What is the probability that one random number is greater than other random numbers?

Given N normal distributions with mean m(1), m(2), ... m(N), and respective standard deviations sig(1), sig(2), ... sig(N), let r(1), r(2), ... r(N) be random numbers drawn from each distribution. What is the probability that r(1) is the largest number?





If there is no solution for general N, how about for N=3?What is the probability that one random number is greater than other random numbers?
What you didn't add was whether the random variables are independent or not. Assuming that the distributions are all independent of each other you can approach it like this.





If r(1) is the largest number then it is larger than r(2), r(3), r(4), etc. so





Pr{max[r(1),r(2),...r(N)] = r(1)} = 螤Pr[r(1)%26gt;r(j)], 1%26gt;j鈮





First define





Zi = [m(1) - r(i)]/sig(1) for i=2,3,...,N then





Pr[r(1)%26gt;r(i)] = Pr[Zi%26lt;Z] is the probability that r(1)%26gt;r(i)





Pr{max[r(1),r(2),...r(N)] = r(1)} = 螤Pr[Zi%26lt;Z]





鈭?br>

鈭玣(x)dx = Pr[Zi%26lt;Z] where f(x)= e^(-陆x虏)


Zi





鈭? 鈭?br>

螤 鈭玣(x)dx = Pr{max[r(1),r(2),...,r(N)] = r(1)}


i=2 ZiWhat is the probability that one random number is greater than other random numbers?
Unfortunately this isn't correct. The easiest way you can tell is that if the distributions are all identical, then the probability that the first random number is largest must be 1/3 (each number has an equal chance of being the largest). The answer provided here yields (1/2)*(1/2). Report Abuse

Well... if it is r(1) is the largest then it is 1/N.
r(1) will be the largest when





r(x) %26lt; r(1) for all 1%26lt;x%26lt;=N





or prob( r(2) %26lt; r(1) ) * prob ( r(3) %26lt; r(1) ) *... prob( r(N) %26lt; r(1) ).





suggest reference Wikipedia entry on Normal distribution for prob( a %26lt; b) as its too difficult to reproduce here.