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I specialize in computer science so statistics isn't my forte but I need some statistics knowledge for my current project.

I have a list of variables whose values have a roughly standard normal probability distribution. The variables are compared and the maximum value is saved for later use. I want to know what the probability distribution of the max value is.

Rationally it should be something like P = N x (∫N)^(v-1) where N is a standard normal distribution ∫N is the integral of a standard normal distribution and v is the number of variables being compared. This theory is however a little hard to test as, if I understand correctly, the integral of a standard normal distribution cannot be calculated in Cartesian space.

Is there more known about this problem? Is the resultant distribution a normal distribution? if not can it be expressed as a Cartesian function? if not can the Mean be calculated easily? Any information that would narrow down the behavior of the max value would be helpful.

The number of variables being compared can be any value, but the most common, most important case is v=4 if that helps.

user1492428
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  • Duplicate: https://math.stackexchange.com/questions/89030/expectation-of-the-maximum-of-gaussian-random-variables – David G. Stork Sep 19 '18 at 22:39
  • @DavidG.Stork is correct that the link answers your question. But I wonder whether the level of it is accessible for you. // The answer depends on $n$ because as you take more observations you are more likely to get extreme max observations. The max of 4 std normal observations averages about 1.03; the max of 20 averages about 1.87; for 100 it's about 2.51. – BruceET Sep 22 '18 at 20:00

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Comment continued: I got the answers in the first part of my Comment by simulation. Specifically, for each sample size, I looked at 100,000 samples, found the max of each, and averaged the 100,000 max values.

Simulations using R statistical software:

mx = replicate(10^5, max(rnorm(4)));  mean(mx)
## 1.029221

mx = replicate(10^5, max(rnorm(20)));  mean(mx)
## 1.868113

mx = replicate(10^5, max(rnorm(100)));  mean(mx)
## 2.508453

Answers in second simulations for each sample size agreed to two decimal places.

Note: Just to make sure you're not misled looking at means, I'll mention that for samples of size four the maximum can be considerably larger than 1.03, but only 10% of the 100,000 samples had maximum values above 1.94.That part is easy to verify analytically: If $Z \sim \mathsf{Norm}(0,1),$ then $[P(Z < 1.94)]^4 = 0.9738^4 \approx 0.9.$

BruceET
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