# Confuse in Determining Domain of Random Variable Transformation.

Suppose $$Y_1$$ and $$Y_2$$ are two random variable have joint probability density function $$f(y_1,y_2)= \begin{cases} 4y_1y_2&0 If $$Y_1=X_1^2$$ and $$Y_2=X_1X_2$$, determine joint p.d.f. of $$X_1$$ and $$X_2$$.

I use jacobian transformation, so I find the Jacobian determinant as below. $$|J|=\left\vert \begin{matrix} \dfrac{\partial Y_1}{\partial X_1}&\dfrac{\partial Y_1}{\partial X_2}\\ \dfrac{\partial Y_2}{\partial X_1}&\dfrac{\partial Y_2}{\partial X_2}\\ \end{matrix} \right\vert = \left\vert \begin{matrix} 2X_1&0\\ X_2&X_1\\ \end{matrix} \right\vert = 2X_1^2.$$

Now, I find the joint p.d.f. as below. $$\begin{eqnarray} g(x_1,x_2)&=&f(y_1,y_2)\vert J\vert\\ &=& f\left(x_1^2,x_1x_2\right)\vert J\vert\\ &=& 4(x_1^2)(x_1x_2)(2x_1^2)\\ &=& 8x_1^5 x_2 \end{eqnarray}$$

After that, I determine the domain of $$x_1$$ and $$x_2$$.

$$Y_1=X_1^2$$ imply $$X_1=\sqrt{Y_1}$$ and $$Y_2=X_1X_2$$ imply $$X_2=\dfrac{Y_2}{\sqrt{Y_1}}$$.

$$X_1=\sqrt{Y_1}$$ and $$0, it's clear that $$0. Now I'm confused to determining the range of $$Y_2$$.

$$X_2=\dfrac{Y_2}{\sqrt{Y_1}}$$, the denominator range is $$0$$ into $$1$$. the numerator range is $$0$$ into $$1$$. If numerator $$>$$ denominator, the range of $$X_2$$ is $$>1$$. But if denominator $$>$$ numerator, the range of $$X_2$$ is $$0$$ into $$1$$. So, the range of $$X_2$$ is $$X_2>0$$. Is it right?

Based on the result of joint pdf of $$X_1$$ and $$X_2$$, $$g(x_1,x_2)=8x_1^5 x_2$$, the double integral of $$g(x_1,x_2)$$ is $$\infty$$, cannot equal 1. So, the range of $$X_2$$ are wrong.

So how to determine the range of $$X_1$$ and $$X_2$$ on this problem?

• Isn´t the joint pdf of $Y_1$ and $Y_2$ given with $$f(y_1,y_2)= \begin{cases} 4y_1y_2&0<y_1<1,0<y_2<1\\ 0&\text{elsewhere} \end{cases}.$$ ? Very confusing!? – callculus Jul 13 at 16:04
• I guess that the first few lines: Suppose $Y_1$ and $Y_2$ are two random variable have joint probability density function $$f(y_1,y_2)= \begin{cases} 4y_1y_2&0<y_1<1,0<y_2<1\\ 0&\text{for other }x \end{cases}.$$ If $Y_1=X_1^2$ and $Y_2=X_1X_2$, determine joint p.d.f. of $Y_1$ and $Y_2$ would have to be changed to Suppose $X_1$ and $X_2$ are two random variable have joint probability density function $$f(x_1,x_2)= \begin{cases} 4x_1y_2&0<y_1<1,0<x_2<1\\ 0&\text{for other }x \end{cases}.$$ If $Y_1=X_1^2$ and $Y_2=X_1X_2$, determine joint p.d.f. of $Y_1$ and $Y_2$. – zoli Jul 13 at 22:34
• Yes, I mean $4y_1y_2$. Mis-typing – Ongky Denny Wijaya Jul 14 at 2:02

The domain is $${\quad\{\langle x_1,x_2\rangle: (0\leq x_1^2\leq 1)\land (0\leq x_1x_2\leq 1)\}\\={\{\langle x_1,x_2\rangle: ({{(-1\leq x_1<0)\land(-1/x_1\leq x_2\leq 0))}\lor{((0\leq x_1\leq 1)\land(0\leq x_2\leq 1/x_1)}})\}}}$$