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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 $x \in [0,4]$

enter image description here

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 $\mu>0$?

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  • $\begingroup$ 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. $\endgroup$ – Patricio Feb 6 at 15:06
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    $\begingroup$ @Patricio The second derivative is 0 where it changes from convex to concave. $\endgroup$ – John L Feb 6 at 15:08
  • $\begingroup$ 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 $h(0)$? $\endgroup$ – Varun Vejalla Feb 11 at 20:42
  • $\begingroup$ @VarunVejalla NIntegrate[(CDF[NormalDistribution[0, 1], x] - 1/2)^2 PDF[ NormalDistribution[0, 1], x], {x, -Infinity, Infinity}] $\endgroup$ – John L Feb 12 at 16:07
  • $\begingroup$ You have $(\Phi(x)-g(x))^2$ in your question. Did you mean $(\Phi(x)-g(\mu))^2$? $\endgroup$ – Varun Vejalla Feb 12 at 16:13
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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.

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