# Show that $\min_{a \in \mathbb{R}} E \left[ \max \left( (1-a) V, a Z \right) \right]$ is minimized by $a$ such that $0<a<1$

Consider 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? 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}

• can you just do the proofs for contradiction first for a<0 then a>1 separately? why do them at the same time like that Commented Nov 22, 2022 at 3:16
• @sjw Maybe, but I couldn't figure it out. I tried to put my thoughts and what I have tried.
– Boby
Commented Nov 22, 2022 at 3:42
• Isn't $E \left[ \max \left( (1-a) V, a Z \right) \right] = 0$ regardless of $a$? That is, \begin{align}\max \left( (1-a) V, a Z \right) &= \begin{cases} (1-a)V, &(1-a)V \geq aZ \\ aZ, &(1-a)V < aZ\end{cases}\end{align} and so \begin{align}E \left[ \max \left( (1-a) V, a Z \right) \right] &= \begin{cases} E[(1-a)V], &(1-a)V \geq aZ \\ E[aZ], &(1-a)V < aZ\end{cases} \\ &= \begin{cases} (1-a)E[V], &(1-a)V \geq aZ \\ aE[Z], &(1-a)V < aZ\end{cases} \\ &= 0\end{align} since $E[V] = E[Z] = 0$. Commented Nov 24, 2022 at 17:07
• @mhdadk you have a mistake in your second step. These should be conditional expectations. I have added Edit2 which shows that if both $V$ and $Z$ are Gaussian then there is a clear dependence on $a$.
– Boby
Commented Nov 24, 2022 at 18:19

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):

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

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

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

4. 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.

• Thanks. This is really nice proof. I am going to go through it very carefully. I hope you don't mind if I ask you questions. Is this something that you have seen before?
– Boby
Commented Nov 28, 2022 at 18:45
• No, this is a truly original question. I am curious about the circumstances that motivated it.
– whuber
Commented Nov 28, 2022 at 22:30