Transformations and set mappings

Hi all,

I would be grateful if you guys could give me some help with the mapping of the support of the original variables to the support of the new ones.

Let x₁ and x₂ have the joint pdf \(\displaystyle \ \mathrm h(x_1,x_2) = 2e^{-x_1-x_2},\ 0<x_1<x_2<\infty,\text zero\ elsewhere.\)
Find the joint pdf y₁= 2x₁and y₂=x₂-x_1.

What I have done so far is take each inequality separately:

\(\displaystyle \displaystyle 0<x_1\ \Rightarrow 0<y_1,, \ x_1<x_2 \Rightarrow \frac{y_1}{2}<y_2 + \frac {y_1}{2} \Rightarrow 0<y_2.\)

But now, assuming what I have done so far is correct, how can we find the upper bounds? I can see from the solution that they are both infinity but I need to understand the intuition behind it.

In general what is the best way to handle such exercises? Are there any useful rules of thumb one can make use of?

Thanks again.

iocal said:

Hi all,

I would be grateful if you guys could give me some help with the mapping of the support of the original variables to the support of the new ones.

Let x₁ and x₂ have the joint pdf \(\displaystyle \ \mathrm h(x,y) = 2e^{-x_1-x_2},\ 0<x_1<x_2<\infty,\text zero\ elsewhere.\)

Click to expand...

You mean \(\displaystyle h(x_1, x_2)\) don't you?

Find the joint pdf y₁= 2x₁and y₂=x₂-x_1.

What I have done so far is take each inequality separately:

\(\displaystyle \displaystyle 0<x_1\ \Rightarrow 0<y_1,, \ x_1<x_2 \Rightarrow \frac{y_1}{2}<y_2 + \frac {y_1}{2} \Rightarrow 0<y_2.\)

But now, assuming what I have done so far is correct, how can we find the upper bounds? I can see from the solution that they are both infinity but I need to understand the intuition behind it.

In general what is the best way to handle such exercises? Are there any useful rules of thumb one can make use of?

Thanks again.

Click to expand...

Start by solving the transforation equations for \(\displaystyle x_1\) and \(\displaystyle x_2\) so you can substitute for them. if \(\displaystyle y_1= 2x_1\), then \(\displaystyle x_1= \frac{y_1}{2}\). If \(\displaystyle y_2= x_2- x_1= x_2- \frac{y_1}{2}\), \(\displaystyle x_2= y_2+\frac{y_1}{2}\). "\(\displaystyle 0< x_1< x_2< \infty\)" means that we have the lower half of the first quadrant, bounded by the lines \(\displaystyle x_2= 0\) and \(\displaystyle x_2= x_1\). The line \(\displaystyle x_2= 0\) gives \(\displaystyle y_2+ \frac{y_1}{2}= 0\), or \(\displaystyle y_2= -\frac{y_1}{2}\). Draw that line on a \(\displaystyle y_1y_2\) coordinate system. The line \(\displaystyle x_1= x_2\) gives \(\displaystyle \frac{y_1}{2}= y_2+ \frac{y_1}{2}\) or \(\displaystyle y_2= 0\), the "\(\displaystyle y_1\)" axis. The range for \(\displaystyle y_1\) and \(\displaystyle y_2\) is the area between those two lines.

HallsofIvy said:

bounded by the lines \(\displaystyle x_2= 0\) and \(\displaystyle x_2= x_1\).

Click to expand...

Don't you actually mean \(\displaystyle \ x_1=0\) ?

HallsofIvy said:

Start by solving the transforation equations for \(\displaystyle x_1\) and \(\displaystyle x_2\) so you can substitute for them. if \(\displaystyle y_1= 2x_1\), then \(\displaystyle x_1= \frac{y_1}{2}\). If \(\displaystyle y_2= x_2- x_1= x_2- \frac{y_1}{2}\), \(\displaystyle x_2= y_2+\frac{y_1}{2}\). "\(\displaystyle 0< x_1< x_2< \infty\)" means that we have the lower half of the first quadrant, bounded by the lines \(\displaystyle x_2= 0\) and \(\displaystyle x_2= x_1\). The line \(\displaystyle x_2= 0\) gives \(\displaystyle y_2+ \frac{y_1}{2}= 0\), or \(\displaystyle y_2= -\frac{y_1}{2}\). Draw that line on a \(\displaystyle y_1y_2\) coordinate system. The line \(\displaystyle x_1= x_2\) gives \(\displaystyle \frac{y_1}{2}= y_2+ \frac{y_1}{2}\) or \(\displaystyle y_2= 0\), the "\(\displaystyle y_1\)" axis. The range for \(\displaystyle y_1\) and \(\displaystyle y_2\) is the area between those two lines.

Click to expand...

I have tried substitution above but I do not understand how to derive the upper bounds. The line \(\displaystyle \ y_2=-\frac{y_1}{2}\) means that \(\displaystyle y_2\) is negative since there is no intercept, something we have ruled out. Unless you mean that that the range then becomes all positive numbers as there is no upper bound. Is that it? What about the upper bound for \(\displaystyle y_1\) then? I am really confused.