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:

$$ \sqrt(y_1)<1 $$

  • 4
    $\begingroup$ 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
  • 2
    $\begingroup$ @Did That should be an answer, not a comment! +1, of course. $\endgroup$ – Dilip Sarwate Nov 27 '15 at 14:52

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.