Suppose that $(e_n)_{n\geq1}$ are i.i.d. exponential random variables with the parameter $\lambda$ = 1. Let $S_n = e_1+e_2 +\cdots+e_n$ and $R_n = S_{n+1}/S_n$ for $n>1$. Prove that $(R_n)_{n\geq1}$ are independent and $R_n$ has density $nx^{-(n+1)}1_{x\geq1}$.
Hint: Let $R_0$ = $e_1$ and compute the joint density of $(R_0,R_1,\ldots,R_n)$ first.
Could anyone help me on this question? Thank you very much!
For $r_0 \ge 0$ and $r_1 \ge 1$, \begin{align} P(R_0 > r_0, R_1 > r_1) &= P(e_1 > r_0, e_2 > (r_1 - 1) e_1) \\ &= \int_{r_0}^\infty e^{-e_1} \int_{(r_1-1)e_1}^\infty e^{-e_2} \, de_2 \, de_1 \\ &= \int_{r_0}^\infty e^{-e_1} e^{-(r_1-1)e_1} \, de_1 \\ &= \frac{1}{r_1} e^{-r_0 r_1} \end{align} Taking $r_0=0$ and $r_1 \ge 1$ yields $P(R_1 > r_1) = 1/r_1$ so the density is $1/r_1^2$.
Similarly, \begin{align} &P(R_0 > r_0, R_1 > r_1, R_2 > r_2) \\ &= P(e_1 > r_0, e_2 > (r_1 - 1) e_1, e_3 > (r_2-1)(e_1+e_2)) \\ &= \int_{r_0}^\infty e^{-e_1} \int_{(r_1-1) e_1}^\infty e^{-e_2} \int_{(r_2-1)(e_1+e_2)}^\infty e^{-e_3} \, de_3 \, de_2 \, de_1 \\ &= \int_{r_0}^\infty e^{-r_2 e_1} \int_{(r_1-1)e_1}^\infty e^{-r_2 e_2} \, de_2 \, de_1 \\ &= \frac{1}{r_2}\int_{r_0}^\infty e^{-r_1 r_2 e_1} \, de_1 \\ &= \frac{1}{r_1 r_2^2} e^{-r_0 r_1 r_2}. \end{align}
In general it seems that $$P(R_0 > r_0, \ldots, R_n > r_n) = \frac{1}{r_1 r_2^2 \cdots r_n^n} e^{-r_0 r_1 \cdots r_n}.$$ (You need to show this rigorously.) From here, plugging in $r_0=0$ to get rid of $R_0$ shows that the $P(R_1 > r_1, \ldots, R_n > r_n) = P(R_1 > r_1) \cdots P(R_n > r_n)$ so $(R_1, \ldots, R_n)$ are independent, and you can easily obtain the marginal densities.