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Assume to groups, $A$ and $B$, which both are normally distributed with mean $0$ and variance $1$.

What is the probability that $Max_A > Max_B$ if $n_A = n_B$ (say 50)?

What is the probability that $Max_A > Max_B$ if $n_A > n_B$ (say 75 and 25)?

I think my general query is how the maximum distribution depends on $n$ and how that can be used to determine the probability of the maximum values of one group exceeding another.

John12
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    in the first case, the problem is symmetrical, so it must be $1/2$ – Exodd Dec 04 '22 at 14:29
  • Agree @Exodd, the intuition is clear on that. However, what is the formal logic? and how can it possibly be extended to varying $n$? – John12 Dec 04 '22 at 14:32
  • You can derive the pdfs/cdfs of a max of a group of $n$ iid sampled elts (from any dist) as $nF(x)^{n-1}f(x)$ where $F$ and $f$ are the cdf and pdf of the dist respectively. The behavior is pretty easy to get a closed form for when the distribution is uniform but for more complex stuff, and for normal dists in particular, it's hard to get a closed form. See this for some discussion. – Vercingetorix Dec 04 '22 at 14:33
  • That being said for q1, it's simply a matter of symmetry and the answer is $0.5$ – Vercingetorix Dec 04 '22 at 14:34
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    You don't need to determine the pdfs/cdfs. If the distributions are identical (and continuous), then any permutation of group assignment to a sample is size $n_A+n_B$ is equally likely. The probability that group A has the maximum value is $n_A/(n_A+n_B)$. – JimB Dec 04 '22 at 17:06

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If the distributions are identical (and continuous), then any permutation of group assignment to a sample of size $n_A+n_B$ is equally likely. The probability that group A has the maximum value is $n_A/(n_A+n_B)$.

JimB
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