# $P(\min(X_1,\dots,X_n) > t) = P(X_1>t,\dots, X_n>t)$

One step in the my solutions book shows...

$P(\min(X_1,\ldots,X_n) > t) = P(X_1>t, \ldots, X_n>t)$, where $X_1, \ldots, X_n$ are independent and $X_j \sim \mathrm{Expo}(\lambda)$

Why is it that the probability that the minimum of a set of exponential random variables is greater than $t$ is just the intersection of the probability that each exponential random variable is greater than $t$.

• "intersection of the probability" doesn't make sense. Sets have intersections; a probability is a number. I think you meant "intersection of the events". – bof Dec 22 '14 at 20:29

Observe that if the minimum of the random variables is greater than $t$, then all the random variables must be greater than $t$ (the converse is also true). There is nothing more to it.
Of course, $X_1, \ldots, X_n$ being independent allows you to write this as the product of probabilities afterwards.