Order statistics of $n$ i.i.d. exponential random variables How to calculate the density function of $Y_{i}=X_{(i)}-X_{(i-1)}$ where $X_{(i)}$ are ordered i.i.d exponential random variable $X_i$ ?  I thought the calculation via integration is too sophisticated , was their any easy way to do that? Also the pairwise independence between $Y$ is also needed ,thanks!
 A: If the ordered $\{ X_{(i)} \}$ are taken from $n$ i.i.d. exponential random variables each with rate $\lambda$, then you can use the memoryless property to say that after $i-1$ terms have been observed, the interval to the next occurance also has an exponential distribution (i.e. to the minimum of the remaining random variables), with rate $(n-i+1)\lambda$, so the density of $Y_i$ is
$$p(y_i)= (n-i+1)\lambda e^{-(n-i+1)\lambda y_i }$$ for $y_i \ge 0$ and $1 \le i \le n$.    
A: Here is my method, please tell me if I'm wrong.
We know that $X_{(i)}=\sum_{k=1}^i Y_k$, that is to say, 
$$\mathbf{X}=\begin{bmatrix} 
1 &  & &\\
1 & 1 & &\\
\vdots& &\ddots\\
1&\cdots&\cdots &1
\end{bmatrix}\mathbf{Y},$$
where $\mathbf{Y}= \begin{bmatrix}Y_1 \\Y_2 \\ \vdots \\Y_n \end{bmatrix},\mathbf{X}= \begin{bmatrix}X_{(1)} \\X_{(2)} \\ \vdots \\X_{(n)} \end{bmatrix}$.
Since we have the combined density function of $\mathbf{X}$, that is 
$$f_{\mathbf{X}}(x_1,x_2,\cdots x_n)=n! \prod_{k=1}^{n} \lambda e^{-\lambda x_k}.$$
Using change of variable, we know that the combined distribution of $\mathbf{Y}$ is 
$$g_{\mathbf{Y}}(y_1,y_2,\cdots y_n)= \prod_{k=1}^{n} \lambda (n+1-k) e^{-\lambda (n+1-k)x_k}$$
Since $Y_i$ can get value from $(0,\infty)$, it's clear that the distribution of each $Y_i$ is exponential distribution with coefficient $\lambda (n+1-i)$.
