Uniform distributed sequences

Question

If $X_n$ uniformly distributed $U[0,X_{n-1}=x_{n-1}]$ for $n>0, X_0=1$. How do we show the sequence $X_n^a$ is uniform.

Answer 1

$E(S_{n+1}|S_1,S_2,..,S_n)=(1+a)^{n+1} E(X_{n+1}^{a}| X_1,X_2,\dots,X_n)$ since $\sigma (S_1,S_2,\dots,S_n)=\sigma (X_1,X_2,\dots,X_n)$ Hence $E(S_{n+1}|S_1,S_2,\dots,S_n)=(1+a)^{n+1} \frac {\int_0^{X_n} t^{a}dt} {X_n}=(1+a)^{n} X^{a}_n=S_n$.