Suppose that I have 3 normal distributions (I wish to extend this to K), such that $p_k\sim\mathcal{N}(\mu_k,\sigma_k^2)$. Assume that $(\mu_k,\sigma_k^2)$ are known. How would you go about calculating the probability that: $p(p_1<p_2<p_3)$?

I understand that when k=2, we come down to the case of the elo model, where we have to use the CDF of a normal distribution (of the for $\Phi(\frac{\mu_1-\mu_2}{\sqrt{2}\sigma})$. I'm not quite sure how this pans out for more than two players.

Side question, if there is no closed form solution, is anyone aware of approximations to this (eg. Copulas maybe?).

  • $\begingroup$ I would look up the 'Henery Model' e.g. this paper by Lo. The Henery model is an alternative to the Harville model (another term to search for), when the outcomes are normally distributed. $\endgroup$ – shabbychef Jan 25 '19 at 6:32
  • $\begingroup$ It's behind a pay wall. Ugh. Thanks though. $\endgroup$ – sachinruk Jan 25 '19 at 8:22
  • $\begingroup$ Are the normal distributions independent? Do you know something about the relations between them? Maybe it makes sense to view $(X_1, X_2, X_3)$ as a multivariate normal Gaussian Variable and use that the distributions of the single variables are just the marginalization...? I.e. this would boil down to computing an integral in $\mathbb{R}^3$ over some exponential over the set $\{x < y < z, x,y,z \in \mathbb{R}\}$ which sounds “doable”... $\endgroup$ – Fabian Werner Jan 25 '19 at 8:36
  • $\begingroup$ Yes they are independent. Know of any approximation methods? In my case K=10. $\endgroup$ – sachinruk Jan 25 '19 at 8:58
  • $\begingroup$ Here is an ungated link: csclub.uwaterloo.ca/~wwyeung/stats-0.pdf.gz $\endgroup$ – kjetil b halvorsen Jan 25 '19 at 16:36

In the case $K=3$ this probability can be reduced to a one-dimensional integral, which can maybe be solved analytically, but at least numerically. For larger $K$ this is similar to integrals occurring with multinomial probit models, in particular you could look at the references given in the help for mnp in R package MNP. But for $K=3$:

$$\DeclareMathOperator{\P}{\mathbb{P}} \P(p_1 < p_2 < p_3)=\int_{-\infty}^\infty \P(p_1 < p_2 <p_3 \mid p_2=u)\phi(\frac{u-\mu_2}{\sigma_2})\frac1{\sigma_2}\; du$$ and by independence we can simplify these (details not given) to $$ \int_{-\infty}^\infty \Phi(\frac{u-\mu_1}{\sigma_1})(1-\Phi(\frac{u-\mu_3}{\sigma_3}))\phi(\frac{u-\mu_2}{\sigma_2})\frac1{\sigma_2}\; du $$ Here $\phi, \Phi$ are standard normal density and cdf respectively.

A numerical example with R:

s1 <- s3 <- 1.;  mu1 <- mu2 <- 10; mu3 <- 9; s2 <- 2
integrand <- function(u) pnorm((u-mu1)/s1)*pnorm((u-mu3)/s3, lower.tail=FALSE)*dnorm((u-mu2)/s2)/s2

integrate(integrand, lower=-Inf, upper=Inf)
0.03572245 with absolute error < 4.3e-05
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