How do I calculate an area of joint uniform distribution with domain

[mattclgn]

This technically isn't a coursework or homework problem:

I have a uniform Joint density function for the lifetimes of two components, let's call them T1 and T2. They have a uniform joint density function, both are positive it follows, and the region is 0<t1<t2<L and L is some positive constant.

So it said the domain has area (L^2)/2 following from this, Density is 1/(((L^2)/2)-1) or uniform density. I have no idea how to get that and area. Can someone help explain? I think I get that since joint I treat one as y and other x i multiply both densities but not really sure.

 

[RUber]

From that fact that 0 < T1 < T2 < L, this defines a triangular region (let T1 = x, T2 = y) that is half a L x L square. So the area of this region is L^2/2.
The density is uniform, and must sum to one over the region.

The density you posted $\frac{ 1}{ \frac{L^2}{2} -1} = \frac { \frac1A}{ 1 - \frac 1A }.$ where A = L^2/2, the area of the region.

I know that this is related to the condition that x < y, but I will need to look into it a little bit more to explain why.

 

[mattclgn]

From that fact that 0 < T1 < T2 < L, this defines a triangular region (let T1 = x, T2 = y) that is half a L x L square. So the area of this region is L^2/2.
The density is uniform, and must sum to one over the region.

The density you posted $\frac{ 1}{ \frac{L^2}{2} -1} = \frac { \frac1A}{ 1 - \frac 1A }.$ where A = L^2/2, the area of the region.

I know that this is related to the condition that x < y, but I will need to look into it a little bit more to explain why.

Awesome, thank you, and i apologize for the lateness of this reply...so by that rationale if it was 0 < T1 < T2 <T3< L, do I divide by three?

 

[RUber]

Awesome, thank you, and i apologize for the lateness of this reply...so by that rationale if it was 0 < T1 < T2 <T3< L, do I divide by three?

You will be in a 3D space, so you would expect L^3 divided by something. Drawing it out helps, but may lead to the wrong guess for the fraction of the cube in your feasible region.
You can also integrate to see the volume of the 3D shape.
$\int_0^L\int_0^{T3}\int_0^{T2} 1 dT1 dT2 dT3$
Note that if you did the same integral in xy, you get L^2/2.