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Been crushing my head with this exercise. 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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Obviously $X/(X+Y)$ is between $0$ and $1$. Let (lower-case) $w$ be a number between $0$ and $1$.

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

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

The probability we seek is then the expected value of that: \begin{align} \operatorname{E} e^{-(1-w)X/w} & = \int_0^\infty e^{-(1-w)x/w} \Big(e^{-x}\,dx\Big) \\[10pt] & = \int_0^\infty e^{-x/w} \, dx \\[10pt] & = w. \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} & \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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    $\begingroup$ Can we say the same for identically distributed random variables with the exponential distribution with a parameter $\alpha \neq 1$? Following the second method I think it shouldnt hold true, am I right? $\endgroup$ – Albanian_EAGLE Nov 28 '13 at 15:53
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    $\begingroup$ It should still be true in that case. $\endgroup$ – Michael Hardy Nov 28 '13 at 18:51
  • $\begingroup$ but then how does the coefficient $\alpha$ simplify in the integration so that we get still just $w$? (your second method) $\endgroup$ – Albanian_EAGLE Nov 29 '13 at 3:09
  • $\begingroup$ If $U$ and $V$ are independent exponentially distributed random variables with expected value $\alpha$, then $X=U/\alpha$ and $Y=V/\alpha$ are independent exponentially distributed random variables with expected value $1$. And $X/(X+Y)$ is equal to $U/(U+V)$. So $\alpha$ just cancels out. $\endgroup$ – Michael Hardy Nov 29 '13 at 5:27
  • $\begingroup$ very helpful, thanks. $\endgroup$ – Albanian_EAGLE Nov 29 '13 at 5:29
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Another way is to do it is to perform the following transformation. Let \begin{align*} U&=\frac{X}{X+Y} \\ V&=X+Y.\end{align*} Then \begin{align*} x(u,v)&=uv\\y(u,v)&=v-uv. \end{align*} The determinant of the Jacobian of the transformation is $$\left| \begin{matrix} v & u\\ -v & 1-u \end{matrix} \right|=v. $$ Then, the joint pdf of $U$ and $V$ is $$f_{U,V}(u,v)=f_{X,Y}(uv,v-uv)|v|=ve^{-v}, \hspace{2mm} \text{for } v\geq0, 0\leq u\leq1. $$ Finally, "integrate out" the $v$ to obtain the marginal pdf of $U=X/(X+Y)$ as $$f_U(u)=\int_0^{\infty}f_{U,V}(u,v)dv=\int_0^{\infty}ve^{-v}dv=1, \hspace{2mm} \text{for } 0\leq u\leq 1.$$ Hence, as Michael Hardy showed, $X/(X+Y)$ is uniformly distributed on $[0,1].$

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In other words, for each $a > 0$, you want to compute $P\left(\frac{X}{X+Y} < a \right)$.

Outline: Find the joint density of $(X,Y)$, and integrate it over the subset of the plane $\left\{ (x,y) : \frac{x}{x+y} < a \right\}$.

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  • $\begingroup$ Sorry, hard for me to follow as english's not my native language. What do you mean with "find the joint density", as in the joint distribution? f(x,y)? That should be easy as x and y are independent variable. Now, after i do that, what do you mean by "the subset of the plane"? I'm sorry, but i'm a business major with a less than perfect english, so i'm having a hard time understanding what should i do. $\endgroup$ – entourager Jun 6 '13 at 19:57

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