Random Variables and Transformations

[Artusartos]

In the last question in this link:

http://pages.uoregon.edu/csinclai/teaching/Fall2009/files/hw8.pdf

1) I did not understand how they got the region for y1, y2, and y3...

2) How would the solution be different (or not possible) if X1, X2, and X3 were not iid?

Thanks in advance

 

[haruspex]

1) I did not understand how they got the region for y1, y2, and y3...

Clearly each is >0. Y1 = X1/(X1+X2), with X1 and X2 both > 0. So X1 < X1+X2, so Y1 < 1. Y1 approaches 1 as X2 approaches zero. Etc.

The proof omits the derivation of the exp[-(y1...] term. If you dig into that you'll find it needed the joint pdfs of the X vars, and it assumes they're iid in the process.

 

[Artusartos]

Clearly each is >0. Y1 = X1/(X1+X2), with X1 and X2 both > 0. So X1 < X1+X2, so Y1 < 1. Y1 approaches 1 as X2 approaches zero. Etc.

The proof omits the derivation of the exp[-(y1...] term. If you dig into that you'll find it needed the joint pdfs of the X vars, and it assumes they're iid in the process.

Thanks a lot :)

I have one more question (if you don't mind)...

Let f(x₁, x₂, x₃) = e^{-(x₁+x₂+x₃)}, 0<x_1,2,3<infinity, zero elsewhere be a joint pdf of X₁, X₂, X₃.

Compute P(X₁= X₂< X₃)

Determine the joint mgf.

So this is the integral that we need to integrate, right (for the mgf)?

\int_0^{\infty} \int_0^{\infty} \int_0^{\infty} e^{x_1(t_1-1)} e^{x_2(t_2-1)} e^{x_3(t_3-1)} dx_1 dx_2 dx_3

But I'm having some trouble with this, because this is what I get after integrating...

\frac{1}{t_1-1}e^{x_1(t_1-1)} |_0^{\infty} \frac{1}{t_2-1}e^{x_2(t_2-1)} |_0^{\infty} \frac{1}{t_3-1}e^{x_3(t_3-1)} |_0^{\infty}...but if we plug in infinity to \frac{1}{t_1-1}e^{\infty(t_1-1)}...don't we get infinity, since e increases forever?

 

[haruspex]

if we plug in infinity to \frac{1}{t_1-1}e^{\infty(t_1-1)}...don't we get infinity, since e increases forever?

Not if t1 < ... what?

 

[Artusartos]

Not if t1 < ... what?

If t1<1...

 

[haruspex]

If t1<1...

Right. It's quite usual that the mgf would only be defined for a range of the parameter, such as |t| < 1. To make use of it, you only need all the derivatives to exist at 0.

 

[Artusartos]

Right. It's quite usual that the mgf would only be defined for a range of the parameter, such as |t| < 1. To make use of it, you only need all the derivatives to exist at 0.

Thanks...