# Probabilities involving Mins and Maxes

1. Sep 16, 2011

### muzihc

I know for sure that P(Max(X, Y) < 1) = P(X < 1)P(Y < 1) if the two RVs are independent, but what does P(Min(X, Y) < 1) equal?

Would it be P(Min(X, Y) < 1) = P(X < 1) + P(Y < 1)?

Thanks

2. Sep 16, 2011

### Stephen Tashi

The probability of "X < 1 or Y < 1" involves the union of two sets. Have you sudied the formula for $P(A \cup B)$ ?

3. Sep 16, 2011

### chiro

Hey muzihc and welcome to the forums.

Have you ever come across order statistics?

http://en.wikipedia.org/wiki/Order_statistic

4. Sep 16, 2011

### muzihc

Hi,

I've studied P(A or B) = P(A U B) - I had a class in probability/statistics. I've never formally studied order statistics, though maybe I've overlapped with it at some point.

The maximum case is kind of intuitive - if the max is less than t, everything else is. We can treat it as P(A and B), in the independent case. On the other hand, if the minimum is less than t, the other random variable isn't necessarily.

I guess we could use P(Min(X, Y) < t) = 1 - P(Min(X, Y) >= t), in which case I assume it's 1 - P(X >= t)P(Y >= t).

5. Sep 17, 2011

### chiro

Hey muzihc.

Take a look at order statistics. Order statistics helps you calculate the distribution of the minimum and maximum of a value given the number of samples that you have. It will help you in the first part of your question.

6. Sep 17, 2011

### Stephen Tashi

Let $A =$ the events where X < t [/itex]
Let $B =$ the events where Y < t [/itex]

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

So for independent events $A$ and $B$

$P(A \cup B) = P(A) + P(B) - P(A) P(B)$

I agree.

$1 - (1 - P(A)) (1-P(B)) = 1 - (1 - P(A) - P(B) + P(A)P(B)) = P(A) + P(B) - P(A)P(B)$

As chiro says, "order statistics" are the topic you should study if you want to tackle more complicated versions of the situation. For example: "In a sample of 4 values drawn from 4 independent random variables A,B,C,D, what is the probability that the second largest value is less than 1?"