Signing a Derivative of an Expectation

Question

Let $$Z=E[F(x)]=\int_{-\infty}^{\infty} F(x) \frac{1}{y} \phi\left(\frac{x-a}{y}\right)\,{\rm d}x$$

where $F(x)$ is a convex function of $x$, $\phi$ is the standard normal PDF and $a$ is some (finite) constant.

I want to show that $$\frac{\partial Z}{\partial y}>0$$

Here's my proof so far:

$$\frac{\partial Z}{\partial y}=\int_{-\infty}^{\infty} F(x) \frac{1}{y^2} \phi\left(\frac{x-a}{y}\right)\left(\frac{(x-a)^2}{y^2}-1\right)\,{\rm d}x$$

The idea of the rest of the proof I have in my mind is that since $\phi(x)=\phi(-x)$ and F is convex, it should be that any increase in the derivative for $x>0$ is greater than a corresponding value for $x<0$. [For example, I'm trying to get something analogous to $\int_{0}^{\infty} (F(x)-F(-x)) \frac{1}{y^2} \phi\left(\frac{x}{y}\right)\,{\rm d}x>0$] But, I'm unable to complete the proof.

Answer 1

With $u=(x-a)/y$, we have

$$ \int_{-\infty}^{\infty} F(x) \frac{1}{y} \phi\left(\frac{x-a}{y}\right)\,\mathrm dx = \int_{-\infty}^{\infty} F(yu+a)\phi(u)\,\mathrm du\;, $$

and

$$ \frac\partial{\partial y}\int_{-\infty}^{\infty} F(yu+a)\phi(u)\,\mathrm du =\int_{-\infty}^{\infty}F'(yu+a)u\phi(u)\,\mathrm du\;. $$

Now integrate by parts; the boundary term vanishes, $F''$ is positive and the indefinite integral of $u\phi(u)$ is negative (proportional to $\phi(u)$) and thus cancels the sign from the integration by parts.

The derivative $\frac{\partial Z}{\partial y}$ in the question seems to contain several errors.

Answer 2

Substituting $u=(x-a)/y$ and differentiating:

$$\frac{\partial Z}{\partial y}=\frac\partial{\partial y}\int_{-\infty}^{\infty} F(yu+a)\phi(u) du=\int_{-\infty}^{\infty}F'(yu+a)u\phi(u)=\int_0^{\infty}(F'(yu+a)-F'(-yu+a))u\phi(u)du$$ Note that $F'$ is non-decreasing to conclude $\frac{\partial Z}{\partial y}\geq 0$. Equality is achieved iff $F$ is linear.