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Let $X$ and $Y$ be independent exponential random variable with rates $\lambda_1$, $\lambda_2$, respectively. What is the cumulative distribution function of $\frac{X}{Y}$?

I know that the joint density function is $f(x,y) = (\lambda_1 e^{-\lambda_1 x})(\lambda_2 e^{-\lambda_2 x})$

can I just take the integral of $\frac{(\lambda_1 e^{-\lambda_1 x})} {(\lambda_2 e^{-\lambda_2 x})}$ to find the c.d.f of $\frac{X}{Y}$?

user59036
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    No, you cannot take the integral that you propose. For $a > 0$, you need to find $P{X/Y \leq a} = P{X \leq aY}$. Can you find the region in the first quadrant where the coordinates $(x,y)$ satisfy $x \leq ay$? Can you integrate the joint density function over this region? – Dilip Sarwate Mar 18 '13 at 00:34
  • $P(X/Y \leq a) = P(X \leq aY) = $$\iint_{x < ay} (\lambda_1 e^{-\lambda_1 x})(\lambda_2 e^{-\lambda_2 x}) dx dy$ = $\int_{y=0}^{\infty} \int_{x=0}^{ay} (\lambda_1 e^{-\lambda_1 x})(\lambda_2 e^{-\lambda_2 x}) dx dy$ Am I on the right track? – user59036 Mar 18 '13 at 01:05
  • You are (except the two $e^{-\lambda_2x}$ which should read $e^{-\lambda_2y}$). Can you carry on? – Did Mar 18 '13 at 20:12
  • @Did I have asked a similar question here. I have used the approach proposed here and have tried to calculate the CDF in my case. Could you please check it and give me your feedback (or even you can answer the question there). Thank you! – din Dec 20 '16 at 10:52
  • @din See there. – Did Dec 20 '16 at 16:28

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