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.

  • $\begingroup$ Better to show whatever you have tried. $\endgroup$ – StubbornAtom Apr 30 '18 at 12:13
  • $\begingroup$ 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 $\frac{X_1-X_2}{X_1-X_2}$? $\endgroup$ – user12321 Apr 30 '18 at 14:40
  • $\begingroup$ Among other ways, one can try to find this using the transformation method involving jacobians. Haven't you done this before? $\endgroup$ – StubbornAtom Apr 30 '18 at 14:56
  • $\begingroup$ I have never heard my lecturer utter the word 'Jacobian' before in a lecture? My method is wrong? $\endgroup$ – user12321 Apr 30 '18 at 14:57
  • $\begingroup$ 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. $\endgroup$ – 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 $|t|<1$.

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