# How to integrate this one P(x1<x2<x3<1)

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1. Oct 15, 2014

### Jon08

1. The problem statement, all variables and given/known data
Let f(x1, x2, x3) = e-(x1+x2+x3), 0<x1,2,3<infinity, zero elsewhere be a joint pdf of X1, X2, X3. The variables are all independent to each other

Compute P(X1< X2< X3|X3<1 )

2. Relevant equations

P(X1< X2< X3|X3<1 )
3. The attempt at a solution
P(X1< X2< X3|X3<1 )=P(X1< X2< X3<1 )/ P(X3<1)
=triple integral of f(x1, x2, x3) dx1dx2dx3 as x1 goes from 0 to 1,x2 goes from x1 to x3, and x3 goes from x2 to 1.

2. Oct 15, 2014

### Simon Bridge

Welcome to PF;
How come x1 does not go from 0 to x2?

Do you not know how to do $\int e^x\;dx$?

3. Oct 15, 2014

### Jon08

wait let me see my solution again.thanks

4. Oct 15, 2014

### Ray Vickson

Hint 1: what is $P( \max(X_1,X_2,X_3) < 1)$?
Hint 2: what is the relation of the quantity in Hint 1 to the desired quantity $P(X_1 < X_2 < X_3 < 1)$?

5. Oct 15, 2014

### Jon08

x1 goes from 0 to x2,x2 goes from x1 to x3, and x3 goes from x2 to 1. Is my limits for the integration correct? Thanks

6. Oct 15, 2014

### haruspex

Because, as stated in the same line, x2 goes from x1 to x3.
Jon08, when you list the ranges in a multiple integral, you need them in a sequence such that no integral range refers to the variable of an earlier range. I.e. if you integrate in the order x3, x2, x1 then the range for x1 cannot refer to x2 or x3, and that for x2 cannot refer to x3.
Other than that, you can generally integrate in any order you like, but the conditional x3 < 1 makes some orders better than others here.

7. Oct 15, 2014

### RUber

The x2<x3 is covered by putting x3 as the upper limit of integration on x2. You do not need to restate x2 as the lower limit of integration for x3. Same argument for x1. You are okay to use 0 as the lower limit on all three integrals.

8. Oct 15, 2014

### haruspex

... so long as the order is x1, x2, x3, right?

9. Oct 15, 2014

### RUber

Correct, I assumed the order of smallest to largest.