Joint Probability Density Functions under change of variables when one change is identity - meaning of it?

Question

Suppose random variables $X_{1}$ and $X_{2}$ have some joint probability density function (pdf) $f(x_{1}, x_{2})$. I wish to find probability density of $Y_{1} = \frac{X_{1}}{X_{2}}$. The textbooks say I can define an additional variable $Y_{2} = X_{2}$ and proceed.

My question, what if I define an additional variable in some other way, i.e. $Y_{2} = g(X_{1}, X_{2})$ for some function $g$? Then the pdf for $Y_{1}$ would be different, i.e. in general it is not unique...Why do we have this freedom in choosing an additional variable? What is the meaning of this?

Answer 1

Surely these 2 integrals corresponding to marginal densities will be different,

Sure that it is false.

Given that you are not responding, I will show you how to do assuming a coherent support.

Let's assume that your joint density is the following

$$f_{XY}(x,y)=8xy$$

where $0<x<y<1$

You have to derive the density of $Z=\frac{X}{Y}$. First observe that $Z \in (0;1)$

• Using the auxiliary variable as $U=X$ you will find the marginal Z in the following way

$$f_Z(z)=\int_0^z \frac{8u^3}{z^3}du=2z$$

• Using the auxiliary variable as $U=Y$ you will find the marginal Z in the following way

$$f_Z(z)=\int_0^18zu^3du=2z$$

as you can see the result is the same... I let you to understand the different integral bounds as an exercise.

Another useful exercise is to write down the joint density $f_{UZ}(u,z)$ in both cases, writing down also its joint support.

Answer 2

what about this example. Let $f(x_{1}, x_{2}) = 8x_{1}x_{2}$. Define $Y_{1} = \frac{X_{1}}{X_{2}}$ and $Y_{2} = X_{2}$. Then Jacobian of transformation is simply $y_{2}$ and we have

$f_{Y_{1}, Y_{2}}(y_{1}, y_{2}) = 8y_{1}y_{2}^{3},$ and marginal density

$f_{Y_{1}}(y_{1}) = \int 8y_{1}y_{2}^{3} dy.$

Now suppose instead I take $Y_{1} = \frac{X_{1}}{X_{2}}$ and $Y_{2} = X_{1}$. Then Jacobian is $\frac{y_{2}}{y_{1}}$, and we have

$f_{Y_{1}, Y_{2}}(y_{1}, y_{2}) = 8\frac{y_{2}^{3}}{y_{1}^{3}}$, and marginal density

$ f_{Y_{1}}(y_{1}) = \int \frac{y_{2}^{3}}{y_{1}^{3}} dy.$

Where integrals are taken over the domain of definition of $f$. Surely these 2 integrals corresponding to marginal densities will be different, regardless of the domain of definition of $f$?

Answer 3

Extended comment on the accepted answer for dummies (consider upvoting the actual accepted answer if anything):

The change of random variables is $Z=\frac X Y$ and $U=X$ in the joint pdf $f_{XY}(x,y)=8xy$ with $0<x<y<1$ results in

$$f_{ZU}(z,u)=\frac{f_{XY}(x,y)}{\text{Jacobian}}=\frac{f_{XY}(x,y)}{\begin{vmatrix}\partial z/\partial x&\partial z/\partial y\\\partial u/\partial x& \partial u/\partial y\end{vmatrix}} {\text{where }x=u\text{ and }y=\frac u z}$$

The Jacobian is

$$\begin{vmatrix}\partial z/\partial x&\partial z/\partial y\\\partial u/\partial x& \partial u/\partial y\end{vmatrix}=\begin{vmatrix}\frac 1 y &-\frac x{y^2} \\1& 0\end{vmatrix}=\frac x {y^2}$$

It follows that

$$f_{ZU}(z,u) = \frac{8 u \frac u z}{\frac{u}{\left(\frac{u}{z}\right)^2}}=\frac{8u^3}{z^3}$$

and then it goes on to getting the marginal pdf of $Z$ from this new joint pdf as

$$f_Z(z)=\int_0^z \frac{8u^3}{z^3}du=2z$$

---

Likewise if the change of random variables is $Z=\frac X Y$ and $U=Y$

$$\begin{vmatrix}\frac{1}{y}&-\frac x{y^2}\\0& 1\end{vmatrix}=\frac 1 y$$

and $f_{ZU}(z,u)=\frac{8zu^2}{1/u}=8zu^3.$