.e., zero). Otherwise, the solution is independent of the sign of $a$ as pointed out in one of the answers.
Edit 2: Just to give an example I worked out. That I think is cool. Let $V$ and $Z$ be standard Gaussian.  Then, using that $\max(a,b)=\frac{a+b+|a-b|}{2}$ we have that
\begin{align}
E \left[ \max \left( (1-a) V, a Z \right) \right]= \frac{1}{2}E \left[  \left| (1-a) V- a Z \right| \right]= \frac{1}{2} \sqrt{ \frac{2}{\pi}} \sqrt{ (1-a)^2+a^2}
\end{align}
 A: It is not possible to limit the solution to the open interval $a \in (0,1),$ because when for instance $(X,Y)$ is standard Normal, the global minimum is attained on the entire closed interval $[0,1]$ and, as you can readily compute, when $X$ is a nontrivial mixture of two Normals the global minimum is attained only at $a=0$ or $a=1.$  I will show that all global minima are attained for $0 \le a \le 1.$
The key idea is that when $Z$ is any random variable with finite expectation and distribution function $F_Z,$
$$E[Z] = \int_{-\infty}^0 -F_Z(z)\,\mathrm dz + \int_0^\infty 1 - F_Z(z)\,\mathrm dz.$$
This solution makes (far) weaker assumptions about the variables $X$ and $Y$ than assumed in the question. I will highlight the assumptions needed as we go along.

For any bivariate random variable $(X,Y)$ and real numbers $a,b,t,$ define
$$g_{X,Y}(a,b) = E[\max(aX, bY)].$$
When $a \gt 0,$
$$E[\max(aX, bY)] = E\left[a\max\left(X, \frac{b}{a}Y\right)\right] = a E\left[\max\left(X, \frac{b}{a}Y\right)\right] = a\,g_{X,Y}\left(1,\frac{b}{a}\right).$$
When $X$ and $Y$ are independent with marginal distribution functions $F_X$ and $F_Y,$ and $a$ and $b$ are positive,
$$\begin{aligned}
\Pr(\max(aX, bY) \le t) &= \Pr(aX\le t,\ bY\le t)  \\&= \Pr\left(X\le \frac{t}{a}\right)\Pr\left(Y\le\frac{t}{b}\right) \\&= F_X\left(\frac{t}{a}\right)F_Y\left(\frac{t}{b}\right).
\end{aligned}$$
Thus
$$\begin{aligned}
g_{X,Y}(a,b) &= a\,g_{X,Y}\left(1,\frac{b}{a}\right) \\&= a\left[\int_{-\infty}^0 -F_X\left(t\right)F_Y\left(\frac{at}{b}\right)\mathrm dt + \int_0^\infty \left(1 - F_X\left(t\right)F_Y\left(\frac{at}{b}\right)\right)\mathrm dt\right].
\end{aligned}$$
Provided $g_{X,Y}(a_0,b_0)$ is defined and finite for some $a_0\gt 0$ and $b_0\gt 0,$ $g_{X,Y}(a,b)$ is defined and finite for every $a\gt 0$ and $b \gt 0$ because $\max\left(aX, bY\right) \le \max(a/a_0,b/b_0) \max(a_0X,b_0Y)$ implies $$g_{X,Y}(a,b)\le \max\left(\frac{a}{a_0}, \frac{b}{b_0}\right)\,g_{X,Y}(a_0,b_0) \lt \infty.$$
Comparable relations hold in the other three quadrants (where $a\lt 0$ and $b\gt 0,$ $a\lt 0$ and $b \lt 0$, or $a\gt 0$ and $b\lt 0$).
Assume now that $F_Y$ is everywhere differentiable with derivative $f_y.$ $g$ is differentiable with derivatives $Dg = (D_1 g, D_2 g)$ given by differentiating under the integral signs.  After doing so, substitute $t = by/a$ to obtain
$$\begin{aligned}
a\, D_1 g_{X,Y}(a,b) &= g_{X,Y}(a,b) - b\int F_X\left(\frac{by}{a}\right) y f_Y(y)\,\mathrm dy;\\
b\, D_2 g_{X,Y}(a,b) &= b\int F_X\left(\frac{by}{a}\right) y f_Y(y)\,\mathrm dy.
\end{aligned}$$
Consequently, for $a\gt 0$ and $b \gt 0,$

$$a D_1 g_{X,Y}(a,b) + b D_2 g_{X,Y}(a,b) = g_{X,Y}(a,b).\tag{*}$$

Because
$$g_{X,Y}(a,b) = g_{-X,Y}(-a,b) = g_{X,-Y}(a,-b) = g_{-X,-Y}(-a,-b),$$
relationships analogous to $(*)$ hold in all four quadrants.
It follows that the function $a\to g_{X,Y}(1-a, a)$  is increasing  for $a \gt 1$ and decreasing  for $a \lt 0.$  Here is an illustration where $Y$ has a standard Normal distribution and $X$ is a mixture of Uniform$(-3,-1)$ and Uniform$(0, 1)$ distributions (with zero mean):

This contour plot for the same $(X,Y)$ is characteristic of the situation:

The bold line plots the locus of $(1-a,a).$  When this line exits the first quadrant into the second (upper left) or fourth (lower right) quadrants, its component in the direction of the $(a,b)$ vector field described by $(*)$ is positive: thus, $g$ increases along the rays parameterized by $(a\mid a \gt 1)$ and $(-a\mid a \gt 0).$
The following conclusions are immediate from the foregoing assumptions (namely, that $(X,Y)$ is independent, $Y$ has an absolutely continuous distribution, and there exist points in each of the four $(a,b)$ quadrants where $g$ is finite):

*

*$g$ is continuous everywhere and differentiable on the set $\{(a,b)\mid a\ne 0, b \ne 0.\}$


*$g$ increases in all directions away from the origin.


*Therefore the restriction of $g$ to any line attains at least one global minimum,


*and that minimum occurs in whatever quadrant in which that line is bounded.
Because the line parameterized by $(1-a,a)$ is bounded in the first quadrant and its intersection with that quadrant corresponds to the interval $a\in [0,1],$ we are done.
