I am having trouble finding the joint distribution of the following.

Joint distribution of $Y_1=X_1/X_2, \quad Y_2=X_2$ when $h(x_1,x_2) = 8x_1x_2$ when $0 < X_1 < X_2 < 1$.

I think I am having trouble with the support for the Ys.

I am thinking that the joint pdf itself is $$f(y_1,y_2)=8y_1y_2^3$$

and I am confident that $0 < Y_1 < 1$.

However, I am having trouble with the region where $Y_2$ should be.

So far I know that $$0 < Y_1Y_2 < Y_2 < 1$$

but I am not sure how to manipulate this to isolate $Y_2$.

My ultimate goal is to find the marginal distributio of $Y_2$, but I am stuck.

I would really appreciate your input.

  • $\begingroup$ $0<y_1<1,0<y_2<1$. No other restrictions on $y_1,y_2$ so the density is $8y_1y_2^{3}$ for $0<y_1<1,0<y_2<1$. $\endgroup$ – Kavi Rama Murthy May 16 at 6:21
  • $\begingroup$ I am not really confident why it works like that. For example, if I want to find the marginal of $X_2$ then I would have to integrate from $x_1$ to 1, but how does $0 < Y_1Y_2 < Y_2 < 1$ simply turns into $0 < Y_2 < 1$? To be more precise, what happened to the $Y_1$ ? $\endgroup$ – hyg17 May 16 at 6:34

When $y_2=x_2$ and $y_1=\frac {x_1} {x_2}$ the inequalities $0<x_1<x_2<1$ and $0<y_1<1,0<y_2<1$ are equivalent. You can verify that each set of inequalities implies the other.

For example, when the second set of inequalities are satisfied note that $x_1=y_1 y_2$ and $y_2=x_2$ so we have $0<x_1<x_2<1$.


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.