How to compute the following formula?

$$ \int_{-\infty}^{+\infty} \Phi(x) N(x\mid\mu,\sigma^2) \, dx $$

$$ \int_{-\infty}^{+\infty} \Phi(x) N(x\mid\mu,\sigma^2) x\,dx $$

where $\Phi(x)=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}} \exp{(-t^2/2)} \, dt$, namely, the cumulative distribution function of normal distribution $N(0,1)$.
$N(x\mid\mu,\sigma^2) $ means the probability density function of Gaussian distribution with mean $\mu$ and variance $\sigma^2$

  • $\begingroup$ Thank you very much. It's my mistake and I have fixed it. $\endgroup$ – tankeco Oct 25 '13 at 17:31
  • $\begingroup$ math.stackexchange.com/questions/74770/… I found the answer of my first formula... $\endgroup$ – tankeco Oct 27 '13 at 14:37
  • $\begingroup$ I have solved the second formula, using $\varphi(x)'=-x\varphi(x)$ $\endgroup$ – tankeco Oct 28 '13 at 2:32

The first integral is $$I_{\mu,\sigma^2}=\int_\mathbb R\Phi(x)\varphi_{\mu,\sigma^2}(x)dx=E(\Phi(\mu+\sigma X))$$ that is, $$I_{\mu,\sigma^2}=P(Y\leqslant \mu+\sigma X)=P(Z\leqslant\nu)=\Phi(\nu)$$ where $(X,Y)$ is i.i.d. standard normal, $$\nu=\frac{\mu}{\sqrt{\sigma^2+1}}$$ and $Z$ is also standard normal since $$Z=\frac{Y-\sigma X}{\sqrt{\sigma^2+1}}$$ Thus,

$$I_{\mu,\sigma^2}=\int_\mathbb R\Phi(x)\varphi_{\mu,\sigma^2}(x)dx=\Phi\left(\frac{\mu}{\sqrt{\sigma^2+1}}\right)$$

The second integral is $$J_{\mu,\sigma^2}=\int_\mathbb R\Phi(x)\varphi_{\mu,\sigma^2}(x)xdx=\int_\mathbb R\Phi(\mu+\sigma x)\varphi(x)xdx$$ Since $x\varphi(x)=-\varphi'(x)$, an integration by parts yields $$J_{\mu,\sigma^2}=\int_\mathbb R\sigma\varphi(\mu+\sigma x)\varphi(x)dx$$ Now, there exists $(\tau,\kappa,\lambda)$ such that $$x^2+(\mu+\sigma x)^2=(\tau x+\kappa)^2+\lambda$$ hence $$J_{\mu,\sigma^2}=\frac\sigma{\sqrt{2\pi}}e^{-\lambda/2}\int_\mathbb R\varphi(\kappa+\tau x)dx=\frac\sigma{\sqrt{2\pi}}e^{-\lambda/2}\frac1{\tau}$$ By identification, $$\tau^2=1+\sigma^2\qquad\kappa=\frac{\mu\sigma}\tau\qquad\lambda=\mu^2-\kappa^2=\frac{\mu^2}{1+\sigma^2}$$ hence

$$J_{\mu,\sigma^2}=\frac{\sigma e^{-\mu^2/(2(1+\sigma^2))}}{\sqrt{2\pi(1+\sigma^2)}}$$

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  • $\begingroup$ things are flipped on 2nd line $\endgroup$ – air May 25 '18 at 6:00
  • 1
    $\begingroup$ @air Quite so. Thanks. $\endgroup$ – Did May 25 '18 at 15:34

The first integral is solved in Distribution of the normal cdf.

So we know that $$ \int_{-\infty}^{\infty} \Phi \left (x \right ) N \left ( x|\mu,\sigma^2 \right ) dx = \Phi \left (\frac{\mu}{\sqrt{1+\sigma^2}} \right ) $$

Now, if we differentiate both sides w.r.t. $\mu$, $$\int_{-\infty}^{\infty} \frac{x-\mu}{\sigma^2} \Phi \left (x \right ) N \left ( x|\mu,\sigma^2 \right ) dx = \frac{1}{\sqrt{1+\sigma^2}} N \left (\frac{\mu}{\sqrt{1+\sigma^2}} \right ) $$

Notice the first term on the left hand side is what we want. Rearranging, $$\int_{-\infty}^{\infty} \Phi \left (x \right ) N \left ( x|\mu,\sigma^2 \right ) x dx = \mu \Phi \left ( \frac{\mu}{\sqrt{1+\sigma^2}} \right ) + \frac{\sigma^2}{\sqrt{1+\sigma^2}} N \left (\frac{\mu}{\sqrt{1+\sigma^2}} \right ) $$

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