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I'm trying to solve a distribution problem. As part of it, I need to find $\int\limits_{-\infty}^\infty\phi(z)\Phi^2(z+q)dz$ and $\int\limits_{-\infty}^\infty\phi(z)\Phi(z)\Phi(z+q)dz$

So far, I've worked out that $\int\phi(z)\Phi^2(z)dz=\frac{1}{3}$ and that $\int\limits_{-\infty}^\infty\phi(z)\Phi(z+q)dz = \Phi\left(\frac{q}{\sqrt2}\right)$, but I can't get any further.

Edit: to get the first identity: if $u = \Phi(z)$ then $du = \phi(z)dz$
$\int\phi(z)\Phi^2(z)dz$ becomes $\int u^2du = \left[\frac{u^3}{3}\right]_{-\infty}^\infty= \frac{\Phi^3(\infty)-\Phi^3(-\infty)}{3}=\frac{1}{3}$

The second comes from here: https://mathoverflow.net/questions/101469/integration-of-the-product-of-pdf-cdf-of-normal-distribution

They are standard normal distributions.

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  • $\begingroup$ No self-study here. I'm trying to figure out the Studentized Range Distribution and this is one of the steps when k=3. $\endgroup$ – Kevin Nowaczyk Sep 23 '16 at 18:44
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    $\begingroup$ Are these arbitrary cdf and pdf? $\endgroup$ – Alex R. Sep 23 '16 at 19:18
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    $\begingroup$ Your first integral can be found here, in terms of Owen's T function: en.wikipedia.org/wiki/List_of_integrals_of_Gaussian_functions $\endgroup$ – David R Sep 23 '16 at 19:33
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$$\int_{-\infty}^\infty \phi(z)\Phi^2(z+q)\,\mathrm dz=\int_{-\infty}^\infty \phi(y-q)\Phi^2(y)\,\mathrm dy.$$ Integrals of this type have no known analytical solution in terms of other well-known functions, but see David Robertson's comment re special functions.

The value of the first integral that you seek is the probability that a $N(q,1)$ random variable is larger than both of two other independent $N(0,1)$ random variables. See this answer of mine for more discussion of the case when $\Phi^2(y)$ is replaced by $\Phi^n(y), n \geq 2.$ Indeed, the value of the first integral as a probability makes the interpretation of your derivation of $\int\phi(z)\Phi^2(z)dz=\frac{1}{3}$ straightforward: the probability that one specific random variable among $3$ iid random variables has the largest value is $\frac 13$ (since all three random variables are equally likely to be the largest.)

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