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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
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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
  • $\begingroup$ is there a solution for finite bound on the integral? For example, what if the integration was from 0 to u (for upper bound)? $\endgroup$ – Lewkrr Jan 28 '15 at 20:20
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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.

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