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I'm a business major, so really distribution theory is not my strong suit. I know how to get the distribution of a ratio of exponential variables and of the sum of them, but i can't piece everything together.

The exercise goes as this:

If $X,Y$ are independent exponentially distributed with $\beta=1$ (parameter of the exponential distribution $= 1$) then what is the distribution of $X/(X+Y)$

Any ideas?

Thanks a lot.

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marked as duplicate by Did, Dilip Sarwate, Start wearing purple, Lord_Farin, Micah Jun 10 '13 at 18:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ OP: Why are you reposting exactly the same question again? $\endgroup$ – Did Jun 10 '13 at 18:37
  • $\begingroup$ Couldn't find old post, thought it had been deleted or something! My bad, sorry! $\endgroup$ – entourager Jun 10 '13 at 18:43
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    $\begingroup$ The "old" post was posted 4 days ago and is your only other question on the site. Both are listed on your user page. $\endgroup$ – Did Jun 10 '13 at 18:46
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Obviously $X/(X+Y)$ is between $0$ and $1$. Let (lower-case) $x$ be a number between $0$ and $1$.

$$ \frac{X}{X+Y} \le x \text{ if and only if } (1-x)X \le xY. $$ We will find this probability.

The conditional probability that $Y\ge\dfrac{(1-x)}{x}X$, given the value of $X$, is $$ e^{-(1-x)X/x}. $$

The probability we seek is then the expected value of that: \begin{align} \mathbb E e^{-(1-x)X/x} & = \int_0^\infty e^{-(1-x)w/x} \Big(e^{-w}\,dw\Big) \\[10pt] & = \int_0^\infty e^{-w/x} \, dw \\[10pt] & = x. \end{align}

In other words, this random variable is uniformly distributed between $0$ and $1$.

Second method: The equality $Y=\frac{1-w}{w}X$ is a straight line through the $(X,Y)$-plane, passing through $(0,0)$ and having positive slope.

We can let $y$ go from $0$ to $\infty$ and then for any fixed value of $y$, let $x$ go from $0$ to $\frac{w}{1-w}y$ \begin{align} & {}\qquad\int_0^\infty \left(\int_0^{wy/(1-w)} e^{-y} e^{-x} \,dx \right) \, dy \\[10pt] & = \int_0^\infty e^{-y}\left(1-e^{-wy/(1-w)}\right) \, dy \\[10pt] & = \int_0^\infty e^{-y} - e^{-y/(1-w)} \, dy \\[10pt] & = 1 - (1-w) \\[10pt] & = w. \end{align}

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