Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi,

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!

Eleanor

share|improve this question

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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