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$X$ and $Y$ are random variables with joint pdf $$\frac{1}{x^2y^2}\qquad,\, x\ge1,y\ge1$$

Set $$U=XY, V=X/Y$$

Explain why the joint range of $U$ and $V$ is given by:

$$\{(u,v):v\in(0,1),u\ge1/v\} \cup \{(u,v):v∈[1,\infty),u\ge v\}$$

I don't understand what this question wants. Can anyone explain to me how to approach this? In response to another bit of the same question, I have successfully shown that the joint pdf of $U$ and $V$ is $$\frac{1}{2u^2v}$$ by going through and subbing in the new variables and obtaining the Jacobian, and then gone on to obtain the marginal of $V$ (I think $1/2v$) but this aspect of the question in relation to range is stumping me.

Any general advice (given in simple terms!) would be welcomed.

PS the question also asks for a sketch.

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  • $\begingroup$ The joint support of $(U,V)$ is precisely what's needed to write the joint density of $(U,V)$ in the first place. $\endgroup$ – StubbornAtom Dec 1 '18 at 22:03
  • $\begingroup$ You might want to add the self-study tag. $\endgroup$ – StubbornAtom Dec 2 '18 at 12:14
  • $\begingroup$ Done and thank you all for the help. It was very useful. $\endgroup$ – StatisticsPersonInTraining Dec 2 '18 at 12:17
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Basic idea:

Draw two graphs, one for X vs Y, another for U and V.

Find the support of X and Y joint distribution (which is $(x\ge 1) \cap (y \ge 1)$).

Convert the points in X-Y support into U-V plain by the relation $U=XY$ and $V=X/Y$, you get the support of U-V joint distribution. For example, (X,Y) = (1,1) ==> (U,V) = (1,1), (X,Y) = (2,2) ==> (U,V) = (4,1).

In your situation, there are two lines $(1,1)$ to $(1,\infty)$ and $(1,1)$ to $(\infty,1)$ in X-Y graph, and you need to convert these two lines from X-Y graph to U-V graph, then you get the answer.

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