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Suppose $\phi(\cdot)$ and $\Phi(\cdot)$ are the density function and cumulative distribution function of the standard normal distribution. $a<b$ are finite real numbers.

How can I calculate the integral:

$$\int^{b}_{a}\Phi\left(\frac{x-\mu}{\sigma}\right)\phi(x)\,\mathrm dx$$

Note that the question is similar to this one. There, the comments to the accepted answer suggest that the solution can be generalized to definite integrals, but I haven't been able to figure it out.

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    $\begingroup$ It would be worthwhile for you to actually prove the result in your linked stats.stackexchange page. $\endgroup$
    – TenaliRaman
    Commented Dec 14, 2018 at 11:35
  • $\begingroup$ I went through the proof and tried extending it along several lines but failed. One complication I ran into is that the derivatives with respect to $\mu$ and $\sigma$ feature truncation terms that also depend on $\mu$ and $\sigma$, so it isn't clear to me how to do the integration in the next step. In the comments it was also suggested that all I need to do is to change the starting point of the integration, but i don't really see how. Sorry if I'm missing anything obvious here, but can you provide a bit more details how to do the generalization? $\endgroup$
    – Bob
    Commented Dec 14, 2018 at 11:40
  • $\begingroup$ Because a clear, correct answer has appeared in a new thread, I have elected to close this one--which has no adequate answer--rather than close the new one. $\endgroup$
    – whuber ♦
    Commented Dec 1, 2020 at 16:03

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Partial answer. If I have some time, I will work on tightening up the result.

In the other (non accepted) answer to the question cited by the OP, it is shown that if $X\sim N(\mu,\sigma^2)$ and $Y\sim N(0,1)$ are independent normal random variables, then \begin{align} P\{X \leq Y \mid Y = w\} &= P\{X \leq w \mid Y = w\}\\ &= \int_{-\infty}^w f_{X\mid Y}(t\mid Y=w) \,\mathrm dt\\ &= \int_{-\infty}^w f_{X}(t) \,\mathrm dt &\scriptstyle{\text{since $X$ and $Y$ are independent}}\\ &= \Phi\left(\frac{w-\mu}{\sigma}\right). \end{align} Consequently, \begin{align} \int_a^b \Phi\left(\frac{w-\mu}{\sigma}\right)\phi(w) \,\mathrm dw &= \int_a^b \left[\int_{-\infty}^w f_{X}(t) \,\mathrm dt\right]f_Y(w)\,\mathrm dw\\ &= \int_{w=a}^{w=b}\int_{-\infty}^w f_{X,Y}(t,w)\,\mathrm dt \,\mathrm dw. \end{align} The value of this double integral is the probability that $(X,Y)$ lies in a horizontal strip of width $b-a$ extending from $-\infty$ to the points on the line segment with end-points $(a,a)$ and $(b,b)$ in the $t$-$w$ plane. The computation of the exact value is not easy but at least we can bound the value fairly easily. Since $\Phi\left(\frac{w-\mu}{\sigma}\right)$ increases from $\Phi\left(\frac{a-\mu}{\sigma}\right)$ to $\Phi\left(\frac{b-\mu}{\sigma}\right)$ as $w$ increases from $a$ to $b$, we get the bounds $$\Phi\left.\left.\left(\frac{a-\mu}{\sigma}\right)\right(\Phi(b)-\Phi(a)\right) < \int_a^b \Phi\left(\frac{w-\mu}{\sigma}\right)\phi(w) \,\mathrm dw < \Phi\left.\left.\left(\frac{b-\mu}{\sigma}\right)\right(\Phi(b)-\Phi(a)\right)$$ which, depending on the values of $a,b,\mu$ and $\sigma$ might be tight enough to be satisfactory for gummint purposes if not for stats.SE purposes.

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  • $\begingroup$ Another way of looking at this is to divide the bounds in the last expression by $\Phi(b)-\Phi(a)$ to obtain: $\Phi\left(\frac{a-\mu}{\sigma}\right) < E[ \Phi\left(\frac{w-\mu}{\sigma}\right) | a < w < b ] < \Phi\left(\frac{b-\mu}{\sigma}\right)$. Essentially, the random variable $\Phi\left(\frac{w-\mu}{\sigma}\right)$ is bounded by the lower and upper bound on its support, respectively. $\endgroup$
    – Bob
    Commented Dec 15, 2018 at 15:24
  • $\begingroup$ @Bob Yes, of course, or replace $\Phi\left(\frac{w-\mu}{\sigma}\right)$ by its lower/upper value in the integral, pull out the constant term, and note that what's left integrates to $\Phi(b)-\Phi(a)$ to get the lower/upper bound. I am hoping to work a little more on the problem to get a better answer -- maybe it can be done more easily in the special case $\sigma=1$ .... $\endgroup$
    – Dilip Sarwate
    Commented Dec 15, 2018 at 15:34

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