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Looking for an analytic solution to the integral below: $$ \int_{-\infty}^\infty \Phi\left(\frac{x - a}{\tau}\right) \phi\left(\frac{x - b}{\sigma}\right)dx $$ where $\Phi(\cdot)$ and $\phi(\cdot)$ are, respectively, the standard normal CDF and PDF.

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3 Answers 3

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$\newcommand\si\sigma$Let $I$ denote the integral in question. Then $$I=\si\Phi\Big(\frac{b-a}{\sqrt{\tau^2+\si^2}}\Big). $$

Indeed, using the substitution $x=b+\si u$ and letting $$A:=\frac\si\tau,\quad B:=\frac{b-a}\tau,$$ we have $$I=\si\int_{-\infty}^\infty\Phi(Au+B)\phi(u)\,du =\si P(V\le AU+B),$$ where $U,V$ are independent standard normal random variables. So, for $W:=V-AU$ we have $W\sim N(0,1+A^2)$ and hence \begin{align*} I=\si P(V\le AU+B)&=\si P(W\le B) \\ &=\si\Phi\Big(\frac B{\sqrt{1+A^2}}\Big) =\si\Phi\Big(\frac{b-a}{\sqrt{\tau^2+\si^2}}\Big),\tag{1}\label{1} \end{align*} as claimed.


There has been a (strange to me) assertion in the discussion of my answer that somehow the value of $I$ should be $\le1$ -- and actually a number of users seem to have been receptive to this assertion. Neither my answer itself nor the arguments in my comments about this assertion seem to have had any effect; on the other hand, no mistake in my arguments has been indicated.

So, I am now giving another argument, involving Mathematica. Strangely, Mathematica cannot evaluate the integral in general, even in the simple case when $a=b$ but $\tau$ is an arbitrary positive real number. However, Mathematica can evaluate the integral when $a=b=0$ and $\tau=1$:

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Of course, this is fully consistent with formula \eqref{1}. In particular, it follows that $I>1$ if $\si>2$ (given that $a=b=0$ and $\tau=1$).

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  • $\begingroup$ Interesting! This is exactly the solution I expected to find, up to the scaling by $\sigma$. I will think about this. $\endgroup$
    – user489812
    Aug 18 at 2:45
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    $\begingroup$ I was expecting the result to live within the interval [0, 1] $\endgroup$
    – user489812
    Aug 18 at 2:55
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    $\begingroup$ @user489812 : The reasoning here is quite simple. You are welcome to check it. As for the scaling, think e.g. of this: Let $\sigma\to\infty$. Then $\phi(\frac{x - b}{\sigma})\to\phi(0)>0$ and hence the integral goes to $\infty$ -- by, say, the Fatou lemma. So, of course, the integral cannot remain in the interval $[0,1]$. $\endgroup$ Aug 18 at 4:20
  • $\begingroup$ Just noting that, up to the scaling by $\sigma$, this result is consistent with what is reported in the paper here: eduardomazevedo.github.io/papers/azevedo-et-al-ab-gaussian.pdf (see page 20) $\endgroup$
    – user489812
    Aug 18 at 6:14
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    $\begingroup$ @user489812 : I do not see anything reported on p. 20 of that paper that is relevant to your question. Also, the function $g$ seems undefined in the paper (as well as other things). However, the formula at the top of p. 21 of that paper corresponds to $\sigma=1$ and is fully consistent with everything in my answer. More importantly, have you checked the very simple reasoning in my answer? Have you considered my comment above, involving the Fatou lemma? Have you considered my comment to Bill Bradley's answer? $\endgroup$ Aug 18 at 11:56
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These notes are just intended as an extended comment on Iosif Pinelis' answer. In my initial version of this post, I made a mistake, but Iosif pointed out the error in the comments below; it should now be correct (and in agreement with Iosif's accepted answer).

Let $I$ be the original poster's integral. Let $C\sim \mathcal{N}(a,\,\tau^{2}), D\sim \mathcal{N}(b,\,\sigma^{2})$ with $C,D$ independent. Then $$ \Pr[C<D] = \int_{-\infty}^\infty \Phi\left(\frac{x - a}{\tau}\right) \left[\frac{1}{\sigma} \phi\left(\frac{x - b}{\sigma}\right)\right]dx=I/\sigma$$ However, since $C$ and $D$ are independent, we have that $C-D \sim \mathcal{N}(a-b,\,\sigma^2 + \tau^{2})$. Then $$ \Pr[C<D] = \Pr[C-D<0] = \Phi\Big(\frac{b-a}{\sqrt{\tau^2+\sigma^2}}\Big)$$ Combining the above, $$ I = \sigma \Phi\Big(\frac{b-a}{\sqrt{\tau^2+\sigma^2}}\Big)$$

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    $\begingroup$ In your second displayed formula, the factor $\frac1\sigma$ is missing on the right-hand side: for the pdf $f_D$ of $D$ and all real $x$, we have, not $f_D(x)=\phi(\frac{x - b}{\sigma})$, but $f_D(x)=\frac1\sigma\,\phi(\frac{x - b}{\sigma})$. Otherwise, your calculation is almost the same as mine. $\endgroup$ Aug 18 at 5:18
  • $\begingroup$ Thanks for the additional follow up! This is very helpful. I will give it some thought (Still puzzled by the scaling by $\sigma$. My expectation was that, as $\sigma \rightarrow \infty$, this integral should go to 0.5. $\endgroup$
    – user489812
    Aug 18 at 5:52
  • $\begingroup$ Dear @IosifPinelis, ah! I see now (and also why I couldn't find the error in your original post, as there was no error). I'll correct my post accordingly. $\endgroup$ Aug 18 at 20:42
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\begin{align} & \sigma \int_{-\infty}^{+\infty} \Phi\left(\frac{x - a}{\tau}\right) \overbrace{ \varphi\left(\frac{x - b}{\sigma}\right) \frac{dx} \sigma }^{ \begin{smallmatrix} & \text{This is the} \\ & \text{$\operatorname N(b,\sigma^2)$ distribution.} \end{smallmatrix} } \\[10pt] = {} & \sigma\operatorname E\left( \Phi\left( \frac{X-a}\tau \right) \right) \text{ where } X\sim\operatorname N(b,\sigma^2) \\[10pt] = {} & \sigma\operatorname E\left( \Pr\left(Z\le \frac{X-a} \tau \right) \,\Bigg\vert\, X \right) \\ & \text{where } Z\sim\operatorname N(0,1) \text{ and } X,Z \text{ are independent} \\[10pt] = {} & \sigma\Pr\left( Z\le \frac{X-a} \tau \right) \\ & \text{(law of total probability)} \\[10pt] = {} & \sigma\Pr\left( Z - \frac{X-a} \tau \le0 \right) \\[10pt] = {} & \sigma\Pr(W\le 0) \text{ where } W\sim\operatorname N\left( \frac{a-b}\tau, 1 + \frac{\sigma^2}{\tau^2} \right) \\[10pt] = {} & \sigma\Pr\left( \frac{W- \frac{a-b}\tau}{\sqrt{1+\frac{\sigma^2}{\tau^2}}} \le \frac{0- \frac{a-b}\tau}{\sqrt{1+\frac{\sigma^2}{\tau^2}}} \right) \\[10pt] = {} & \sigma \Pr\left( Z \le \frac{0- \frac{a-b}\tau}{\sqrt{1+\frac{\sigma^2}{\tau^2}}} \right) = \sigma \Pr\left( Z\le \frac{b-a}{\sqrt{\tau^2+\sigma^2}} \right) \\[10pt] = {} & \sigma \Phi\left( \frac{b-a}{\sqrt{\tau^2+\sigma^2}} \right). \end{align}

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