# How do I find the limits of a joint density function and calculate the inequalities?

The r.v.'s ${X_1}$ and ${X_2}$ are independent and equidistributed with density function $$f_X(x)=4x^3, 0 \le x \le 1,$$ and equal to zero otherwise. Set ${Y_1=X_1\sqrt(X_2)}$ and ${Y_2=X_2\sqrt(X_1)}$. Determine the joint density function of ${Y_1}$ and ${Y_2}$.

I have managed to calculate the joint density function using the transformation THM (change of variable) and got that it is equal to $$f_Y{_1},_Y{_2}(y_1,y_2)=[indpt.]=f_X{_1}((y_1^4/y_2^2)^{1/3})*f_X{_2}((y_2^4/y_1^2)^{1/3})*|{\mathbf J}|$$ where the Jacobian is $$|{\mathbf J}|=4/3*y_1^{-1/3}*y_2^{-1/3}$$ which gives me the joint density $$f_Y{_1},_Y{_2}(y_1,y_2)=64/3*(y_1*y_2)^{5/3}.$$ This is correct. However I am confused regarding how to find the "limits" where ${f_Y{_1},_Y{_2}(y_1,y_2)}$ is not zero. The answer is supposed to be $$0 < y_1^2<y_2<\sqrt(y_1)<1$$ But I do not know how to calculate it. How do I find this? Any help much appreciated.

Edit: Thanks for the comments. Could someone please just clarify by showing how to get that:

$$\sqrt(y_1)<1$$

• Note that $$x_1=y_1^{4/3}y_2^{-2/3}\qquad x_2=y_2^{4/3}y_1^{-2/3}$$ hence the domain of integration is $$0<y_1^{4/3}y_2^{-2/3}<1\qquad 0<y_2^{4/3}y_1^{-2/3}<1$$ Now simplify these inequalities to $$0<y_1^2<y_2\qquad 0<y_2^2<y_1$$ and conclude. – Did Nov 27 '15 at 14:30
• @Did That should be an answer, not a comment! +1, of course. – Dilip Sarwate Nov 27 '15 at 14:52

$$0<y_1^2<y_2$$ and $$0<y_2^2<y_1.$$ By rewriting the second equation we get $$0<y_2<\sqrt(y_1)$$ which we can add to the first equation above. This gives us: $$0<y_1^2<y_2<\sqrt(y_1)$$
Now, the relationship that ${ y_1^2<\sqrt(y_1)}$ implies that ${ y_1<1}$ which means that ${\sqrt(y_1)}$ can never really become equal to 1, hence ${\sqrt(y_1)<1}$. And thus we have the sought inequality, $$0 < y_1^2<y_2<\sqrt(y_1)<1.$$ Hope this helps someone else.