# How to prove the variance of this function of a normal random variable is decreasing?

Partial answers found here and here.

Suppose $$\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$ and $$\Phi(x)=\int_{-\infty}^x\phi(t)dt=\frac{1}2 \text{Erfc}[-\frac{x}{\sqrt{2}}]$$ where Erfc is the complementary error function. These are the density and distribution function of a standard normal random variable.

Define the function $$g(\mu)=\int_{-\infty}^{\infty}\Phi(x) \phi(x-\mu)dx$$.
$$g(\mu)$$ is the expected value of $$\Phi(X)$$ when $$X$$ has a normal distribution with mean $$\mu$$ and variance 1.
Differentiate under the integral to observe that $$g'(\mu)=\int_{-\infty}^{\infty}\Phi(x)(x-\mu) \phi(x-\mu)dx=\frac{\phi\left(\frac{\mu}{\sqrt{2}}\right)}{\sqrt{2}}$$
Since $$g'(\mu)$$ is always positive, $$g(\mu)$$ is an increasing function.

Furthermore, define the function $$h(\mu)=\int_{-\infty}^{\infty}(\Phi(x)-g(\mu))^2 \phi(x-\mu)dx$$.
$$h(\mu)$$ is the variance of $$\Phi(X)$$ when $$X$$ has a normal distribution with mean $$\mu$$ and variance 1.

I can shown that:
$$h(0)=\frac{1}{12}$$
$$h'(0)=0$$
$$h''(0)=-\frac{1}{2 \pi}+\frac{1}{2 \sqrt{3}\pi} \approx -0.06727$$
$$\lim_{\mu\rightarrow \infty}h(\mu)=0$$
$$h(\mu)=h(-\mu)$$ so that $$h$$ is an even function.

I can draw $$h(x)$$ for $$x \in [0,4]$$ I can also find all the terms of the Taylor series for $$h(\mu)$$ at $$\mu=0$$. But, I cannot show the derivative is never $$0$$ for $$\mu \ne 0$$. Is it possible to prove that $$h(\mu)$$ is decreasing for $$\mu>0$$?

• Your graph suggests that the function goes from concave to convex and I believe this implies that the derivative needs to be zero at some point. – Patricio Feb 6 at 15:06
• @Patricio The second derivative is 0 where it changes from convex to concave. – John L Feb 6 at 15:08
• Using $g'(\mu)$, you can find that $g(\mu) = \frac{1+\text{erf}\left( \frac{\mu}{2} \right)}{2}$. And numerically, Mathematica says that $h(0) \approx 0.00356 \not=\frac{1}{12}$. I might have made a mistake, but I don't think so. Can you show your work for evaluating $h(0)$? – Varun Vejalla Feb 11 at 20:42
• @VarunVejalla NIntegrate[(CDF[NormalDistribution[0, 1], x] - 1/2)^2 PDF[ NormalDistribution[0, 1], x], {x, -Infinity, Infinity}] – John L Feb 12 at 16:07
• You have $(\Phi(x)-g(x))^2$ in your question. Did you mean $(\Phi(x)-g(\mu))^2$? – Varun Vejalla Feb 12 at 16:13

Computing the derivative of $$h$$ we can readily see that

$$h'(\mu)=2\int_{-\infty}^{\infty}dx~\Phi(x+\mu)\phi(x+\mu)\phi(x)-2g(\mu)g'(\mu)$$

However it is true that

$$\phi(x)\phi(x+\mu)=\phi(\frac{\mu}{\sqrt{2}})\phi(x\sqrt{2}+\frac{\mu}{\sqrt{2}})$$

whereupon we can simplify the above integral to the form

$$h'(\mu)=2g'(\mu)\left(\int_{-\infty}^{\infty}dt~\Phi(\frac{t}{\sqrt{2}}+\mu)\phi(\frac{\mu}{\sqrt{2}}+t)-g(\mu)\right)\equiv2g'(\mu)\Lambda(\mu)$$

It suffices to show that $$\Lambda(\mu)\leq 0$$ for $$\mu\geq 0$$. We know that $$\Lambda(0)=0, \Lambda(\infty)=0$$ and we compute (the derivation can be done similarly as above)

$$\Lambda'(\mu)=\frac{\phi(\mu/\sqrt{6})}{\sqrt{6}}-\frac{\phi(\mu/\sqrt{2})}{\sqrt{2}}$$

It is a trivial exercise in algebra to find the sign of this function and in fact we see that $$\Lambda$$ possesses a single minimum at $$\mu=\sqrt{3\ln3}$$ and a maximum at $$\mu=-\sqrt{3\ln3}$$. We are only interested in positive values however, so from the above data we conclude that $$\sup_{x\geq 0}\Lambda(x)=0$$ and thus for all $$x\geq 0$$ the result follows.

We conclude that $$h$$ is decreasing on the positive x-axis.