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January 26th, 2018, 02:19 AM   #1
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probability to be largest of three normal distributed random variables

Suppose there are three random variable x1...x3, all of the normal distributed with means m1...m3 and standard deviation 1. I want to know the probability that x1>x2 & x1>x3, i.e. that x1 is the largest of x1...x3.

Simple case: let us consider only 2 random variables, x1 and x2. I can then build a new random variable delta = x2 - x1. delta is normal distributed with mean m2-m1 and standard deviation sqrt(2). I can calculate the probability tha delta < 0, which is equivalent to x2 < x1.

But what should I do to calculate the probability given more than two random variables? I would be happy to know a solution for 3.

False solution: just multiply the probabilities that x1>x2 and x1>x3. But... they are not independent. Given that x1>x2, x1 will be more often large than small, and this is a good precondition for x1>x3. Example: m1=m2=m3=0. The probability of x1>x2 = 0.5, and so is the probability for x1>x3. The probability for x1>x2 & x1>x3 is, however, larger than 0.5*0.5=0.25, it is 1/3, as is obvious from symmetry reasoning.
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January 26th, 2018, 09:38 PM   #2
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You don't specify that $x_1, ~x_2, ~x_3$ are independent but you only give their marginal densities so I'll assume they are.

Thus their joint density is the product of the their marginals.

$f_{X_1,X_2,X_3}(x_1,x_2,x_3) = \dfrac{\exp \left(-\frac{1}{2} \left(x_1-m_1\right){}^2-\frac{1}{2} \left(x_2-m_2\right){}^2-\frac{1}{2} \left(x_3-m_3\right){}^2\right)}{2 \sqrt{2} \pi ^{3/2}}$

$P[X_1>X_2 \wedge X_1 > X_3] =

\displaystyle \int_{-\infty}^\infty \int_{-\infty}^{x_1}\int_{-\infty}^{x_1}~f_{X_1,X_2,X_3}(x_1,x_2,x_3)~dx_3~dx _2~dx_1$

This can be evaluated once you select values for the means.
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January 27th, 2018, 11:57 PM   #3
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Thank you, romsek, for your reply. What I was looking for is a solution without the need to calculate integrals. Consider for instance the two dimensional case. In this case there would also be the possibility to take the integrals... just like you do. But by transforming the problem to a one-dimensional problem (creating a new random variable which is just the difference of the two normally distributed random variables... and which is also a normally distributed random variable) and assuming that error function or cumulative normal distribution are implemented (so there is no need to do a integral in the one-dimensional case) one ends with a solution that gives the probability by simply calling normcdf (in Matlab) or the like. What I was looking for is a similarly simple solution for the three-dimensional case. I want to call it inside an optimization loop, so cpu time matters.
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