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I am considering using transformation U=X/Y and V=Y (X,Y are iid distributions defined on real line). Almost all the textbooks I have read did not consider the case when Y=0, at which U is undefined.

Also, sometimes we have U=XY and V=Y. Its inverse mapping is X=U/V and Y=V. And X is undefined when V=0.

So why can we ignore these cases?

I know the Jacobian method is derived from real analysis. But my analysis textbook only consider the transformation on a closed area. And the transformation mentioned above is defined on ($-\infty$,$+\infty$) . Is this the reason?

Thanks a lot.

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  • $\begingroup$ You cannot ignore these cases: you still have to make sure that they occur with zero probability, for otherwise the transformation is undefined. Usually--depending on the joint distribution of $(X,Y)$--that analysis is so simple and obvious that it is not even mentioned. $\endgroup$ – whuber May 24 '19 at 12:29
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@whuber's answer is right. I have thought a little bit more and have the following proof.

As in the question,

X,Y~iid and $-\infty<X,Y<+\infty$ and $U=\frac{X}{Y}$ , $V=Y$.

So loosely speaking,

$$ \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f(x,y)dydx = $$

$$ \lim_{\epsilon\to0}\bigg(\int_{-\infty}^{+\infty}\int_{-\infty}^{0-\epsilon}f(x,y)dydx + \int_{-\infty}^{+\infty}\int_{0-\epsilon}^{0+\epsilon}f(x,y)dydx+ \int_{-\infty}^{+\infty}\int_{0+\epsilon}^{+\infty}f(x,y)dydx\bigg)= $$

$$ \lim_{\epsilon\to0}\bigg(\iint_{\{(u,v):u=x/y,v=y\leqslant-\epsilon\}}f(u,v)|J|dudv+0+\iint_{\{(u,v):u=x/y,v=y\geqslant\epsilon\}}f(u,v)|J|dudv\bigg)= $$

$$ \lim_{\epsilon\to0}\bigg(\iint_{\{(u,v):u=x/y,v=y\leqslant-\epsilon\}\cup\{(u,v):u=x/y,v=y\geqslant\epsilon\}}f(u,v)|J|dudv\bigg) $$

The middle term is zero because the inner integration is 0. This is because $f(x,y)$ is continuous with respect to y and its integral exists on a closed interval $[-\epsilon,\epsilon]$. Thus, its original function $F(x,y)$ exists and $\lim_{\epsilon\to0}\big(\int_{0-\epsilon}^{0+\epsilon}f(x,y)dy\big) = \lim_{\epsilon\to0}\big(F(\epsilon)-F(-\epsilon)\big)=0$.

After taking the limit, the rest term is exactly the transformation without considering the case when Y=0. And in the probabilistic language, the inner integration equals to zero means the probability of that event is zero.

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