Random variables problem

[player]

Hello all. I'm having a lot of trouble with several of the problems on my latest stat homework. Up until now I've understood almost everything, but this is new assignment is going over my head. Here's the one that's giving me the most trouble.

Let X1, X2, and X3 be exponentially distributed random variables with expectations 2, 3, and 4 respectively. If the three random variables are independent, calculate

1. P(min{X1,X2,X3} > 5)

2. P(max{X1,X2,X3} > 5)

3. E[min{X1,X2,X3}]

I realize that this site asks for an attempt first, but I have been trying to figure this one out for several hours now and I just don't know what to do with it. I think my biggest issue is the max and mins. I searched through the entire chapter for anything that resembles this and came up empty. I really want to learn how these work because I'm positive that something like this will be on the final. Thanks for any help!

[Dason]

For question 1 it helps to realize that for the min to be greater than 5 that ALL of the random variables need to be greater than 5. Since they're independent the joint probability of that occurring is just the product of the marginal probabilities of that event.

[player]

Thank you, that makes perfect sense. By using that logic, would the second one just be 1 - P(max{X1,X2,X3} <= 5)? It also seems to me that the 3rd one should just be the min expectation of X1, X2, and X3 which would be 2.

[BGM]

The minimum function is not linear; you cannot exchange the order of
operation with expectation in general. i.e.



For question 3, you may wish to obtain the probability density function
of first, or more directly, use



as the support of the minimum of exponentials are also positive.

[player]

I tried the suggestion for part 3. If I ran the calculation correctly, the answer is 0.923. That seems wrong to me since the expectation of all three variables is above 2.

[Dason]

But shouldn't you expect that the expectation of the minimum of a bunch of random variables is less than the minimum of the expectations?

Also the result you get seems to be reasonable according to some quick simulation in R

> n<-100000
> dat<-data.frame(rexp(n,1/2),rexp(n,1/3),rexp(n, 1/4))
> mean(apply(dat,1,min))
[1] 0.921456

[BGM]

I tried the suggestion for part 3. If I ran the calculation correctly, the answer is 0.923. That seems wrong to me since the expectation of all three variables is above 2.

Actually if you know the Jensen's Inequality, you can prove the inequality
as well.