Ordered iid Exp(1) Random Variable: Conditional Density Function

Question

I'm working on the following exercise from Achim Klenke's "Probability Theory: A Comprehensive Course" (3rd Ed, Exercise 15.1.3):

Let $n \in \mathbb N$ and let $X_1, \ldots, X_n$ be i.i.d. exponentially distributed random variables with parameter $1$. Let $Y_1, \ldots, Y_n$ be independent exponentially distributed random variables with $\mathbf P_{Y_k} = \exp_k$. That is, $$(Y_1, \ldots, Y_n) \stackrel{\mathcal D}{=} (X_1, X_2/2, X_3/3, \ldots, X_n/n)$$ where $\stackrel{\mathcal D}=$ denotes equivalently distributed. Finally, sort the values of $X_i$ by size $X_{(1)} > X_{(2)} > \cdots > X_{(n)}$. Show that $$ \left( X_{(n)}, X_{(n-1)}, \ldots, X_{(1)}\right) \stackrel{\mathcal D}= \left( Y_n, Y_{n-1} + Y_n, \ldots, Y_1 + Y_2 + \cdots + Y_n\right). $$ Hint: First check that $X_{(n)} \stackrel{\mathcal D}= Y_n$. Show that the conditional distribution $\mathcal L\left[\left(X_{(1)} - X_{(n)}, \ldots, X_{(n-1)} - X_{(n)}\right) | X_{(n)}\right]$ does not depend on $X_{(n)}$ and that it equals the (unconditional) distribution of the ordered values of $X_1, \ldots, X_{n-1}$.

Here Distribution of ordered independent exponential random variables there is a theoretical solution.

Since I am not able to finish it, I think I am doing something wrong in the conditional law's density calculation.

Indeed, for the distribution of the ordered values $X_{(1)},...,X_{(n-1)}$ I get, by integrating out the minumim (please, mind that $X_{(1)}$ is the maximum in the notation used by prof. Klenke) from the joint distribution of the ordered statistics, this density:

$f_{X_{(1)},...,X_{(n-1)}}(x_1,...,x_{(n-1)})= n! f_{X_1}(x_1)...f_{X_1}(x_{n-1})F_{X_1}(x_{n-1})=n!e^{-(x_1+...+x_{n-1})}(1-e^{x_{n-1}})$.

By applying the change of variables theorem I get for the conditional density above:

$$\frac{f_{X_{(1)}-x_n,...,X_{(n-1)}-x_n,X_{(n)}}(x_1,...,x_{n-1},x_n)}{f_{X_{(n)}}(x_n)}= \\\quad \frac{f_{X_{(1)},...,X_{(n-1)},X_{(n)}}(x_1+x_n,...,x_{n-1}+x_n,x_n)}{f_{X_{(n)}}(x_n)}= \\\quad \frac{n!f_{X_1}(x_1+x_n)...f_{X_1}(x_{n-1}+x_n)f_{X_1}(x_{n})}{{f_{X_{(n)}}(x_n)}}= \\\quad \frac{n!e^{-(x_1+...+x_{n-1})}e^{-nx_n}}{ne^{-nx_n}}$$

This latter should be equal to the unconditional density that I calculated above. Any suggestions?

Edit

Could it be something like the following equations based on the iid and “memoryless” properties of the sample?

$P[X_{(1)}-X_{(n)}<y_1,…,X_{(n-1)}-X_{(n)}<y_{n-1}| X_{(n)}=x_n]= \\= n! P[X_{1}-X_{n}<y_1,…,X_{n-1}-X_{n}<y_{n-1}| X_{n}=x_n]= \\=n!P[X_{1}<y_1+x_n,…,X_{n-1}<y_{n-1}+x_n| X_{n}=x_n]=\\=n!P[X_1<y_1]…P[X_{n-1}<y_{n-1}]$

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 $X_{(n)}$. Before I tried to divide the joint density of the ordered statistics by the density of the minimum. Is my reasoning right?

Thank you.

Answer 1

Firstly we draw a sample with size $n-1$ from exponential distribution with parameter $1$, say $(Z_1, Z_2, \cdots Z_{n-1})$ and the order statistics are $(Z_{(1)}, \cdots, Z_{(n-1)})$ where $Z_{(1)} > \cdots > Z_{(n-1)}$.

Actually in your OP (without editting), the result is right since the CDF of $(X_{(1)} - X_{(n)}, ..., X_{(n-1)}-X_{(n)} ) | X_{(n)}$ $$ f_{X_{(1)} - X_{(n)}, \cdots,X_{(n-1)} - X_{(n)} | X_{(n)} = x_n}(x_1, \cdots, x_{n-1}) = (n-1)!e^{-(x_1 + \cdots + x_{n-1})} $$ equals to the CDF of the order statistics of $(Z_1, ..., Z_{n-1})$.

But in your edit, the reasoning is wrong. We should show that $$ P(X_{(1)} - X_{(n)} < y_1, \cdots, X_{(n-1)} - X_{(n)} < y_{n-1}| X_{(n)} = x_n) = P(Z_{(1)} < y_1, \cdots, Z_{(n-1)} < y_{n-1}) $$ It is a little subtle that why we should not show $$ P(X_{(1)} - X_{(n)} < y_1, \cdots, X_{(n-1)} - X_{(n)} < y_{n-1}| X_{(n)} = x_n) = P(X_{(1)} < y_1, \cdots, X_{(n-1)} < y_{n-1}) $$ The straight above equation does not hold under the notation in the OP.

Anyway the two methods (calculate the CDF or reason by probability) lead to the same result.

The next step is show that the conditional distribution of $(X_{(1)}-X_{(n)}, \cdots, X_{(n-1)} - X_{(n)}) | X_{(n)}$ and $(Y_{n-1}+Y_{n} - Y_n, \cdots ,Y_1+Y_2+\cdots + Y_n - Y_n) | Y_n = (Y_{n-1}, \cdots, Y_1 + Y_2 + \cdots Y_{n-1})$ since $Y_i$s are indenpent of each other.

It suffices to prove that the distribution of $(Z_{n-1}, \cdots, Z_{1})$ is the same as $(Y_{n-1}, Y_{n-2} + Y_{n-1}, \cdots, Y_{1}+Y_2 + \cdots +Y_{n-1})$ by applying the induction proof.

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Update:

Regarding the notation problem, do you mean that if I substitute the $y_i$ s with $z_i$ would hold? In my head I named the $Y$ the difference of the order statistics with the minimum. $Z$ would have been more correct

The notation $y_i$ and $z_i$ don't matter. It is just similar to $f(y_1, y_2, \cdots, y_n)$ and $f(z_1, z_2, \cdots, z_n)$ where $f$ is some density function. Replacing $y_i$ by $z_i$ don't help much.

Also, you are considering a sample of $n−1$ observation. I thought it was all referred to same original sample of $n$ observation. So I calculated the unconditional distribution of the ordered sample by integrating the joint density of the ordered statistic $X_{(i)}$.

A new sample sized $n-1$ (say $Z$) is slightly (even negligibly) different from the first $n-1$ observations of the old sample $X$ with size $n$. I just use another notation $Z$ to denote the order statistic of the first $n-1$ observations of $X$ with size $n$.

Integrating the density is your first method which is correct to show that the density of $(X_{(1)} - X_{(n)}, \cdots, X_{(n-1)} - X_{(n)})|X_{(n)}$ is the same as the density of the order statistic of the first $n-1$ observations $X_1, \cdots, X_n$ in the form $f(x_1, \cdots, x_{n-1}) = (n-1)! e^{-(x_1 + \cdots x_n)}$.(Surely you can replace $x_i$ by $y_i$).

But in your edit:

$$ \begin{align} &P[X_{(1)}−X_{(n)} < y_1,…,X_{(n−1)}−X_{(n)}< y_{n−1}|X_{(n)}=x_n] \\ &= n!P[X_1−X_n<y_1,…,X_{n−1}−X_n<y_{n−1}|X_n=x_n] \\ &=n!P[X_1<y_1+x_n,…,X_{n−1}<y_{n−1}+x_n|X_n=x_n] \\ &=n!P[X_1<y_1]…P[X_{n−1}<y_{n−1}] \\ \end{align} $$

The first equation is wrong since $$ P[X_{(1)}−X_{(n)} < y_1,…,X_{(n−1)}−X_{(n)}< y_{n−1}|X_{(n)}=x_n] \\ \neq n!P[X_1−X_n<y_1,…,X_{n−1}−X_n<y_{n−1}|X_n=x_n] \\ $$

It should be $$ \begin{align} &P[X_{(1)}−X_{(n)} < y_1,…,X_{(n−1)}−X_{(n)}< y_{n−1}|X_{(n)}=x_n] \\ &= \frac{P[X_{(1)}−X_{(n)}<y_1,…,X_{(n−1)}−X_{(n)}<y_{n−1}, X_{(n)}=x_{n}]}{P[X_{(n)}=x_{n}]} \\ &= \lim_{h\to 0^+}\frac{P[X_{(1)}<y_1+x_n,…,X_{(n−1)}<y_{n−1}+x_n, x_n \le X_{(n)}<x_{n}+h]}{P[x_n \le X_{(n)}<x_{n}+h]} \end{align} $$ By integrating the density it leads to the same result as your first method.