Let $X_{1}, \ldots, X_{n}$ be independent and identically distributed random variables, with exponential distribution of the parameter $\lambda$. Consider $Y_{(1)}, \ldots, Y_{(n)}$ the associated order statistics and define $$Z_{1} = Y_{(1)} \quad \text{and} \quad Z_{j} = Y_{(j)} – Y_{(j – 1)}, j = 2, \ldots , n$$ Show that $Z_{1},\ldots, Z_{n}$ are independent and $Z_{j} \sim \textbf{Exp} (\lambda (n -j +1)), j= 1,\ldots, n$.
My approach: Let, $X_{i}\overset{i.i.d}{\sim}$ with $i=1,2,\ldots,n$ and $X_{i}\sim \textbf{Exp}(\lambda)$. So, since that $X_{i}'s$ are independent and indentically distributed, then $f$ is the common density function (this mean that $X_{i}'s$ has the same density).
Since $Y_{(1)},\ldots, Y_{(n)}$ are the associated order statistic to $X_{1},\ldots,X_{n}$, so the joint density is $$f_{Y_{(1)},\ldots,Y_{(n)}}(y_{1},\ldots,y_{n})=n!f(y_{1})\cdots f(y_{n}), \quad y_{1}<y_{2}<\cdots<y_{n}$$
How can I continue or solve this problem?
Really, I don't know how to continue.
I think I can use some transformation theorem, but I don't sure how can use that.
I was studying a book and I found similar problem where using that ${n!}$ =$\prod_{1}^{n}(n-i+1)$ and $\sum_{1}^{n}Y_{i}$ = $\sum_{1}^{n}(n-i+1)(Y_{i}-Y_{i-1})$ ,so the joint p.d.f of the $Y_{i}$
${n!}\lambda^nexp(-{\lambda}\sum_{1}^{n}y_{i})$ may be expressed as $\prod_{1}^{n}{\lambda}(n-i+1)exp(-{\lambda}(n-i+1)z_{i})$ ($z_{0}=0$ )
But I don't know how to adapt that to the solution of my problem.
Best Answer
I answered quite the same question here. The only differences are that in the original post the distribution was $Exp(1)$ and the requeste distribution was the law of $Z_j=Y_{(j+1)}-Y_{(j)}$
Unfortunately this forum is an Italian forum...but I think you can understand the solution because formulas are formulas....in any language.
On the contrary please advice and I will repost the solution in English
Solution details
Just to simplify the notation, let's set $Y$ and $X$ the two order statistics and $Z=Y-X$ the rv we are looking for distribution ($Y>X$).
Using the known result on the joint density of two Order Statistics we get
$$f_{XY}(x,y)=\frac{n!}{(j-2)!(n-j)!}(1-e^{-\lambda x})^{j-2}e^{-\lambda(n-j)y}\lambda e^{-\lambda x}\lambda e^{-\lambda y}\cdot\mathbb{1}_{(x;\infty)}(y)$$
In order to get the CDF of the rv $Z=Y-X$ we have to integrate the joint density over the desired support, say:
$$F_Z(z)=\frac{n!}{(j-2)!(n-j)!}\int_0^{\infty}(1-e^{-\lambda x})^{j-2}\lambda e^{-\lambda x} \Bigg[ \int_x^{x+z}\lambda e^{-\lambda(n-j+1)y} dy\Bigg] dx$$
The integral is not difficult...it is enough to sove the inner integral in $dy$ (very easy) and get out of the sign of integral all what does not depend on x.
You get
$$\frac{n!}{(j-2)!(n-j+1)!}[1-e^{-\lambda(n-j+1)z}]\int_0^{\infty}(1-e^{-\lambda x})^{j-2}\lambda (e^{-\lambda x})^{n-j+2}dx$$
Now substitute
$1-e^{-\lambda x}=u$
and you get
$$F_Z(z)=\frac{n!}{(j-2)!(n-j+1)!}[1-e^{-\lambda(n-j+1)z}]\underbrace{\int_0^1 u^{j-2}(1-u)^{n-j+1}du}_{=Beta(j-1;n-j+2)}=1-e^{-\lambda(n-j+1)z}$$
As
$$Beta(j-1;n-j+2)=\frac{\Gamma(j-1)\Gamma(n-j+2)}{\Gamma(n+1)}=\frac{(j-2)!(n-j+1)!}{n!}$$
That is exactly the CDF they ask you to show.