I'm having trouble seeing why for RVs $X_{1}, \ldots, X_{n}$ it is true that:

$$Pr(min(X_{1}, \ldots, X_{n}) > x ) = Pr(X_{1} > x, \ldots, X_{n} > x)$$

In other words: Why is the event that the minimum of a set of RVs is greater than x equivalent to the event that all RVs are greater than x?

Is this because the minimum itself is a random variable and if one of the $X_{i}$ $\leq x$ it could happen that $X_{i}$ is the minimum?

  • 4
    $\begingroup$ If the smallest of a set of things is greater than $x$, then so are the rest of them! $\endgroup$
    – Glen_b
    Dec 15, 2014 at 11:43

1 Answer 1


Yup, the minimum is a r.v. too. Now, break the equation down into two parts:

  • First, suppose that $\min(X_1, \dotsc, X_n) > x$. Since $X_i \geq \min(X_1, \dotsc, X_n)$ for each $i$, it follows that $X_i > x$ for each $i$.

  • Second, suppose that $X_i > x$ for each $i$. Then $\min(X_1, \dotsc, X_n) > x$

So it must be that the event "$\min(X_1, \dotsc, X_n) > x$" is equivalent to the event "$X_i > x$ for all $i$". And since they're the same event, they must have the same probability.

  • $\begingroup$ If $\min(X_1, \dotsc, X_n) > x$, then we have $X_i \geq \min(X_1, \dotsc, X_n) > x$ for each $i$. So $X_i > x$ for each $i$. $\endgroup$
    – Andy Jones
    Dec 15, 2014 at 10:17

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