As usual, let $\Phi$ and $\varphi$ denote the cumulative density function and the density function of a standard normal random variable.

On the wiki page "List of integrals of Gaussian functions", I have found an expected value integral involving a standard normal r.v. and its cdf,


for which I do not know how to do the last step in my solution:

My ansatz is to introduce a parameter integral, $I:=I(a)$, and finding its derivative:

$$ \begin{align*} \frac{\partial I}{\partial a}&=\int_{-\infty}^{\infty}x\varphi(x)\varphi(a+bx)dx\\ &=\int_{-\infty}^{\infty}x\frac{e^{-\frac{1}{2}\left(x^2(1+b^2)+2abx+a^2\right)}}{2\pi}dx\\ &=a\frac{e^{-\frac{1}{2}\frac{a^2}{1+b^2}}}{\sqrt{1+b^2}\sqrt{2\pi}}\\ &=a\frac{\varphi\left(\frac{a}{\sqrt{1+b^2}}\right)}{\sqrt{1+b^2}} \end{align*} $$

Integrating the derivative, we obtain:

$$ \begin{align} I&=\int \frac{\partial I}{\partial a}da + C\\ &=\frac{b}{\sqrt{1+b^2}}\varphi\left(\frac{a}{\sqrt{1+b^2}}\right)+C, \end{align} $$

which equals the solution on the wiki page plus a constant term $C$. From here on, I do not know how to get rid of the integration constant, i.e. how to show that $C=0$.

I do know that for $b=0$ it holds that $I=0$. Is this be sufficient to pin down $C$ to zero? Or do I miss something completely?

  • $\begingroup$ I haven't thought much about it, and this goes in the opposite direction of your proof, but would it be fruitful to write $\Phi(x) = \int_{-\infty}^x \phi(y) dy$ so that $I$ is a double integral with an easier (probabilistic) interpretation? $\endgroup$ – user217285 Dec 7 '17 at 9:47
  • $\begingroup$ I have tried this as well, but it did not bring me closer to a solution. $\endgroup$ – Kermittfrog Dec 7 '17 at 12:07
  • $\begingroup$ You can set $b=0$ to get $C=0$. The reason why that is allowed is because the equality must hold for all $b$. So choosing $b=0$ gives you the result. $\endgroup$ – Shashi Dec 9 '17 at 16:05
  • $\begingroup$ @Shashi, thank you very much. I'd consider this an answer. $\endgroup$ – Kermittfrog Dec 11 '17 at 6:27

Apart from parameter integral, we can solve the problem from the perspective of probability theory. Notice that

$$I = \int_{-\infty}^{+\infty}x\Phi(a+bx)\varphi(x)dx = \mathrm{E}[X\Phi(a+bX)]$$

Let's say X and Z are independent standard normal variables, then

$$\Phi(a+bX) = P(Z\leq a+bX|X) = \mathrm{E}(1_{\{Z\leq a+bX\}}|X)$$

and therefor

$$\mathrm{E}[X\Phi(a+bX)] = \mathrm{E}[X\cdot\mathrm{E}(1_{\{Z\leq a+bX\}}|X)]=\mathrm{E}[\mathrm{E}(X\cdot1_{\{Z\leq a+bX\}}|X)]$$

Recall from the Tower Property for conditional expectation that, for any two random varivbles $\xi$ and $\eta$

$$\mathrm{E}\xi = \mathrm{E}[\mathrm{E(\xi|\eta)}]$$

Using the upper equality for $\xi=X\cdot1_{\{Z\leq a+bX\}}$ and $\eta=X$, we obtain that

$$\mathrm{E}(X\Phi(a+bX))=\mathrm{E}(X\cdot1_{\{Z\leq a+bX\}})$$

Next, all we need to do is to calculate the double integral, for convenience, let's say $a>0$ and $b>0$(details as follows) \begin{split} \mathrm{E}(X\cdot1_{\{Z\leq a+bX\}})&=\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}\int_{\frac{z-a}{b}}^{+\infty}\frac{1}{\sqrt{2\pi}}xe^{-\frac{x^2}{2}}dxdz\\ &=\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}\cdot\frac{1}{\sqrt{2\pi}}e^{-\frac{(z-a)^2}{2b^2}}dz\\ &=\frac{b}{\sqrt{1+b^2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\frac{a^2}{(1+b^2)}}\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}\cdot\frac{b}{\sqrt{1+b^2}}}e^{-\frac{1}{2}\frac{(z-\frac{a}{1+b^2})^2}{\frac{b^2}{1+b^2}}}dz\\ &=\frac{b}{\sqrt{1+b^2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}\frac{a^2}{(1+b^2)}}\\ &=\frac{b}{\sqrt{1+b^2}}\varphi(\frac{a}{\sqrt{1+b^2}}) \end{split} Note that, for other values of a and b, the deduction above also applies.

  • $\begingroup$ +1 very elegant solution $\endgroup$ – Aaron Hendrickson Mar 29 at 21:53

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