Integral involving normal pdf and cdf

[MikeGT]

Suppose φ(x) and Φ(x) denote the familiar standard normal pdf and cdf. I am interested in an expression for the integral:

S φ(ax+b) Φ(x) dx, from ZERO to INFINITY.

Many thanks

 

[MikeGT]

I forgot to mention something potentially interesting. The integral

S φ(ax+b) Φ(x) dx, from -INFINITY to INFINITY

has been obtained by Gupta and Pillai in a technical report of the 60's as equal to

Φ( b/sqrt(1+a^2) ).

They mention this in a lemma but I cannot see how it was obtained. I checked it numerically and it is true.

Can that be useful?

Best

M

 

[Mute]

Written explicitly, your expression is

$$\int_0^\infty dx~\frac{e^{-(ax+b)^2/2}}{\sqrt{2\pi}}\left[\int_{-\infty}^x dt~\frac{e^{-t^2/2}}{\sqrt{2\pi}}\right]$$

Have you tried playing around with the double integral?

 

[mathman]

The exponent of the integrand is -[x² + (ax+b)²)]/2. This can then be converted into a form -(cx+d)² + f, where c,d,f are constants depending on a and b. The integration should give you the described result.

 

[MikeGT]

I don't see how they did it when x runs from -INF to +INF. You say the exponent is not -(ax+b)^2 but it is: -[x^2+(ax+b)^2]. Apparently you're taking the extra x^2 term from the dt - integral. But how the combination can produce Φ when they integrate from -inf to +inf? They should get some constant or 1 but not Φ. To get b/sqrt(1+a^2) they do need what you describe above and an integral from 0 to +inf. Am I missing something?



Many thanks for the replies

M

 

[mathman]

Sorry. I misread your original question. I confused the capital and lower case phi's.

 

[mistergo_2000]

You can solve this by differentiating the integrand with respect to b, then re-integrating.