Bivariate Transformation

Question

Why can I not let $V=X$ in this transformation as opposed to $V=Y$? I have tried it with $V=X$ and i get a different joint pdf.

Answer 1

With $U=X/Y$ and $V=X$, you have $X=V$ and $Y=V/U$. The different inverse transformation should lead you to expect a different joint pdf, but the resulting calculation is essentially unchanged. Specifically:

$$J(U,V)=\begin{vmatrix}0&1\\-vu^{-2}&u^{-1}\end{vmatrix}=vu^{-2}$$

and hence

$$ f_{U,V}(u,v) =f_{X,Y}(v,vu^{-1})\left|vu^{-2}\right| =\frac1{2\pi}\exp\left\{-\frac12(v^2+v^2u^{-2}))\right\}\left|vu^{-2}\right|. $$

Now, for any real $u$,

\begin{align*} f_U(u)=\int_{-\infty}^\infty f_{U,V}(u,v)dv &=\int_{-\infty}^\infty\frac1{2\pi}\exp\left\{-\frac12(v^2(1+u^{-2}))\right\}\left|vu^{-2}\right| dv \\ &=\int_0^\infty\frac1\pi\exp\left\{-\frac12(v^2(1+u^{-2}))\right\}vu^{-2} dv \qquad\text{as integrand is even function}\\ &=\frac1\pi\left[ \frac{-\exp\{-\frac12(v^2(1+u^{-2}))\}}{u^2(1+u^{-2})} \right]_0^\infty \qquad\text{by direct integration}\\ &=\frac1\pi\left[ \frac1{u^2(1+u^{-2})} \right] =\frac1{\pi(1+u^2)}. \end{align*}

In the end, the only real difference is that you have an annoying $(1+u^{-2})$ that hangs around until the last step.