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I don't quite understand what the min and max functions do in terms of probability and random variables or how they're able to have PDFs and CDFs. If $V = max(X_1, X_2 ... X_n) $ , what is getting returned? Is it supposed to be the max value of each rv at some point or is it giving me the maximum value that can be found within all of the rv? A max is a single value, so how can there be a changing probability that would necessitate a pdf or cdf?

AWW
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    It's just the same as the maximum or minimum of a sequence of real numbers in usual mathematical parlance. – Angina Seng Oct 01 '17 at 10:13
  • The value of $V$ changes with different realizations of the $X_{i}$'s; i.e., $V$ itself is a random variable (this can be proved more rigorously) – nemo Oct 01 '17 at 10:20
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    If $X_1$ is the number of spades in a bridge hand, $X_2$ the number of hearts, $X_3$ the number of diamonds, $X_4$ the number of clubs, then $V=\max(X_1,X_2,X_3,X_4)$ is the length of your longest suit. – bof Oct 01 '17 at 10:44

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Recall that a random variable is a measurable function $X : \Omega \to \mathbb{R}$ where $\Omega$ is some probability space. So the complete definition is, for random variables $X_1, X_2, \ldots, X_n : \Omega \to \mathbb{R}$:

$$V(\omega) = \max \{ X_1(\omega), X_2(\omega), \ldots, X_n(\omega) \}$$

so $V$ is itself a random variable $V : \Omega \to \mathbb{R}$.

For instance, suppose you are throwing $4$ dice. Each throw is a random variable with range $\{ 1, 2, 3, 4, 5, 6 \}$ where different throw results are assigned corresponding numbers. Now the maximum of the four numbers is a single number, but it is somehow random: it depends on the results of all four throws, thus it's a random variable.

Of course the distribution above is discrete, so it doesn't have a valid PDF, but for (absolutely) continuous distributions the rule is the same.

Adayah
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