I was reading this section about the minimum and maximum of a series of random variables:

Suppose that $X_1, \dots, X_n$ are independent variables with cdf's $F_1, \dots, F_n$, respectively.

The cdf $F_m(\cdot)$ of $\max{\{ X_1, \dots, X_n \}}$ is then given by

$$\begin{align} F_M(x) &= P(\max{\{X_1, \dots, X_n \}} \le x) = P(X_1 \le x, \dots, X_n \le x) \\ &= P(X_1 \le x \cap \dots \cap X_n \le x) = P \left( \cap_{i = 1}^n X_i \le x \right) \\ &= \prod_{i = 1}^n P(X_i \le x) = \prod_{i = 1}^n F_i(x) \end{align}$$

The cdf $F_m(\cdot)$ of $\min{\{ X_1, \dots, X_n \}}$ is given by

$$F_m(x) = 1 - \prod_{i = 1}^n (1 - F_i(x))$$

If I'm interpreting this correctly, the first part says that the probability that the maximum of a set of random variables is less than or equal to some value $x$ is equal to the probability that each of the random variables in that set is less than or equal to the value $x$? I'm struggling to understand why this must be true. After all, the former only considers a single value (the maximum of the set), and disregards the rest, whereas the latter considers whether every random variable is less than or equal to $x$, and so can have multiple random variables that satisfy this condition, no?

And with regards to the $\min$, where did the double $1 -$ in $F_m(x) = 1 - \prod_{i = 1}^n (1 - F_i(x))$ come from? Intuitively, I can see why you would require one $1 -$ in there to get the $\min$ from the $\max$, but I don't see why it does this twice?

Thank you.

  • 1
    $\begingroup$ Consider a set of numbers $\{a_1 ... a_n\}$ and their maximum, denoted as $a^*$. If $a^* \leq x$, then of course all the elements of a set are also $\leq x$ because no element can be greater than maximum $a^*$. On the other hand, if $a_1 \leq x, ..., a_n \leq x$, then maximum $a^*$ is also $\leq x$. This establishes $a_1 \leq x, ..., a_n \leq x \iff a^* \leq x$. Using this with probabilities establishes the result. $\endgroup$ – Evgeny Mar 2 at 23:30

They are just using two basic facts : $max \{a_1,a_2,...,a_n\} \leq x$ iff $a_i \leq x$ for each $i$ and $min \{a_1,a_2,...,a_n\} > x$ iff $a_i > x$ for each $i$. This gives $P(min \{X_1,X_2,...,X_n\}\leq x) =1-P(min \{X_1,X_2,...,X_n\} > x)=1-\prod P(X_i >x)=1-\prod [1-P(X_i \leq x)]=1-\prod (1-F_i(x))$.

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