Conditional density of ordered iid Exp(1) I am trying to apply the change of variable theorem for the calculation of the distribution, via the conditional density, of this transformation of joint iid $X_i$ that are Exp(1) variables with no success:
$$f_{X_{(2)}-X_{(1)},...,X_{(n)}-X_{(1)}|X_{(1)}}(x_2,...,x_n)$$
where $X_{(1)}$ is the minimum, $X_{(2)}$ is the second lowest variable and so on (ordered statistic).
Any suggestion? This distribution, according to Exercise 15.1.3 of Probability Theory by A. Klenke (3rd version), should be equal to the unconditional distribution of the ordered sample $(X_{2},...,X_{n})$. (For the ones checking the book, I reversed the notation).
Edit
Could it be something like the following equations based on the iid and “memoryless” properties of the sample?
$P[X_{(2)}-X_{(1)}<y_2,…,X_{(n)}-X_{(1)}<y_n| X_{(1)}=x_1]= \\= n! P[X_{2}-X_{1}<y_2,…,X_{n}-X_{1}<y_n| X_{1}=x_1]= \\=n!P[X_{2}<y_2+x_1,…,X_{n}<y_n+x_1| X_{1}=x_1]=\\=n!P[X_2<y_2]…P[X_n<y_n]$
Basically I am considering in the $n!$ permutation also the fact that the lowest observed value could come from any observation in the sample. Thus I condition on the (arbitrarily) first observation to be the lowest and no more on the minimum $X_{(1)}$. Before I tried to divide the joint density of the ordered statistics by the density of the minimum. Is my reasoning right?
Edit
https://math.stackexchange.com/a/4626276/1073326
Here @xzm solved it very clearly by suggesting that the unconditional distribution is related to ordering a sample of dimension $n-1$, if I understood properly.
 A: For iid $(X_i)_{i=1}^n$ with $X_{i}\sim\text{Exponential}\left(i\right)$ according to this answer we can write:
$$\left(X_{\left(1\right)},X_{\left(2\right)},X_{\left(3\right)},\dots,X_{\left(n\right)}\right)=\left(Y_{n},Y_{n}+Y_{n-1},Y_{n}+Y_{n-1}+Y_{n-2},\dots,Y_{n}+Y_{n-1}+Y_{n-2}+\dots+Y_{1}\right)$$
where the $(Y_{i})_{i=1}^n$ are independent and $Y_{i}\sim\text{Exponential}\left(i\right)$.
Then we have: $$\left(X_{\left(2\right)}-X_{\left(1\right)},X_{\left(3\right)}-X_{\left(1\right)},\dots,X_{\left(n\right)}-X_{\left(1\right)}\right)=\left(Y_{n-1},Y_{n-1}+Y_{n-2},\dots,Y_{n-1}+Y_{n-2}+\dots+Y_{1}\right)$$
Note that according to the same reasoning the random vector on RHS has the same distribution as: $$\left(Z_{\left(1\right)},Z_{\left(2\right)},\dots,Z_{\left(n-1\right)}\right)$$
if the $\left(Z_{i}\right)_{i=1}^{n-1}$ are iid with $Z_{i}\sim\text{Exponential}\left(1\right)$.
Moreover there is independence with respect to random variable $Y_n=X_{\left(1\right)}$ so that any conditions on $X_{\left(1\right)}$ do not affect the distribution of the random vector.
