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Let $$ X_1 , ..., X_n$$ be independent random variables, all if which have the same exponential distribution with pdf $$ x \mapsto e^{-x} ,x \ge 0 $$ How do I find the distribution for $$ Y = {X_1\over X_1 +... + X_n}? $$

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    $e^{-x} x$ is not Exponential ... it is Gamma – 2017-01-21
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    I mean $e^{-x}$ if $x>=0$ – 2017-01-21
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    Just as a piece of information.Just for help. You can use "\leq"(less or equal), to indicate $\leq$, or "\geq" (greater or equal), for $\geq$. – 2017-01-21

2 Answers 2

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Observe we can rewrite this as $$ Y=\frac{X}{X+Z} $$ where $X=X_1$ is a standard exponential and $Z=\sum_{k=2}^nX_k$ is a Gamma($n-1,1$) independent of $X.$ We can write the CDF for $Z$ as $$ \begin{eqnarray}P(Y\le y) &=& P\left(\frac{X}{X+Z}\le y\right) \\&=& P\left(X\le \frac{y}{1-y}Z\right)\\&=& \int_0^\infty \frac{z^{n-2}e^{-z}}{\Gamma(n-1)}P\left(X\le \frac{y}{1-y}z\right)dz\\&=&\int_0^\infty \frac{z^{n-2}e^{-z}}{\Gamma(n-1)}\left(1-\exp\left(-\frac{y}{1-y}z\right)\right)dz\\&=&1-\int_0^\infty \frac{z^{n-2}e^{-z}}{\Gamma(n-1)}e^{-\frac{y}{1-y}z}dz\\&=&1-\int_0^\infty \frac{z^{n-2}}{\Gamma(n-1)}e^{-\frac{1}{1-y}z}dz\\&=& 1-(1-y)^{n-1}\end{eqnarray}$$

ADDENDUM

We can see this answer in a slightly more sophisticated way by remembering some facts about Poisson processes and their relationship to the exponential distribution.

Recall $T_n\equiv\sum_{i=1}^n X_i$ is the wait time to the $n$-th jump and $T_1 \equiv X_1$ is the wait time to the first. So $Y=T_1/T_n$ is the ratio of the wait time to the first to the wait time to the $n$-th. If you look at the time interval $[0,T_n],$ the $n-1$ jumps will be distributed uniformly in that interval, since they happen at a constant rate, independently of each other (this can be shown rigorously). Thus since $Y=T_1/T_n$ is just the time of the first jump, normalized by the length of the interval, it will be distributed like the smallest of $n-1$ independent uniform $[0,1]$ variables.

To get the distribution for this smallest uniform $U_{min}$, observe $$ P(U_{min} \ge y) = P(U_1\ge y, U_2 \ge y,\ldots, U_{n-1}\ge y) = P(U\ge y)^{n-1} = (1-y)^{n-1}$$ which is identical to the distribution we got by direct calculation.

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    Why gamma distribution ? How does it help for the exercise ? – 2017-01-21
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    @callculus The sum of $n-1$ exponential RVs is Gamma(n-1). I computed the CDF of the RV in question, so I'd say it solves the problem. There's a better way to think of it in terms of Poisson processes, but the mechanical approach was easier to write down. – 2017-01-21
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    You haven´t mentioned that $Z=\sum_{i=2}^n X_i$. Thus I was little bit confused. – 2017-01-21
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    @callculus Ahh, yeah I see now I also made a stupid notation choice of making my $Z$ OP's Y and having another Y in there. Edited. – 2017-01-21
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Hint: $Z:=\sum_{i=2}^n X_i$ has a Gamma distribution, you can work out the distribution of $\ln Z -\ln X_1$ (with characteristic functions) and hence of $W:=\frac{Z}{X_1}$. Since $Y=\frac{1}{1+W}$, you can get the distribution of that too.