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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.

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  • $\begingroup$ $0<y_1<1,0<y_2<1$. 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
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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 $0<x_1<x_2<1$.

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