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$\phi(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}$ is the pdf of a standard normal distribution.

$\Phi(x) = \int_{- \infty}^x \phi(t) dt$ is the cdf of a standard normal distribution.

How does one calculate the following: $$ T = \int\limits_0^{\infty} \Phi^2(bx) \phi(x) dx $$

In fact, I can find the result in wiki, https://en.wikipedia.org/wiki/List_of_integrals_of_Gaussian_functions#CITEREFOwen1980

But I would like to know how to do it.

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closed as off-topic by RP_, Neil Strickland, Jan-Christoph Schlage-Puchta, Ben McKay, Mateusz Kwaśnicki Jul 22 '18 at 22:17

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  • 1
    $\begingroup$ a closed-form answer will not be forthcoming for any $b$; for $b=1$ one has $T=7/24$. $\endgroup$ – Carlo Beenakker Jul 18 '18 at 18:30
  • $\begingroup$ Mathematica confirms $T=7/24$ for $b=1$ but cannot solve the general case analytically. $\endgroup$ – David G. Stork Jul 18 '18 at 23:23
  • $\begingroup$ In fact, there is closed form solution, please see en.wikipedia.org/wiki/… $\endgroup$ – Magica Jul 19 '18 at 10:15
  • $\begingroup$ did you know there was a closed-form solution when you first posted the question? $\endgroup$ – Carlo Beenakker Jul 19 '18 at 10:43
  • $\begingroup$ no, after posting, I find it in wiki $\endgroup$ – Magica Jul 19 '18 at 11:10
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Maybe this proof is too large, but you can deduce the result you want, sorry for that.

First of all:

  • $\phi(x)= \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$
  • $\Phi(x)=\int_{-\infty}^{x}\phi(t)dt=\frac{1}{2}(1+erf(\frac{x}{\sqrt{2}}))$

To prepare the proof, start with this facts:

  1. $\int_{0}^{\infty}xe^{-x^2a}dx=\frac{1}{2a}$ (Proof by direct integral)
  2. $erf'(x)=\frac{2}{\sqrt{\pi}}e^{-x^2}$ (Proof by definition of $erf$ function)
  3. $$\int_0^{\infty}xe^{-\alpha^2x^2}erf(\beta x)dx=\frac{\beta}{2\alpha^2 \sqrt{\alpha^2+\beta^2}} $$ Hint for 3. Integrating by parts, with \begin{equation}u=erf(\beta x) \rightarrow du=\frac{2\beta}{\sqrt{\pi}}e^{-(x\beta)^2}dx\end{equation} $$dv=xe^{-\alpha^2x^2} \rightarrow v=-\frac{e^{-x^2\alpha^2}}{2\alpha^2}$$ the result follows really easy.

Now, you have all the ingredients to calculate this monster: $$T(b)=\int_{0}^{\infty}\Phi^2(bx)\phi(x)dx$$ Derivating $T(b)$ respect $b$, changing $\Phi(bx)$ by the definition in terms of $erf$ function, and using the fact 2, will lead you to:

$$T'(b)=...=\frac{1}{2\pi}\Big(\int_0^{\infty}{xe^{-x^2(b^2+1)/2}}dx + \int_0^{\infty}{x \cdot erf\big(\frac{bx}{\sqrt{2}}\big)e^{-x^2(b^2+1)/2}}dx \Big) =\frac{1}{2\pi}(A+B)$$

Compute A using the fact 1 to get: $$A=\frac{1}{b^2+1}$$ and compute B using the fact 3 with $\alpha=\sqrt{\frac{b^2+1}{2}}$ and $\beta= \frac{b}{\sqrt{2}}$ to get: $$B=\frac{b}{(b^2+1)\sqrt{2b^2+1}}.$$

It remains to integrate $T'(b)=\frac{A+B}{2\pi}$ respect $b$: \begin{align}T(b)=&\frac{1}{2\pi}\Big(\int_{0}^{b}\frac{1}{t^2+1}dt+\int_{0}^{b}\frac{t}{(t^2+1)\sqrt{2t^2+1}}dt\Big)=\\ =&\frac{1}{2\pi}\Big(arctan(b)+arctan\big(\sqrt{2b^2+1}\big)\Big)+C. \hspace{0.5cm}\square \end{align}

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  • $\begingroup$ Here is a shorter version: First, differentiate under the integral sign with respect to $b$. Second, integrate the result explicitly with respect to $x$: wolframalpha.com/input/…. Third, apply the limits of integration at $x=0$ and $\infty$. Fourth, integrate the result with respect to $b$. $\endgroup$ – Matt F. Jul 19 '18 at 21:19
  • $\begingroup$ Thanks a lot, at last, the integral should be indefinite integral? not from 0 to b? $\endgroup$ – Magica Jul 20 '18 at 11:53
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By a rather circuitous route I get $$T = \dfrac{\arctan\left(\sqrt{2b^2+1}\right) + \arctan(b)}{2\pi}$$

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  • 1
    $\begingroup$ impressive result! $\endgroup$ – Carlo Beenakker Jul 19 '18 at 7:49
  • $\begingroup$ Yeah, it is the right result, but how to obtain? $\endgroup$ – Magica Jul 19 '18 at 10:14
  • 1
    $\begingroup$ I found a linear recurrence for the Maclaurin series of $T(b)$, converted that to a differential equation, and solved the differential equation. $\endgroup$ – Robert Israel Jul 19 '18 at 17:11

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