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I have a task. $X$ and $Y$ are independent random variables with exponential distribution and $\mathbb{E}X=1$, $\mathbb{E}Y=\frac{1}{2}$. Calculate $\mathbb{E}\big(\frac{X}{X+Y}\big)$.

I tried to calculate it as $\int_{0}^{\infty}\int_{0}^{\infty} \frac{x}{x+y}2e^{-2y-x}\mathrm{d}x\mathrm{d}y$, but it is hard.

Is the easier way to calculate this task?

Bob
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John1357
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  • You could always simulate, perhaps a little less than $0.614$ – Henry May 19 '22 at 10:00
  • I wrote that I tried but it is too complicated. This task is from actuarial exams and for this task is 10 minutes, so It is really hard to calculate this in this period of time. – John1357 May 19 '22 at 10:10
  • Apparently if you manage the integration, you get $2-\log_e(4) \approx 0.61370563888$ but I do not see an easy way of doing it – Henry May 19 '22 at 10:14

1 Answers1

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Note that $$ \mathsf{E}\frac{X}{X+Y}=\int_0^1\mathsf{P}\!\left(\frac{X}{Y}>\frac{t}{1-t}\right)\, dt $$ because $\mathsf{P}(X/(X+Y)>t)=\mathsf{P}(X>t(X+Y))=0$ for $t\ge 1$. Since the cdf of the ratio $X/Y$ is given by $F(x)=1/(1+2/x)$ for $x\ge 0$, $$ \mathsf{E}\frac{X}{X+Y}=2\int_0^1\frac{1-t}{2-t}\,dt=2-2\ln(2). $$