A subtle modification to the $t$ distribution Let $X$ and $Y\,$ be independent normal random variables with zero mean. What are the differences between the distributions of
$$A\triangleq \frac{X+Y}{X-Y} \quad\text{ and }\quad B\triangleq \frac{X+Y}{|X-Y\,|}\quad?$$
 A: 1. If the variances of $X$ and $Y$ differ, the distributions of $A$ and $B$ differ
To see this, note that $[B\gt0]=[X+Y\gt0]$ has probability $\frac12$ since the distribution of $X+Y$ is centered normal, hence symmetric. On the other hand, 
$$
[A\gt0]=[X^2\gt Y^2]=[\sigma^2U^2\gt \tau^2V^2],
$$
where $\sigma^2$ is the variance of $X$, $\tau^2$ is the variance of $Y$ and $U$ and $V$ are i.i.d. standard normal random variables. Hence $[A\gt0]=[U^2\gt (\tau^2/\sigma^2)V^2]$ has probability $\lt\frac12$ if $\tau^2\gt \sigma^2$ and $\gt\frac12$ if $\tau^2\lt \sigma^2$, but never exactly $\frac12$.
2. If the variances of $X$ and $Y$ coincide, the distributions of $A$ and $B$ coincide
To see this, note that, when the variances of $X$ and $Y$ coincide, $X+Y$ and $X-Y$ are i.i.d. hence $A$ is distributed as $U/V$ and $B$ as $U/|V|$, where $U$ and $V$ are i.i.d. standard normal random variables. Thus $A=SB$ where $S=\mathrm{sign}(V)$. Now, $S$ is independent on $(U,|V|)$ while $B$ is $(U,|V|)$-measurable, hence $S$ and $B$ are independent. Since the distribution of $B$ is symmetric, this implies that the distributions of $A=SB$ and $B$ coincide.
3. The distributions of $A$ and $B$
In the second case, the distribution of $A$ and $B$ is standard Cauchy. In the first case, introduce the parameter
$$
\varrho=\frac{\sigma^2-\tau^2}{\sigma^2+\tau^2}.
$$
Then, the same arguments as above show that $A$ and $B$ can be realized as 
$$
A=\varrho+\sqrt{1-\varrho^2}C,\qquad B=SA,
$$ 
where $S$ and $C$ are independent, $S=\pm1$ is a symmetric sign and $C$ is standard Cauchy. This explains why $A$ and $B$ are i.i.d. if and only if the distribution of $A$ is symmetric if and only if the distribution of $A$ is centered if and only if $\varrho=0$ if and only if $X$ and $Y$ have the same variance.
A: They have the same distribution if $X$ and $Y$ have the same variance.  It is not too difficult to see that $X>Y$ implies $A(X,Y) = A(-X,-Y) = B(X,Y) = B(Y,X)$ and $A(Y,X) = A(-Y,-X) = B(-X,-Y) = B(-Y,-X)$ and conversely $X<Y$ implies $A(X,Y) = A(-X,-Y) = B(-X,-Y) = B(-Y,-X)$ and $A(Y,X) = A(-Y,-X) = B(X,Y) = B(Y,X)$.  Since $(X,Y), (Y,X), (-X,-Y),$ and $(-Y,-X)$ all have the same probability distribution, $A$ and $B$ have the same distribution.
