Suppose we have the density and distribution of the standard normal. How can one calculate the integral:

$\int_{-\infty}^{\infty} \Phi (a + bX) \phi (c + eX) dx$

Note this is not included in the Wikipedia list of integrals of Gaussian functions.

Further, this is not the same as How can I calculate $\int^{\infty}_{-\infty}\Phi\left(\frac{w-a}{b}\right)\phi(w)\,\mathrm dw$ because that solution only works for a standard normal second term, i.e. $\phi (X)$ whereas this problem includes a coefficient and additional term. One cannot put into the same form as that case because $c$ and $e$ cannot be brought out of the normal density due to their involvement in the exponential component.

  • 1
    $\begingroup$ Not sure if this helps, but we know that $$\int_{-\infty}^{\infty}\Phi(a+bX) \phi( X ) dX=\Phi\left(\frac{a}{\sqrt{1+b^2}}\right)$$ $\endgroup$
    – user30490
    Commented Nov 11, 2015 at 5:29
  • $\begingroup$ @ZERO Thanks, I am familiar with that integral and have seen short proofs on here, but unfortunately dealing with the second linear term is still unclear. $\endgroup$
    – SG2013
    Commented Nov 11, 2015 at 5:40
  • $\begingroup$ Are you sure it has an analytical closed form? $\endgroup$
    – Zhanxiong
    Commented Nov 11, 2015 at 6:01
  • $\begingroup$ It was suggested to me that it does, but I am not sure. $\endgroup$
    – SG2013
    Commented Nov 11, 2015 at 6:12
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    $\begingroup$ @Glen_b In which case it really ought to be migrated to Mathematics, because it's about changing variables in integration. :-) $\endgroup$
    – whuber
    Commented Nov 12, 2015 at 0:03

1 Answer 1


But if you know the form discussed in comments

$$\int_{-\infty}^{\infty}\Phi(a+bx) \phi( x ) dx=\Phi\left(\frac{a}{\sqrt{1+b^2}}\right)$$

it's quite straightforward!

Just do the substitution $Z = c+eX$ (don't forget the Jacobian), and use the fact there that you already know (but in $z$ rather than $x$).


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