1
$\begingroup$

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.

$\endgroup$
  • 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
2
$\begingroup$

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 $P(min \{X_1,X_2,...,X_n\}\leq x) =1-P(min \{X_1,X_2,...,X_n\} > x)=1-\prod P(X_i >x)=1-\prod [1-P(X_i \leq x)]=1-\prod (1-F_i(x))$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.