Show that: $$\int_{y=0}^\infty \frac{8y}{\sigma}\phi\left(\frac{y}{\sigma}\right)\Phi\left(\frac{y}{\sigma}\right)\ dy\ -\int_{y=0}^\infty \frac{4y}{\sigma}\phi\left(\frac{y}{\sigma}\right)\ dy\ =\frac{2}{\sqrt \pi}\sigma $$ where $\phi(.)$ and $\Phi(.)$ i.e. normal pdf and cdf.

I tried by letting $t=\frac{1}{\sigma}$ then $y=t\sigma$ and $dt=\frac{1}{\sigma}dy.$ The equation became $$\int_{0}^\infty 8t\sigma\phi(t)\Phi(t)\ dt\ -\int_{0}^\infty 4t\sigma\phi(t)\ dt\ =\frac{2}{\sqrt \pi}\sigma $$ First I worked with $$\int_{0}^\infty 8t\sigma\phi(t)\Phi(t)\ dt =8\sigma\int_{0}^\infty t\Phi(t)d(\Phi(t))$$

Let $u=t$ $du=dt,$ $dv=\Phi(t)d(\Phi(t))$ $$v=\frac{1}{2}\Phi^2(t)=t\frac{1}{2}\Phi^2(t)\rvert_{0}^\infty-\int_{0}^\infty \frac{1}{2}\Phi^2(t)\ dt\\ =\frac{t}{2}[\Phi^2(\infty)-\Phi^2(0)]-\int_{0}^\infty \frac{1}{2}\Phi^2(t)\ dt$$ But then I'm stuck here. Please help me. Perhaps there's a different way to prove this, either by analytic or numerical. thanks

  • 1
    $\begingroup$ With some cream on top? $\endgroup$
    – Did
    Commented Jan 3, 2018 at 6:56
  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ Commented Jan 3, 2018 at 7:01
  • $\begingroup$ actually i did some effort by trying to use partial integral but still don't get the answer, @JoséCarlosSantos thanks for your suggestions $\endgroup$
    – R.gya
    Commented Jan 3, 2018 at 7:22
  • 1
    $\begingroup$ We can't see the effort if you don't post it. $\endgroup$ Commented Jan 3, 2018 at 11:31

1 Answer 1


0. To be proven: $$\int_{0}^\infty 8t\varphi(t)\Phi(t)dt-\int_{0}^\infty 4t\varphi(t)dt =\frac2{\sqrt \pi} $$ that is,



$$I=\int_{0}^\infty t\varphi(t)\left(\Phi(t)-\tfrac12\right)\ dt$$

1. By definition of $\varphi$ as the PDF of a standard normal random variable $X$, $$I=E\left(X\left(\Phi(X)-\tfrac12\right):X>0\right)$$ 2. For every $x>0$, for some standard normal random variable $Y$ independent of $X$, $$\Phi(x)-\tfrac12=P(x>Y>0)$$ hence $$I=E\left(X:X>Y>0\right)$$ 3. For every $x$, $$x\varphi(x)=-\varphi'(x)$$ hence, for every $y$, $$E(X:X>y)=\int_y^\infty x\varphi(x)dx=\varphi(y)$$ which yields $$I=E(\varphi(Y):Y>0)=\int_0^\infty\varphi^2(y)dy$$ 4. Finally, $$\varphi^2(y)=\frac1{\sqrt{2\pi}}\varphi(\sqrt2 y)$$ hence the change of variable $z=\sqrt2 y$ yields $$I=\frac1{\sqrt{2\pi}}\int_0^\infty\varphi(z)\frac{dz}{\sqrt2}=\frac1{2\sqrt\pi}P(X>0)=\frac1{4\sqrt\pi}$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .