# bivariate transformation when U=X/Y and Y=0

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.

• 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. – whuber May 24 '19 at 12:29

@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 $$-\infty and $$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.