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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.

Enrico
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  • $(X_{(2)}-X_{(1)},\dots,X_{(2)}-X_{(1)})$ and $X_{(1)}$ are independent because the distribution is "memoryless". – drhab Jan 19 '23 at 08:34
  • I know the $X_i$ are Exp(1) and so are "memoryless". But I don't understand why this property holds for the ordered variables. Could you explain a little bit? Thank you. – Enrico Jan 19 '23 at 10:14
  • The answer to this question might help. – drhab Jan 19 '23 at 13:14
  • I’m sorry but I can’t see a solution. I get for the unconditional distribution of $X_{(2)},…,X_{(n)}$ this density: $n!F_{X_1}(x_2)f_{X_1}(x_2),…,f_{X_1}(x_n)$. At this point I don’t know either if this is right. Suggestions? Thanks. – Enrico Jan 19 '23 at 19:39
  • This answer might help. Also have a look at this question. – drhab Jan 20 '23 at 12:26
  • Thank you but I see that these answers are about "spacing", the difference between two consecutive ordered observations. Here instead I have the difference from the minimum of the sample for every observation and I am stuck. – Enrico Jan 20 '23 at 14:03
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    https://math.stackexchange.com/questions/2690357/show-that-x-1-minx-is-independent-of-ancillary-y-i-x-n-x-i – user51547 Jan 22 '23 at 20:48
  • Could you help me with calculations? I think we are quite near the solution, especially if we can show the equivalence with the unconditional distribution of the exercise – Enrico Jan 23 '23 at 20:22

1 Answers1

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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.

drhab
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  • Thank you l. Do you think there is a way to solve it also by using only the change of variable theorem? That would be very interesting as it does not depend on “seeing” the equivalence of those distributions involving $Y_i$. – Enrico Jan 20 '23 at 17:44