Suppose $\phi(\cdot)$ and $\Phi(\cdot)$ are density function and distribution function of the standard normal distribution.

How can one calculate the integral:

$$\int^{\infty}_{-\infty}\Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw$$

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    $\begingroup$ This is all fine. An early reference to a more general result which includes this one is Ellison (1964, J.Am.Stat.Assoc, 59, 89-95); see Corollary 1 of Theorem 2. $\endgroup$ – user27178 Jun 22 '13 at 12:31

A more conventional notation is

$$y(\mu, \sigma) = \int\Phi\left(\frac{x-\mu}{\sigma}\right)\phi(x) dx = \Phi\left(\frac{-\mu}{\sqrt{1+\sigma^2}}\right).$$

This can be found by differentiating the integral with respect to $\mu$ and $\sigma$, producing elementary integrals which can be expressed in closed form:

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

$$\frac{\partial y}{\partial \sigma}(\mu, \sigma) = \frac{\mu\sigma }{\sqrt{2 \pi } \left(\sigma ^2+1\right)^{3/2}}e^{-\frac{1}{2}\frac{\mu ^2}{\sigma ^2+1}}.$$

This system can be integrated, beginning with the initial condition $y(0,1)$ = $\int\Phi(x)\phi(x)dx$ = $1/2$, to obtain the given solution (which is easily checked by differentiation).

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    $\begingroup$ I double-checked the answer via numeric integration and contouring the ratios for $-2 \le \mu \le 2$, $0 \lt \sigma \le 2$: agreement was to eleven significant figures throughout this range. $\endgroup$ – whuber Jun 6 '13 at 19:02
  • $\begingroup$ wow, clever solution. $\endgroup$ – Cam.Davidson.Pilon Jun 6 '13 at 19:03
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    $\begingroup$ I think this one can be done almost by inspection. The first term under the integral is a uniform[0,1] random variable. Since the normal pdf is symmetric, the integral should be $1 \over 2$ $\endgroup$ – soakley Jun 6 '13 at 19:04
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    $\begingroup$ @soakley Your approach works for $y(0,1)$, but it's not clear how it would apply to other arguments of $y$. $\endgroup$ – whuber Jun 6 '13 at 19:05
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    $\begingroup$ @whuber Sorry for not understanding, but once we have the two closed forms for the derivative and the initial condition, how do we go from there to the final solution? In other words, what did you do with the closed form expressions for the derivatives and the initial condition? $\endgroup$ – user106860 Nov 7 '18 at 4:30

Let $X$ and $Y$ be independent normal random variables with $X \sim N(a,b^2)$ and $Y$ a standard normal random variable. Then, $$P\{X \leq Y \mid Y = w\} = P\{X \leq w\} = \Phi\left(\frac{w-a}{b}\right).$$ So, using the law of total probability, we get that $$P\{X \leq Y\} = \int_{-\infty}^\infty P\{X \leq Y \mid Y = w\}\phi(w)\,\mathrm dw = \int_{-\infty}^\infty \Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw.$$ Now, $P\{X \leq Y\} = P\{X-Y \leq 0\}$ can be expressed in terms of $\Phi(\cdot)$ by noting that $X-Y \sim N(a,b^2+1)$, and thus we get $$\int_{-\infty}^\infty \Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw = \Phi\left(\frac{-a}{\sqrt{b^2+1}}\right)$$ which is the same as the result in whuber's answer.

  • $\begingroup$ Just wanted to mention this was generalized to the multivariate case in Lemma 2.1 of "On fundamental skew distributions" by Arellano-Valle, Genton $\endgroup$ – bringingdownthegauss May 11 '20 at 18:50
  • $\begingroup$ @ Dilip Sarwate: Would you mind to tell how did you get the second expression of P{X≤Y}=∫∞−∞P{X≤Y∣Y=w}ϕ(w)dw ? To my understanding, the integrand of the integral should be the product of P{X≤Y∣Y=w} with the pdf of w, not with ϕ(w). Am I wrong ? $\endgroup$ – user295357 Sep 6 '20 at 19:59
  • $\begingroup$ @user295357 $w$ is a real number and does not have a pdf. I am multiplying the conditional probability $P\{X \leq Y\mid Y=w\}$ by the "probability" $\phi(w)$ of the conditioning event $\{Y=w\}$ and summing these by integrating. Remember that an integral is basically a sum, and remember also that the total probability law is the sum $$P(A)=\sum_i P(A\mid B_i)P(B_i).$$ All I am doing is using the continuous version of the law of total probability. Compare $$f_X(x)=\int f_{X\mid Y}(x\mid y)f_Y(y) \mathrm dy$$ which you can understand by noting that $ f_{X\mid Y}(x\mid y)f_Y(y)=f_{X,Y}(x,y)$ $\endgroup$ – Dilip Sarwate Sep 6 '20 at 21:32
  • $\begingroup$ @ Dilip Sarwate: Thank for your reply. I understand that if Y (as well as w) is a discrete RV, then the integrand should be the product of P{X≤Y∣Y=w} with the probability of w. However, here Y is a continuous random variable of normal distribution, so I think the integrand should be the product of P{X≤Y∣Y=w} with the pdf of w. Furthermore, even you consider Y (as well as w) as a discrete RV, is there any reason to assume the probability that Y=w being ϕ(w) ? $\endgroup$ – user295357 Sep 6 '20 at 23:25
  • $\begingroup$ Dilip Sarwate's elegant answer only works because of the independence between $X$ and $Y$. Otherwise we don't have $P(X \le Y | Y=w) \ne P(X \le w)$ $\endgroup$ – user67724 Mar 26 at 21:43

Here is another solution: We define \begin{align*} I(\gamma) & =\int_{-\infty}^{\infty}\Phi(\xi x+\gamma)\mathcal{N}(x|0,\sigma^{2})dx, \end{align*} which we can evaluate $\gamma=-\xi\mu$ to obtain our desired expression. We know at least one function value of $I(\gamma)$, e.g., $I(0)=0$ due to symmetry. We take the derivative wrt to $\gamma$ \begin{align*} \frac{dI}{d\gamma} & =\int_{-\infty}^{\infty}\mathcal{N}((\xi x+\gamma)|0,1)\mathcal{N}(x|0,\sigma^{2})dx\\ & =\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{1}{2}\left(\xi x+\gamma\right)^{2}\right)\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left(-\frac{x^{2}}{2\sigma^{2}}\right)dx. \end{align*} and complete the square \begin{align*} \left(\xi x+\gamma\right)^{2}+\frac{x^{2}}{\sigma^{2}} & =\underbrace{\left(\xi^{2}+\sigma^{-2}\right)}_{=a}x^{2}+\underbrace{-2\gamma\xi}_{=b}x+\underbrace{\gamma^{2}}_{=c} \\ &=a\left(x-\frac{b}{2a}\right)^{2}+\left(c-\frac{b^{2}}{4a}\right) \left(c-\frac{b^{2}}{4a}\right)\\ & =\gamma^{2}-\frac{4\gamma^{2}\xi^{2}}{4\left(\xi^{2}+\sigma^{-2}\right)}\\ &=\gamma^{2}\left(1-\frac{\xi^{2}}{\xi^{2}+\sigma^{-2}}\right)\\ &=\gamma^{2}\left(\frac{1}{1+\xi^{2}\sigma^{2}}\right) \end{align*} Thus, \begin{align*} \frac{dI}{d\gamma} & =\frac{1}{2\pi\sigma}\exp\left(-\frac{1}{2}\left(c-\frac{b^{2}}{4a}\right)\right)\sqrt{\frac{2\pi}{a}}\int_{-\infty}^{\infty}\sqrt{\frac{a}{2\pi}}\exp\left(-\frac{1}{2}a\left(x-\frac{b}{2a}\right)^{2}\right)dx\\ & =\frac{1}{2\pi\sigma}\exp\left(-\frac{1}{2}\left(c-\frac{b^{2}}{4a}\right)\right)\sqrt{\frac{2\pi}{a}}\\ &=\frac{1}{\sqrt{2\pi\sigma^{2}a}}\exp\left(-\frac{1}{2}\left(c-\frac{b^{2}}{4a}\right)\right)\\ & =\frac{1}{\sqrt{2\pi\left(1+\sigma^{2}\xi^{2}\right)}}\exp\left(-\frac{1}{2}\frac{\gamma^{2}}{1+\xi^{2}\sigma^{2}}\right) \end{align*} and integration yields

$$ \begin{align*} I(\gamma) &=\int_{-\infty}^{\gamma}\frac{1}{\sqrt{2\pi\left(1+\sigma^{2}\xi^{2}\right)}}\exp\left(-\frac{1}{2}\frac{z^{2}}{1+\xi^{2}\sigma^{2}}\right)dz\\ &=\Phi\left(\frac{\gamma}{\sqrt{1+\xi^{2}\sigma^{2}}}\right) \end{align*} $$

which implies

$$ \begin{align*} \int_{-\infty}^{\infty}\Phi(\xi x)\mathcal{N}(x|\mu,\sigma^{2})dx &=I(\xi\mu)\\ &=\Phi\left(\frac{\xi\mu}{\sqrt{1+\xi^{2}\sigma^{2}}}\right). \end{align*} $$


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