Distribution of ordered independent exponential random variables

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 $X_1, \ldots, X_{n-1}$. By an iteration procedure, prove the claim.

Using independence and the fact that $X_{(n)} = \min\{X_1, \ldots, X_n\}$, we have that $$\mathbf P\left[X_{(n)} > x\right] = \mathbf P\left[\mathop\bigcap_{k=1}^n\{X_k > x\}\right] = \mathbf P\left[X_1 > x\right]^n = e^{-nx} = \mathbf P\left[Y_n > x\right],$$ so we have $X_{(n)} \stackrel{\mathcal D}= Y_n$. But I'm having trouble with showing the statement about the conditional distribution. I'm not even sure what the "unconditional distribution of the ordered values of $X_1, \ldots, X_{n-1}$" would be, and why it would be different from the distribution of $X_{(1)}, \ldots, X_{(n-1)}$.

EDIT: I tried getting some insight when looking at it just when $n=3$, and found \begin{align*} \mathbf P\left[X_{(2)} - X_{(3)} > x\right] &= \int_0^\infty \mathbf P\left[X_{(2)}- X_{(3)} > x | X_{(3)} = y\right] \mathbf P_{X_{(3)}}[dy] \end{align*} so I can kind of see how the distribution of the $X_{(k)} - X_{(n)}$ are going to come into play, and then using a linear transformation to get the distributions of each $X_{(k)}$, but I'm still not seeing how to compute the conditional distributions.

Any suggestions?

Answer 1

What Klenke says is true and can be verified by computing conditional densities. But I will suggest general direct approaches to compute both the distribution of $(X_{(n)}, \dots, X_{(1)})$ and of $Z = (Y_1, Y_1 + Y_2, \dots, Y_1 + \dots + Y_n)$ when $X_1, \dots, X_n$ are arbitrary i.i.d. rvs with densities, and $Y_1, \dots, Y_n$ are arbitrary indepndent rvs with densities, and then you can specialize to this specific problem.

It is a consequence of the change of variables theorem that whenever $X_1, \dots, X_n$ are i.i.d. and have density $f_{X_1}$ (wrt Lebesgue measure), the density $f$ of $(X_{(n)}, \dots, X_{(1)})$ is $$f(x_n, \dots, x_1) = n!f_{X_1}(x_n) \dots f_{X_1}(x_1), \hspace{20pt} x_n < x_{n - 1} < \dots < x_1.$$

Suppose $Y_1, \dots, Y_n$ are independent and that each $Y_j$ has density $f_{Y_j}$ wrt Lebesgue measure. We have $Z = AY$, where $A \in GL(n, \mathbb{R})$ and $\det(A) = 1$. So by the change of variables theorem of integration, $Z$ has density $$f_Z(z) = f_Y(y)|\det(A)|^{-1} = f_Y(y), \hspace{20pt} y = A^{-1}z.$$

Edit: To verify Klenke's claim about $\mathcal{L}(X_{(1)} - X_{(n)}, \dots, X_{(n-1)} - X_{(n)} \mid X_{(n)})$, fix $x_n \in \mathbb{R}$. We need to compute the measure $$\mathcal{L}(X_{(1)} - X_{(n)}, \dots, X_{(n-1)} - X_{(n)} \mid X_{(n)} = x_n) = \mathcal{L}(X_{(1)} - x_n, \dots, X_{(n-1)} - x_n \mid X_{(n)} = x_n).$$ Assuming the joint density $f_{X_{(1)} - x_n, \dots, X_{(n-1)} - x_n, X_{(n)}}$ exists (which will be proven below), we get that $\mathcal{L}(X_{(1)} - x_n, \dots, X_{(n-1)} - x_n \mid X_{(n)} = x_n)$ has density $$f(x_1, \dots, x_{n - 1}) = f_{X_{(1)} - x_n, \dots, X_{(n-1)} - x_n, X_{(n)}}(x_1, \dots, x_n)/f_{X_{(n)}}(x_n).$$ The existence and formula for the joint density $f_{X_{(1)} - x_n, \dots, X_{(n-1)} - x_n, X_{(n)}}$ follows from the change of variables theorem as above. The rest is plugging in the formulas for when $X_i$ are i.i.d. exponential(1).