# Distribution of $\frac{X_2-X_1}{X_2+X_1}$ when $X_1$ and $X_2$ are i.i.d exponential variables

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.

• Better to show whatever you have tried. – StubbornAtom Apr 30 '18 at 12:13
• I have attempted the question again. I have that that the $X_1+X_2$ has a gamma (2,$\lambda$) density. Do I now just find the density of $X_1-X_2$ and divide to get the density for $\frac{X_1-X_2}{X_1-X_2}$? – user12321 Apr 30 '18 at 14:40
• Among other ways, one can try to find this using the transformation method involving jacobians. Haven't you done this before? – StubbornAtom Apr 30 '18 at 14:56
• I have never heard my lecturer utter the word 'Jacobian' before in a lecture? My method is wrong? – user12321 Apr 30 '18 at 14:57
• You can find the densities of $X_1+X_2$ and $X_1-X_2$, but you cannot just divide the densities to find the required distribution. – StubbornAtom Apr 30 '18 at 15:01

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 $$|t|<1$$.