Let $X_1,\ldots,X_n$ be a sample of independent random variables. Then let $Y_1, \ldots, Y_n$ be the corresponding ordered statistics. Prove that $Y_1, Y_2 - Y_1, \ldots, Y_n - Y_{n-1}$ are independent random variables:
My work so far:
Define $\overline{Z} = (Z_1,\ldots, Z_n)$, where: \begin{align*} Z_1 &= Y_1 &\quad Y_1 &= h_1(Z)=Z_1\\ Z_2 &= Y_2 - Y_1 &\quad Y_2 &= h_2(Z)= Z_2 + Z_1\\ &\vdots &\vdots\\ Z_n &= Z_n - Z_{n-1} &\quad Y_n&= h_n(Z) =\sum_{i=1}^{n}Z_i \end{align*}
Then by the Transformation Theorem:$$f_{\overline{Z}}(\overline{z})=f_{\overline{Y}}(\overline{h(z)})\cdot |J\overline{h(z)}|$$
It is clear that $|J\overline{h(z)}|= 1$, and because we know that the sample is made of independent random variables, we have that:$$f_{\overline{Y}}(\overline{y}) = n! \prod_{i=1}^{n} f_{\overline{X}}(y_i)$$
Now this is where I get stuck, substituting I get:$$f_{\overline{Z}}(\overline{z}) = n! \cdot f_{\overline{X}}(z_1)\cdot f_{\overline{X}}(z_1 + z_2)\cdot \cdots$$
What step should I take next, or is there a different approach...
Thanks!
It looks as if it may not be true
As a counterexample, take $X_i$ i.i.d. $0$ or $1$ each with probability $\frac12$. If any of the $Y_j-Y_{j-1}$ are $1$ then none of the others are, so are not independent
Similarly for a continuous counterexample $X_i$ i.i.d. uniform on $[0,1]$. If any of the $Y_j-Y_{j-1}$ are greater than $\frac12$ then none of the others are, so are not independent