I was making some computation and I've been stucked in this one.

Let $B$ and $b$ be positive contant. We call $\phi(x)$ standard gaussian distribution and $\Phi(x)$ its cumulative function, i.e. $$ \phi(x) = \frac{1}{\sqrt{2 \pi}}e^{-\frac{x^2}{2}} $$ $$ \Phi(x) = \int_{-\infty}^x \phi(s) ds $$

then compute

$$ \int_{-\infty}^{+\infty}x\Phi(x)\phi(Bx - b) dx $$

If it helps I can proove this result: $$ \int_{-\infty}^{+\infty}x\Phi(x)\phi(x) dx = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-x^2}= \frac{1}{\sqrt{2}} $$

Any suggestion or hint will be appreciated, thank you!



  • $\begingroup$ Is the last integral correct? I'm getting $\int_{-\infty}^{\infty}x\phi(x)\Phi(x)dx=\frac{1}{2\sqrt{\pi}}$. $\endgroup$ – Anne Dec 16 '17 at 5:03
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    $\begingroup$ @Anne I'm sorry, you're right... I've edited the question. Thanks! $\endgroup$ – clarkmaio Dec 16 '17 at 10:02
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    $\begingroup$ the tittle is a clear (+1) $\endgroup$ – tired Dec 16 '17 at 17:10
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    $\begingroup$ Marketing for matematicians...i know something about it XD $\endgroup$ – clarkmaio Dec 16 '17 at 17:11
  • $\begingroup$ btw: i encourage you to scan this paper: nvlpubs.nist.gov/nistpubs/jres/73B/jresv73Bn1p1_A1b.pdf $\endgroup$ – tired Dec 16 '17 at 17:12

We have: \begin{eqnarray} I:= \int\limits_{{\mathbb R}} x \Phi(x) \phi(B x-b) dx = \int\limits_{{\mathbb R}} (\frac{y+b}{B}) \Phi(\frac{y+b}{B}) \phi(y) \frac{dy}{B} \end{eqnarray} Now since $\phi(y)^{'} = -y \phi(y)$ we can treat one of the resulting integrals with integration by parts. We have: \begin{eqnarray} \int\limits_{{\mathbb R}} \underbrace{y \phi(y)}_{-\phi^{'}(y)} \Phi(\frac{y+b}{B}) dy=\int\limits_{\mathbb R} \phi(y) \phi(\frac{y+b}{B}) \frac{1}{B} dy = \phi(\frac{b}{\sqrt{1+B^2}}) \frac{1}{\sqrt{1+B^2}} \end{eqnarray} where the last step is done by completing the integrand to square. The second integral is treated by differentiating it with respect to the $b$ parameter. We have: \begin{eqnarray} f(b)&:=& \int\limits_{\mathbb R} \phi(y) \Phi(\frac{y+b}{B})dy \quad \Rightarrow \quad f^{'}(b) = \int\limits_{\mathbb R} \phi(y) \phi(\frac{y+b}{B}) \frac{1}{B} dy = \phi(\frac{b}{\sqrt{1+B^2}}) \frac{1}{\sqrt{1+B^2}}\\ &\Rightarrow & f(b) = \Phi(\frac{b}{\sqrt{1+B^2}}) \end{eqnarray} where we used the boundary condition $f(-\infty)=0$. Bringing all together we have: \begin{equation} I = \frac{1}{B^2} \left[\phi(\frac{b}{\sqrt{1+B^2}}) \frac{1}{\sqrt{1+B^2}}+b \Phi(\frac{b}{\sqrt{1+B^2}}) \right] \end{equation}

  • $\begingroup$ Nice idea! Thx!! $\endgroup$ – clarkmaio May 18 '18 at 6:48

We have $$ \int_{- \infty}^x ds \ \phi(s) = \int_{- \infty}^0 ds \ \phi(s + x) \ .$$ Using Fubini's theorem, we have $$ \int_{-\infty}^\infty dx \ x \Phi(x) \phi(Bx - b) = \int_{- \infty}^0 ds \int_{-\infty}^\infty dx \ x \phi(s + x) \phi(Bx + b) \ .$$ Computing the inner integral, we have $$ \int_{-\infty}^\infty dx \ x \phi(s + x) \phi(Bx + b) = \frac{(bB - s)e^{- \frac{(Bs + b)^2}{2(1 + B^2)}} }{\sqrt{2 \pi}(1 + B^2)^\frac{3}{2}} \ .$$ We have $$ \int_{- \infty}^0 ds \int_{-\infty}^\infty dx \ x \phi(s + x) \phi(Bx + b) = \int_{0}^\infty ds \ \frac{(bB + s)e^{- \frac{(- Bs + b)^2}{2(1 + B^2)}} }{\sqrt{2 \pi}(1 + B^2)^\frac{3}{2}} \ .$$ From here, I suppose one has a representation in terms of the error function. I do not see any other way to simplify this expression.

If we set $B = 1$ and $b = 0$, then the value of this integral agrees with the value that Anne calculated in the comments.


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