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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:

$$ \sqrt(y_1)<1 $$

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    $\begingroup$ 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. $\endgroup$ – Did Nov 27 '15 at 14:30
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    $\begingroup$ @Did That should be an answer, not a comment! +1, of course. $\endgroup$ – Dilip Sarwate Nov 27 '15 at 14:52
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I think I figured it out.

Since Did's comments below the stated question illustrates how to get

$$ 0<y_1^2<y_2 $$ 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.

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