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Let $X_1, X_2, ...$ are iid exponential random variables, $S_k=\sum_{i=1}^{k}X_i$.

I want to find the distributions of $S_k/S_n$ for $k=1,...,n-1$.

i first used transformation $Y_i=S_k/S_n , (k=1,...,n-1)$ and $Y_n=X_1+...+X_n$ So that i assume Jacobian is $y_n^{n-1}$ since $$x_1=y_1y_n, x_2=y_2y_n-y_1y_n,...,x_{n-1}=y_{n-1}y_n-y_{n-2}y_n,x_n=y_n(1-y_{n-1}).$$

so $$f_{Y_1,...,Y_n}(y_1,...,y_n)=\lambda^ne^{-\lambda y_n}y_n^{n-1},\space y_1\in(0,1),y_2\in(y_1,1),...,y_n\in(0,\infty)$$

Is this right way to calculate the distribution of $S_k/S_n=Y_k$? i found tricky calculating these.. am i missing of mistaking something?

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Each $S_k$ as the sum of iid. exponential $X_i \sim \mathrm{Exp}(\lambda)$ is $\mathrm{Gamma}(k,\lambda)$, regardless of you're using rate parameter of scale parameter.

The fraction is $\displaystyle \frac{S_k}{S_n} = \frac{S_k }{S_k + S'_k}$, with $\displaystyle S'_k \equiv \sum_{i = k+1}^n X_i \sim \mathrm{Gamma}(n-k,\lambda)$ that is also Gamma, and we have independence $S_k \perp S'_k$ since $X_i$ are iid.

Therefore $\frac{S_k}{S_n}$ as a fraction of Gamma over "same Gamma plus another Gamma" follows $\mathrm{Beta}(k,n-k)$.

If you really want to calculate the joint density (and then "integral out the rest" to get the marginals), you should start with $n = 2$ then do $n = 3$ to see the patterns. Don't dive into general $n$ directly unless you're already fairly with the relevant integrals.

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    $\begingroup$ (+1) It is enough to derive the density of $\frac{S_k}{S_k+S_k'}$ using a change of variables since we know the distribution of both $S_k$ and $S_k'$ and they are independent. $\endgroup$ – StubbornAtom Oct 19 '18 at 16:48
  • $\begingroup$ Thank you very much Lee and @StubbornAtom ! $\endgroup$ – Moon Oct 19 '18 at 17:00

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