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I was wondering if there is an intuitive way to understand why it is that if $X_1, \ldots X_n$ are iid, then $P(\max(X_1, \ldots, X_n) \leq x) = P(X_1 \leq x, \ldots X_n \leq x)$?

The standard explanation is that in order for the maximum to be less than something, every element must be less than that something as well. However, from a very rough intuitive standpoint, it seems that while $P(\max(X_1, \ldots, X_n) \leq x)$ cared only about the maximum, $P(X_1 \leq x, \ldots X_n \leq x)$ now cares about each and every variable, so it seems that we are working with more information in the latter. Basically, it feels as if $P(X_1 \leq x, \ldots X_n \leq x)$ has more "going on" than $P(\max(X_1, \ldots, X_n) \leq x)$. Is there a nice way to think about this clearly? Thanks!

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    $\begingroup$ "it seems that while $P(max(X_1, \ldots, X_n) \leq x)$ cared only about the maximum, $P(X_1 \leq x, \ldots X_n \leq x)$ now cares about each and every variable, so it seems that we are working with more information in the latter." But the only way you know that the max is less than x is to know that every $X_i$ is less than $x$, so you still need to know something about every variable if you're interested in max $\endgroup$ – Brenton Nov 24 '15 at 23:08
  • $\begingroup$ The events $(\max\{X_1,\dots,X_m\}\leq x)$ and $(X_1\leq x,\dots,X_n\leq x)$ are the same, no? So not only do they have the same probability but they are in fact equal (and the hypothesis of independence is irrelevant for this= $\endgroup$ – Mariano Suárez-Álvarez Nov 24 '15 at 23:16
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However, from a very rough intuitive standpoint, it seems that while $\mathsf P(\max(X_1 ,…,X_n )≤x)$ cared only about the maximum, $\mathsf P(X_1 ≤x,…,X_n ≤x)$ now cares about each and every variable

While the first probability is only concerned about the maximum, the maximum itself is concerned about the values of the individual variables.

The maximum of a list of variables is less than a value if and only if each and every variable in the list is less than that value.   $\big\{\max\limits_{k=1}^n\{X_k\}\leq x\big\}$ and $\bigcap\limits_{k=1}^n\{X_k\leq x\}$ are equivalent, as they describe the same event.

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  • $\begingroup$ I'm sure sure I get the intersection statement you put. Should it be the union instead? $\endgroup$ – user136503 Nov 24 '15 at 23:43
  • $\begingroup$ No, it's conjunctive; an "and" connection. $\{\max\{X_1, X_2\}\leq x\} \iff \{X_1\leq x\}\cap\{X_2\leq x\}$ $\endgroup$ – Graham Kemp Nov 25 '15 at 0:06
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It has nothing to do with iid, or probability for that matter.

"Each person in the room is at most 80 years old" is equivalent to "The oldest person in the room is at most 80 years old".

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since $X_{i}$'s are identical, we can write this $P(X_{1}\leq X_{2})=P(X_{1}\leq X_{i})$ for any i=2,...,n. It does not matter which value of i you consider in above equality.and, since X_{i}$'s are iid then we can intuitively see why your equality holds.

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