How to find this particular probability?

[dexterdev]

Homework Statement

Suppose there are three statistically i.i.d continuous random variables X1, X2, X3 each are uniformly distributed in the range [0,1]. How to find the probability P(X1+X2<X3)?

Homework Equations

The below given equations are the steps to the solution. But I can't figure out how the limits of integral comes this way.

$\int_0^1 \int_0^{x_3}\int_0^{x_3-x_2} \,dx_1\,dx_2\,dx_3 =\int_0^1 \int_0^{x_3} (x_3-x_2) dx_2\,dx_3 = \int_0^1 x_3^2 - \frac{x_3^2}{2}\,dx_3 = \frac16 = 0.1\overline 6$

The Attempt at a Solution

I tried this using a software called MATLAB by generating three pseudo random variables (1000 samples) and finding X1+X2−X3 and plotting its CDF through a MATLAB tool called dfittool. I got the answer around 0.169. But how do I do this theoretically? Especially how to figure out the limits in those integrals?

 

[Ray Vickson]

Your questions make no sense: you have already obtained the answer theoretically, and you have already written the limits of integration.

 

[dexterdev]

@Ray Vickson : Yes I have got those limits from someone else, but never told how they come?

 

[D H]

They come from two different concerns, that x₁+x₂<x₃ and that each x_i must be between 0 and 1. Those integration limits represent the intersection of those concerns.

 

[HallsofIvy]

You are told that the three variables all lie in [0, 1]. The limits on the outer integral, with respect to $x_3$ must be constants so must be 0 and 1. The next inner integral can have limits depending on $x_3$. Since we have $x_1+ x_2< x_3$ and $x_1$ can be 0, $x_2$ can go from 0 to $x_3$. Finally, $x_1+ x_2< x_3$ means that $x_1< x_3- x_2$ so the inmost integral has limits of 0 to $x_3- x_2$.