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.

  • $\begingroup$ 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
  • 2
    $\begingroup$ Are these arbitrary cdf and pdf? $\endgroup$ – Alex R. Sep 23 '16 at 19:18
  • 3
    $\begingroup$ 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

$$\int_{-\infty}^\infty \phi(z)\Phi^2(z+q)\,\mathrm dz=\int_{-\infty}^\infty \phi(y-q)\Phi^2(y)\,\mathrm dy.$$ 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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.