In this question, $X_1$ and $X_2$ are independent exponentially distributed random variables, each with parameter $\lambda$. I am asked to find the distribution of $\frac{X_2-X_1}{X_2+X_1}$.
I am not very comfortable with these types of questions just yet as you can probably tell. Can somebody explain this problem? I am getting 0 as a numerator but I am not confident in this answer.
Thanks in advance.
A routine approach goes like this:
Write down the joint density of $(X_1,X_2)$.
Transform $(X_1,X_2)\to (Y_1,Y_2)$ where $Y_1=\frac{X_2-X_1}{X_2+X_1}$ and $Y_2=X_2+X_1$.
Calculate the Jacobian of transformation and hence the joint density of $(Y_1,Y_2)$.
Integrate the joint density of $(Y_1,Y_2)$ wrt $y_2$ to find the marginal density of $Y_1$.
Alternatively, you could find the distribution function of $Y_1=\frac{X_2-X_1}{X_2+X_1}$, namely $P(Y_1\leqslant t)$.
For $|t|<1$, that last probability simplifies to
\begin{align} P\left(\frac{X_2}{X_1}<\frac{1+t}{1-t}\right)&=\iint_{\frac{y}{x}<\frac{1+t}{1-t}}\lambda e^{-\lambda x} \lambda e^{-\lambda y}\,\mathrm{d}x\,\mathrm{d}y \\&=\int_0^\infty \lambda e^{-\lambda x}\left(\int_0^{\frac{1+t}{1-t}x} \lambda e^{-\lambda y}\,\mathrm{d}y\right)\mathrm{d}x \\&=\frac{t+1}{2} \end{align}
This is the distribution function of a uniform distribution on $(-1,1)$ for $|t|<1$.
