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Let $X,Y$ be Exponential random variables with parameter alpha and beta.

Appendix: Evaluation of the integral: \begin{align} f_{X/Y}(u) = {} & \alpha\beta \int_0^\infty e^{-(\alpha u+\beta)y} y \, dy \\[8pt] = {} & \frac{\alpha\beta}{(\alpha u+\beta)^2} \int_0^\infty e^{-(\alpha u+\beta)y} (\alpha u+\beta) y \big( (\alpha u + \beta) \, dy \big) \\[8pt] = {} & \frac{\alpha\beta}{(\alpha u+\beta)^2} \int_0^\infty e^{-v} v \,dv \\[8pt] = {} & \frac{\alpha\beta}{(\alpha u+\beta)^2}. \end{align}

I know that $E[X] = \int_{-\infty}^\infty x f(x) dx$. I want to use this formula but I know what $f_{\frac{X}{Y}}$ is without knowing what "$X$" is.

I don't know what $"x"$ is when if I were to find the $E[\frac{X}{Y}]$ by using the formula $E[X] = \int_{-\infty}^\infty x f(x) dx$. What am I letting X and Y be in the integral?

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$$\mathbb E\left[\frac{X}{Y}\right]=\int uf_{\frac{X}{Y}}(u)du$$

drhab
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  • Is the range of t either from x/y to infinity or 0 to x/y? @drhab –  Feb 17 '20 at 19:24
  • What do you mean with "the range of t"? That is not clear to me because I cannot find any "t" anywhere in question or answer. If you wonder about $f_{X/Y}(u) $ then I can say that it takes the value mentioned in your answer if $u>0$ and takes value $0$ otherwise. This because it is the PDF of a random variable that only takes positive values and can take any positive value. – drhab Feb 17 '20 at 20:41
  • I meant the limits of integration for u @drhab –  Feb 17 '20 at 21:04
  • What are the limits of integration for u? Are the limits of integration for u independent of y and x? –  Feb 17 '20 at 21:05
  • In principal there are no limits of integration. If $U$ is a random variable and $f_U$ denotes its PDF (which is a function with domain $\mathbb R$) then $\mathbb EU=\int uf_U(u)du$ or - if you like that better - $\mathbb EU=\int_{-\infty}^{\infty}uf_U(u)$. It might be that $f_U$ takes value $0$ on e.g. $(-\infty,0]$ and in that case the integral can also be written as $\int_0^{\infty}uf_U(u)du$. In your case $U=X/Y$ and its PDF is known. This leads to the equality in my answer. We have nothing to do with any $y$ or $x$. The only thing we need is the PDF of $U=X/Y$. – drhab Feb 18 '20 at 08:38