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

How can one calculate the integral:

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

  • 5
    $\begingroup$ 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).

  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.