What is P(X1>X2,X1>X3,…,X1>Xn)P(X_1>X_2 , X_1>X_3,… , X_1>X_n)?

All X are mutually independent and from normal distributions, each with its own mean and variance. If it’s easier, P(X1Xii{1,...,n}) is fine although I suspect it’s the same. If it matters, n is between 5 and 20.

I found three similar questions:

  1. The answer to this one is for only three random variables.
  2. The answer to this one is for only mean 0 and variance 1.
  3. I’m unsure if this one applies. If it does, I don’t know how to apply it. Its top answer is for three random variables.

(This is not homework.)


For n \gt 2 this needs numeric integration, as indicated in several of the links.

To be explicit, let \phi_i be the PDF of X_i and \Phi_i be its CDF. Conditional on X_1 = t, the chance that X_1 \gt X_i for the remaining i is the product of the individual chances (by independence):

\Pr(t \ge X_i, i=2,\ldots,n) = \Phi_2(t)\Phi_3(t)\cdots\Phi_n(t).

Integrating over all values of t, using the distribution function \phi_1(t)dt for X_1, gives the answer

= \int_{-\infty}^{\infty} \phi_1(t) \Phi_2(t)\cdots\Phi_n(t)dt.

For n=20, the integral takes 5 seconds with Mathematica, given vectors \mu and \sigma of the means and SDs of the variables:

\[CapitalPhi] = MapThread[CDF[NormalDistribution[#1, #2]] &, {\[Mu], \[Sigma]}];
\[Phi] = PDF[NormalDistribution[First[\[Mu]], First[\[Sigma]]]];
f[t] := \[Phi][t] Product[i[t], {i, Rest[\[CapitalPhi]]}]
NIntegrate[f[t], {t, -Infinity, Infinity}]

The value can be checked (or even estimated) with a simulation. In the same five seconds it takes to do the integral, Mathematica can do over 2.5 million iterations and summarize their results:

m = 2500000;
x = MapThread[RandomReal[NormalDistribution[#1, #2], m] &, {\[Mu],\[Sigma]}]\[Transpose];
{1, 1./m} # & /@ SortBy[Tally[Flatten[Ordering[#, -1] & /@ x]], First[#] &]

For instance, we can generate some variable specifications at random:

{\[Mu], \[Sigma]} = RandomReal[{0, 1}, {2, n}];

In one case the integral evaluated to 0.152078; a simulation returned

{{1, 0.152387}, … }

indicating that the first variable was greatest 0.152387 of the time, closely agreeing with the integral. (With this many iterations we expect agreement to within a few digits in the fourth decimal place.)

Source : Link , Question Author : Chris Idzerda , Answer Author : whuber

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