Independence of the spacing of order statistics characterizes exponential distribution?

Question

The question is: Let $Y_1 < Y_2$ be the order statistics of a random sample of size $2$ from a distribution of the continuous type which has p.d.f $f(x)>0$ provided $x \geq 0$, and $0$ elsewhere. Show that the independence of $Z_1 = Y_1$ and $Z_2=Y_2-Y_1$ characterizes the exponential distribution.

This is as far as I went: The joint distribution of the order statistics : $$f(y_1,y_2)=2f(y_1)f(y_2), 0 < y_1 < y_2$$.

The Jacobian of change of variable is $1$, thus $$g(z_1,z_2)=f(z_1,z_1+z_2)=2f(z_1)f(z_1+z_2)$$.

$Z_1$ and $Z_2$ are independent if and only if $$g(z_1,z_2)=g(z_1)g(z_2)$$ Not sure how to continue, can anyone please help?

Edit: I found a similar question: Independence of spacing of order statistics of exponential distribution My question is just the reverse of the problem.

Answer 1

Small corrections to your work so far (being clear about the domains of the densities):

• the joint density is $f_{Y_1, Y_2}(y_1, y_2) = 2 f(y_1) f(y_2)$ only for $0 \le y_1 \le y_2$; it is zero elsewhere.
• then $g_{Z_1, Z_2}(z_1, z_2) = 2 f(z_1) f(z_1+z_2)$ for $z_1, z_2 \ge 0$; it is zero elsewhere.

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Showing that $f(z_1+z_2)$ factorizes.

We have $$g_{Z_1}(z_1) = \int_0^\infty g_{Z_1, Z_2}(z_1, u) \, du = 2 f(z_1) \int_0^\infty f(z_1+ u) \, du, \qquad \forall z_1 \ge 0,$$ but by independence we also have $$g_{Z_1}(z_1) g_{Z_2}(z_2) = g_{Z_1, Z_2}(z_1, z_2), \qquad \forall z_1, z_2 \ge 0$$ as you noted. Combining these two leads to $$g_{Z_2}(z_2)\int_{0}^\infty f(z_1 + u) \, du = f(z_1 + z_2), \qquad \forall z_1, z_2 \ge 0.$$

In particular, there exist functions $h_1, h_2$ such that $$f(z_1 + z_2) = h_1(z_1) h_2(z_2), \quad \forall z_1, z_2 \ge 0.$$

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Exploiting the fact that $f(z_1+z_2)$ factorizes to determine the functional form of $f$.

We have $f(z) = h_1(0) h_2(z) = h_1(z) h_2(0)$ for all $z \ge 0$, so $f$, $h_1$, and $h_2$ are the same up to a multiplicative constant. In particular, if we define $\tilde{f}(z) = f(z)/f(0)$, we have the functional equation $$\tilde{f}(z_1 + z_2) = \tilde{f}(z_1) \tilde{f}(z_2), \qquad \forall z_1, z_2 \ge 0,$$ which can be shown to have solution $\tilde{f}(z) = e^{cz}$ assuming $\tilde{f}$ is continuous. So, $f(z) = ae^{cz}$ for some constants $a$ and $c$. Since $f$ is a density on $[0, \infty)$, this narrows it down to the exponential distributions.