I'm trying to evaluate the following integral:

$\int_{-\infty}^\infty \phi(x)\Phi(a+bx)^2dx$

where Phi is the cdf of a std Normal random variable, and phi is the pdf $(1\sqrt(2pi))exp(-x^2\2)$.

I have that $\int_{-\infty}^\infty \phi(x)\Phi(a+bx)dx = Phi(\frac{a}{\sqrt(1+b^2)})$, so I could integrate the above expression if I could evaluate

$\int_0^y \phi(x)\Phi(cx)dx$

By way of background, I have a discrete choice experiment where individual i has latent utility $u_{ij}$ for item $j$

$u_{ij}=X_{ij}(\beta+b_i) + \epsilon_{ij}$,

$\epsilon_{ij} \sim iid N(0,\sigma^2)$ $b_i \sim N(0,\eta^2)$ are random effects

and I am trying to express the intra-rater agreement as a function of $X$, $\beta$, $\sigma$ and $\eta$.

Any thoughts would be much appreciated!



The equation for $\int_{-\infty}^\infty \phi(x) \Phi(a+b x) \ dx$ comes from interpreting this as $P(X > Y)$ where $X \sim N(0,1)$ and $Y \sim N(-a/b, 1/b^2)$ are independent, so that $X - Y \sim N(a/b, 1 + 1/b^2)$. Similarly $\int_{-\infty}^\infty \phi(x) \Phi(a+bx)^2 \ dx$ would be $P(X > \max(Y,Z))$ where $X \sim N(0,1)$, $Y \sim N(-a/b, 1/b^2) $ and $Z \sim N(-a/b, 1/b^2)$. However, I don't see any easy way to simplify this (except in the case $a=0, b=1$ where symmetry says each of $X$, $Y$ and $Z$ is equally likely to be the greatest), because $\max(Y,Z) - X$ does not have a hormal distribution. There may not be a "closed-form" solution. Of course, series expansions and asymptotics are available.


If a = 0 and b = 1, Mathematica evaluates the integral to 1/3 but otherwise it doesn't make any progress. Perhaps the integral could be evaluated using contour integration techniques.


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