Jointly Complete Sufficient Statistics for Uniform$(a, b)$ Distributions

Question

Let $\mathbf{X}= (x_1, x_2, \dots x_n)$ be a random sample from the uniform distribution on $(a,b)$, where $a < b$. Let $Y_1$ and $Y_n$ be the largest and smallest order statistics. Show that the statistic $(Y_1, Y_n)$ is a jointly complete sufficient statistic for the parameter $\theta = (a, b)$.

It is no problem for me to show sufficiency using factorization.

Question: How do I show completeness? Preferably I would like a hint.

Attempt: I can show $\mathbb E[g(T(x))] = 0$ implies $g(T(x)) = 0$ for the one parameter uniform distribution, but I am getting stuck on the two parameter uniform distribution.

I tried playing around with $\mathbb E[g(Y_1, Y_n)]$ and using the joint distribution of $Y_1$ and $Y_n$, but I am not sure if I am going in the correct direction, as the calculus is tripping me up.

Answer 1

Let's take care of the routine calculus for you, so you can get to the heart of the problem and enjoy formulating a solution. It comes down to constructing rectangles as unions and differences of triangles.

First, choose values of $a$ and $b$ that make the details as simple as possible. I like $a=0,b=1$: the univariate density of any component of $X=(X_1,X_2,\ldots,X_n)$ is just the indicator function of the interval $[0,1]$.

Let's find the distribution function $F$ of $(Y_1,Y_n)$. By definition, for any real numbers $y_1 \le y_n$ this is

$$F(y_1,y_n) = \Pr(Y_1\le y_1\text{ and } Y_n \le y_n).\tag{1}$$

The values of $F$ are obviously $0$ or $1$ in case any of $y_1$ or $y_n$ is outside the interval $[a,b] = [0,1]$, so let's assume they're both in this interval. (Let's also assume $n\ge 2$ to avoid discussing trivialities.) In this case the event $(1)$ can be described in terms of the original variables $X=(X_1,X_2,\ldots,X_n)$ as "at least one of the $X_i$ is less than or equal to $y_1$ and none of the $X_i$ exceed $y_n$." Equivalently, all the $X_i$ lie in $[0,y_n]$ but it is not the case that all of them lie in $(y_1,y_n]$.

Because the $X_i$ are independent, their probabilities multiply and give $(y_n-0)^n = y_n^n$ and $(y_n-y_1)^n$, respectively, for these two events just mentioned. Thus,

$$F(y_1,y_n) = y_n^n - (y_n-y_1)^n.$$

The density $f$ is the mixed partial derivative of $F$,

$$f(y_1,y_n) = \frac{\partial^2 F}{\partial y_1 \partial y_n}(y_1,y_n) = n(n-1)(y_n-y_1)^{n-2}.$$

The general case for $(a,b)$ scales the variables by the factor $b-a$ and shifts the location by $a$. Thus, for $a \lt y_1 \le y_n \lt b$,

$$F(y_1,y_n; a,b) = \left(\left(\frac{y_n-a}{b-a}\right)^n - \left(\frac{y_n-a}{b-a} - \frac{y_1-a}{b-a}\right)^n\right) = \frac{(y_n-a)^n - (y_n-y_1)^n}{(b-a)^n}.$$

Differentiating as before we obtain

$$f(y_1,y_n; a,b) = \frac{n(n-1)}{(b-a)^n}(y_n-y_1)^{n-2}.$$

Consider the definition of completeness. Let $g$ be any measurable function of two real variables. By definition,

$$\eqalign{E[g(Y_1,Y_n)] &= \int_{y_1}^b\int_a^b g(y_1,y_n) f(y_1,y_n)dy_1dy_n\\ &\propto\int_{y_1}^b\int_a^b g(y_1,y_n) (y_n-y_1)^{n-2} dy_1dy_n.\tag{2} }$$

We need to show that when this expectation is zero for all $(a,b)$, then it's certain that $g=0$ for any $(a,b)$.

Here's your hint. Let $h:\mathbb{R}^2\to \mathbb{R}$ be any measurable function. I would like to express it in the form suggested by $(2)$ as $h(x,y)=g(x,y)(y-x)^{n-2}$. To do that, obviously we must divide $h$ by $(y-x)^{n-2}$. Unfortunately, for $n\gt 2$ this isn't defined whenever $y=x$. The key is that this set has measure zero so we can neglect it.

Accordingly, given any measurable $h$, define

$$g(x,y) = \left\{\matrix{h(x,y)/(y-x)^{n-2} & x \ne y \\ 0 & x=y}\right.$$

Then $(2)$ becomes

$$\int_{y_1}^b\int_a^b h(y_1,y_n) dy_1dy_n \propto E[g(Y_1,Y_n)].\tag{3}$$

(When the task is showing that something is zero, we may ignore nonzero constants of proportionality. Here, I have dropped $n(n-1)/(b-a)^{n-2}$ from the left hand side.)

This is an integral over a right triangle with hypotenuse extending from $(a,a)$ to $(b,b)$ and vertex at $(a,b)$. Let's denote such a triangle $\Delta(a,b)$.

Ergo, what you need to show is that if the integral of an arbitrary measurable function $h$ over all triangles $\Delta(a,b)$ is zero, then for any $a\lt b$, $h(x,y)=0$ (almost surely) for all $(x,y)\in \Delta(a,b)$.

Although it might seem we haven't gotten any further, consider any rectangle $[u_1,u_2]\times [v_1,v_2]$ wholly contained in the half-plane $y \gt x$. It can be expressed in terms of triangles:

$$[u_1,u_2]\times [v_1,v_2] = \Delta(u_1,v_2) \setminus\left(\Delta(u_1,v_1) \cup \Delta(u_2,v_2)\right)\cup \Delta(u_2,v_1).$$

In this figure, the rectangle is what is left over from the big triangle when we remove the overlapping red and green triangles (which double counts their brown intersection) and then replace their intersection.

Consequently, you may immediately deduce that the integral of $h$ over all such rectangles is zero. It remains only to show that $h(x,y)$ must be zero (apart from its values on some set of measure zero) whenever $y \gt x$. The proof of this (intuitively clear) assertion depends on what approach you want to take to the definition of integration.

Answer 2

Following @whuber's answer, we have the joint density of $(Y_{(1)},Y_{(n)})$ $$f(y_1, y_n,a,b)=\frac{n(n-1)}{(b-a)^n}(y_n-y_1)^{n-1},\quad\forall a\leq y_1\leq y_n\leq b\text{ and }a,b\in\mathbb{R}$$

For any function $g(x_1,x_n)$ so that $\mathbb{E}_{a,b}\left[g(y_1,y_n)\right]=0,\forall a,b\in\mathbb{R}\text{ and }a<b$, we have $$0=\frac{n(n-1)}{(b-a)^n}\int_{a}^{b}\int_{a}^{y_n}g(y_1,y_n)(y_n-y_1)^{n-2} dy_1dy_n,\forall a,b\in\mathbb{R}\text{ and }a<b.$$ The integral area of $(y_1, y_n)$ is a triangle with vertices $(a,a)$, $(a,b)$ and $(b,b)$. With varying $a\in\mathbb{R}$ $b\in\mathbb{R}$ $a<b$, these triangles generate the Borel $\sigma$-algebra of $\mathcal{B}=\{(x,z)\in\mathbb{R}^2:x\leq z\}$. Thus $$0=\frac{n(n-1)}{(b-a)^n}\int_{A}g(y_1,y_n)(y_n-y_1)^{n-2} d(y_1,y_n),\text{ for any Borel set }A\subset\mathcal{B}.$$ This means $$g(y_1,y_n)(y_n-y_1)^{n-2}\equiv0,a.e.\iff g\equiv 0,a.e.$$ Thus we conclude $(Y_{(1)},Y_{(n)})$ is a complete statistic for $(a,b).$

Answer 3

I think it is worth elaborating a little bit the step of how to reach $g = 0$ a.e. $\lambda$ from the integral equation in @Tan's and @whuber's answer, as it demonstrates some classical measure theory techniques that could be used in other similar completeness proving problems.

Denote $g(x, y)(y - x)^{n - 2}$ by $f(x, y)$, then Tan's answer showed that \begin{align} \int_a^b\int_a^y f(x, y)dxdy = 0\; \text{ for all } a < b. \tag{1} \end{align}

whuber's clever geometric argument shows that $(1)$ implies that \begin{align} \int_I f(x, y)dxdy = 0\; \text{ for all } I \in \mathscr{I}. \tag{2} \end{align} where $\mathscr{I}$ is the class of bounded rectangles in $\mathbb{R}^2$:
\begin{align} \mathscr{I} = \{(a_1, b_1] \times (a_2, b_2]: a_1 < b_1, a_2 < b_2\}. \end{align}

Our goal is to prove by that $f = 0$ a.e. on $\mathbb{R}^2$ under the condition $(2)$ (which obviously implies $g = 0$ a.e. on $\mathbb{R}^2$). This is implied$^\dagger$ by $f = 0$ a.e. on $I_M := (-M, M] \times (-M, M]$ for arbitrary $M > 0$, to which we prove it below.

It is well known that $\mathscr{I}$ generates the Borel $\sigma$-field $\mathscr{R}^2$ on $\mathbb{R}^2$, hence $\mathscr{I} \cap I_M$ generates the $\sigma$-field $\mathscr{R}^2 \cap I_M := \{A \cap I_M: A \in \mathscr{R}^2\}$ in $I_M$ (see Theorem 10.1 in Probability and Measure by Patrick Billingsley for the proof). Hence if we can show that the class \begin{align} \mathscr{A} := \left\{A \in \mathscr{R}^2 \cap I_M: \int_A f(x, y)dxdy = 0\right\} \tag{3} \end{align} is a $\sigma$-field in $I_M$, then $\mathscr{I} \cap I_M \subset \mathscr{A}$ (which is implied by $(2)$) and $\sigma(\mathscr{I} \cap I_M) = \mathscr{R}^2 \cap I_M$ together imply that $\mathscr{R}^2 \cap I_M \subset \mathscr{A}$. Since $f$ is measurable, $A_1 := \{(x, y): f(x, y) > 0\} \in \mathscr{R}^2$ and $A_2 := \{(x, y): f(x, y) < 0\} \in \mathscr{R}^2$, it then follows by $\mathscr{R}^2 \cap I_M \subset \mathscr{A}$ that $A_1 \cap I_M \in \mathscr{A}$, $A_2 \cap I_M \in \mathscr{A}$. Therefore $(3)$ implies that \begin{align} \int_{A_1 \cap I_M}f(x, y)dxdy = 0, \quad \int_{A_2 \cap I_M} -f(x, y)dxdy = 0. \tag{4} \end{align}

Note that $fI_{A_1} = f^+ = \max(f, 0) \geq 0$ and $-fI_{A_2} = f^- = \max(-f, 0) \geq 0$, $(4)$ can be rewritten as \begin{align} \int_{I_M}f^+(x, y)dxdy = \int_{I_M}f^-(x, y)dxdy = 0, \end{align} which implies that $f^+ = f^- = 0$ a.e. on $I_M$. Therefore, $f = f^+ - f^- = 0$ a.e. on $I_M$.

Therefore, to complete the proof, it remains to show that $\mathscr{A}$ is a $\sigma$-field. In fact, because $\mathscr{I} \cap I_M$ is a $\pi$-system, it suffices to show that $\mathscr{A}$ is a $\lambda$-system in view of Dynkin's $\pi$-$\lambda$ theorem. To this end:

1. $I_M \in \mathscr{A}$. This follows from $(2)$ directly.
2. If $A \in \mathscr{A}$, then \begin{align} \int_{I_M - A}f(x)dxdy = \int_{I_M}f(x)dxdy - \int_A f(x)dxdy = 0 - 0 = 0, \end{align} which shows $I_M - A$ lies in $\mathscr{A}$. Therefore $\mathscr{A}$ is closed under the formation of complementation.
3. If $A_n$ lies in $\mathscr{A}$ for all $n$ and $A_1, A_2, \ldots$ are disjoint, then $|fI_{\cup_n A_n}| \leq |f|$ (note that $\int_{I_M}fd\lambda = 0$ implies that $f$ is integrable in $I_M$) and Lebesgue's dominated convergence theorem imply that \begin{align} \int_{\cup_n A_n} fd\lambda = \sum_n \int_{A_n}fd\lambda = 0, \end{align} which shows $\cup_n A_n$ lies in $\mathscr{A}$. Therefore $\mathscr{A}$ is closed under the formation of countable disjoint unions.

This completes the proof.

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$^\dagger$: For each $m \in \mathbb{N}$, define $B_m = \{x \in I_m: f(x) \neq 0\}$. Then the sequence $\{B_m\}$ is increasing and converges to the set $\cup_{m = 1}^\infty B_m = \{x \in \mathbb{R}^2: f(x) \neq 0\}$. It then follows by the continuity from below of $\lambda$ and $f = 0$ a.e. on every $I_m$ that $\lambda(\cup_{m = 1}^\infty B_m) = \lim_{m \to \infty}\lambda(B_m) = 0$, i.e., $f = 0$ a.e. on $\mathbb{R}^2$.