Let $X_1$ and $X_2$ be independent exponential random variables with parameters $λ_1$ and $λ_2$ respectively. What is the probability density of $X_1/X_2$?

I tried to integral form $-\infty$ to $+\infty$ but its seems I cannot work out a results. If it need some other method?

  • $\begingroup$ Sometimes "exponential with parameter $\alpha$" means $\displaystyle e^{-x/\alpha}\,\left( \frac{dx}\alpha \right)$ for $x\ge0,$ so that $\alpha$ is the expected value, and sometimes it means $\displaystyle e^{-\alpha x} (\alpha \, dx),$, for $x\ge0,$ so that $\alpha$ is the rate and $1/\alpha$ is the expected value. Which of those do you have in mind? $\qquad$ $\endgroup$ – Michael Hardy Nov 28 '16 at 15:58
  • $\begingroup$ "I tried to integral form $-\infty$ to $+\infty$ but its seems I cannot work out a results" Excellent! Please show the details of your computations. $\endgroup$ – Did Nov 28 '16 at 16:02
  • $\begingroup$ Note that $Y = X_1/X_2 \le y$ is a sector of the first quadrant bounded by the $x_2$-axis and a diagonal line. Integrating the joint density over that sector gives you the value of the cumulative distribution function at $y.\qquad$ $\endgroup$ – Michael Hardy Nov 28 '16 at 16:02
  • $\begingroup$ Also note that $\Pr(X_1/X_2 >0) = 1$, so you shouldn't expect negative numbers in the region over which you need to integrate. $\qquad$ $\endgroup$ – Michael Hardy Nov 28 '16 at 16:03
  • $\begingroup$ math.stackexchange.com/questions/33778/… $\endgroup$ – StubbornAtom Dec 3 '18 at 17:50

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