Confidence Sets and Inversion — Examples

The full power set $2^{\mathbb{Z}}$ is an algebra of subsets closed under countable unions, so it is a sigma-algebra by [F5]. Explicitly set $P_{\theta }(A)=\frac{1}{3}({\mathbf{1}}_A(\theta -1)+{\mathbf{1}}_A(\theta )+{\mathbf{1}}_A(\theta +1))$ for every $A\subseteq \mathbb{Z}$. These values are nonnegative and have total mass $1$. For a pairwise disjoint sequence $A_m$, ${\sum }_m{P}_{\theta }(A_m)=\frac{1}{3}{\sum }_{z\in \{-1,0,1\}}{\sum }_m{\mathbf{1}}_{A_m}(\theta +z)=\frac{1}{3}{\sum }_{z\in \{-1,0,1\}}{\mathbf{1}}_{{\cup }_m{A}_m}(\theta +z)=P_{\theta }({\cup }_m{A}_m)$. Thus it is countably additive, hence a probability measure, and specifies the experiment family in [F6]. Under $P_{\theta }$, $Q(X;\theta )$ assigns mass $1\mathrm{/}3$ to each of $-1,0,1$, independently of $\theta$; hence $Q$ is a pivot with $P_{\theta }(Q\in B)=1$.

Define $F_Q(t):=P_{\theta }(Q\le t)$ and the generalized quantile $q_r:=\inf \{t\in \mathbb{R}:F_Q(t)\ge r\}$. By the definition of $C$, the event $\{\theta \in C(X)\}$ is $\{\mathrm{\mid }X-\theta \mathrm{\mid }\le 1\}$, which occurs for all three possible observations. Thus its probability and coverage equal $1$; this exceeds $4\mathrm{/}5$, while [F4] also applies since $P_{\theta }(Q\in B)=1\ge 4\mathrm{/}5$. The pivot cdf is $0$ below $-1$, $1\mathrm{/}3$ on $[-1,0)$, $2\mathrm{/}3$ on $[0,1)$, and $1$ from $1$ onward. Thus the generalized $1\mathrm{/}10$ and $9\mathrm{/}10$ quantiles are $-1$ and $1$, and their closed interval contains all atoms.

Let $G$ have the standard normal law. Its density is $\varphi (u)=e^{-u^2\mathrm{/}2}\mathrm{/}\sqrt{2\pi }$, positive everywhere and bounded above by $1\mathrm{/}\sqrt{2\pi }$, with total mass one by [F1]. Thus $\mathrm{\Phi }(t)=P(G\le t)$ satisfies $0<\mathrm{\Phi }(b)-\mathrm{\Phi }(a)={\int }_a^b\varphi (u)du\le (b-a)\mathrm{/}\sqrt{2\pi }$ for every $a<b$, proving strict increase and continuity. The events $\{G\le m\}$ and $\{G>-m\}$ increase to the whole space as integer $m\to \mathrm{\infty }$; [F6] gives $\mathrm{\Phi }(m)\to 1$ and $P(G>-m)\to 1$, hence $\mathrm{\Phi }(-m)\to 0$. Choose finite endpoints bracketing $1-\alpha \mathrm{/}2$; [F5] gives a quantile between them, and strict increase gives uniqueness.

The standard normal density is symmetric and has no atoms. Therefore $\mathrm{\Phi }(-z_{1-\alpha \mathrm{/}2})=\alpha \mathrm{/}2$, and the probability of $[-z_{1-\alpha \mathrm{/}2},z_{1-\alpha \mathrm{/}2}]$ is $1-\alpha$.

For every candidate mean $\eta$, $q_{\eta }(y)=(y-\eta )\mathrm{/}\sigma$ is measurable and has standard normal law under $P_{\eta }$, so it is a pivot. The event $\eta \in [y-z\sigma ,y+z\sigma ]$ is exactly $-z\le q_{\eta }(y)\le z$; applying [F4] and [F3] gives exact coverage $1-\alpha$ for every $\eta$.

For all $x,y\in {\mathbb{R}}^n$, $\mathrm{\mid }\bar{x}-\bar{y}\mathrm{\mid }\le {\max }_i\mathrm{\mid }{x}_i-y_i\mathrm{\mid }\le \mathrm{\parallel }x-y{\mathrm{\parallel }}_2$. The residual map $r(x)=x-\bar{x}\mathbf{1}$ is linear and orthogonal to $\mathbf{1}$, so Pythagoras gives $\mathrm{\parallel }r(x-y){\mathrm{\parallel }}_2\le \mathrm{\parallel }x-y{\mathrm{\parallel }}_2$; the reverse triangle inequality then gives $\mathrm{\mid }s(x)-s(y)\mathrm{\mid }\le \mathrm{\parallel }x-y{\mathrm{\parallel }}_2\mathrm{/}\sqrt{n-1}$. Hence $s$ is continuous, $D_n=\{x:s(x)>0\}$ is open, and its trace Borel sets form a sigma-algebra by [F6]. By [F1], every full law ${\tilde{P}}_{\mu ,\lambda }$ gives $D_n$ probability one, so its restriction $P_{\mu ,\lambda }$ remains a probability measure; [F7] therefore gives the stated experiment for each fixed $\lambda >0$.

Put $c=t_{\nu ,1-\alpha \mathrm{/}2}$, which exists uniquely by [F2]. Symmetry and continuity give $F_{t_{\nu }}(0)=1\mathrm{/}2$; strict increase and $1-\alpha \mathrm{/}2>1\mathrm{/}2$ imply $c>0$. Since $F_{t_{\nu }}(c)=1-\alpha \mathrm{/}2$, symmetry gives $F_{t_{\nu }}(-c)=\alpha \mathrm{/}2$; continuity means neither endpoint is an atom, so $P(-c\le T_{\nu }\le c)=1-\alpha$.

For each candidate mean $\eta \in \mathbb{R}$, define $q_{\eta }(x)=(\bar{x}-\eta )\mathrm{/}(s(x)\mathrm{/}\sqrt{n})$ on $D_n$. Its denominator is positive there, so it is continuous and hence measurable on the subspace $D_n$ by the open-set definition of its trace Borel sigma-algebra [F6]. For every Borel $B\subseteq \mathbb{R}$, the restricted probability satisfies $P_{\eta ,\lambda }\{q_{\eta }\in B\}={\tilde{P}}_{\eta ,\lambda }\{x\in D_n:q_{\eta }(x)\in B\}$, and [F1] gives this value as the common $t_{\nu }$ probability because the full law assigns probability one to $D_n$. Thus $(q_{\eta }{)}_{\eta \in \mathbb{R}}$ is a pivot by [F4].

Inverting $B=[-c,c]$ gives $\{\eta :q_{\eta }(x)\in B\}=[\bar{x}-cs(x)\mathrm{/}\sqrt{n},\bar{x}+cs(x)\mathrm{/}\sqrt{n}]=I(x)$ by solving the two inequalities, and each membership section is measurable because $q_{\eta }$ is measurable. Step 1.2 gives the pivot mass $\nu (B)=1-\alpha$; the pivot-inversion theorem [F5] and confidence-set definition [F3] therefore give exact coverage $1-\alpha$ for every mean and every fixed variance.

The sample sigma-algebra is the Borel sigma-algebra of ${\mathbb{R}}^n$ by [F7]. The maximum map $m(x)={\max }_i{x}_i$ satisfies $\mathrm{\mid }m(x)-m(y)\mathrm{\mid }\le {\max }_i\mathrm{\mid }{x}_i-y_i\mathrm{\mid }$, so it is continuous. The preimages of open sets are therefore open; since the Borel sigma-algebra is generated by open sets, preimages of every Borel set are Borel and $M=m(X)$ is measurable. Also $m\mathrm{/}\theta$ is continuous because $\theta >0$. The coordinate variables are independent by [F1], and [F2] gives their product joint law.

If $0\le u\le 1$, then $P_{\theta }(M\mathrm{/}\theta \le u)=P_{\theta }(X_1\le u\theta ,\dots ,X_n\le u\theta )={\prod }_{i=1}^{n}u=u^n$. For $u<0$ the probability is $0$, and for $u\ge 1$ it is $1$. This law does not depend on $\theta$, so $M\mathrm{/}\theta$ is a pivot by [F3].

The distribution function $u^n$ established in step 1.2 is continuous. Therefore $P_{\theta }({\alpha }^{1\mathrm{/}n}\le M\mathrm{/}\theta \le 1)=1-({\alpha }^{1\mathrm{/}n}{)}^n=1-\alpha$. By [F5], pivot inversion gives exact coverage.

Since $\theta >0$, the inequality ${\alpha }^{1\mathrm{/}n}\le M\mathrm{/}\theta \le 1$ is equivalent to $M\le \theta \le M\mathrm{/}{\alpha }^{1\mathrm{/}n}$. Thus the inverted set is $[M,M\mathrm{/}{\alpha }^{1\mathrm{/}n}]\cap (0,\mathrm{\infty })$, and [F4] identifies its coverage with the probability computed in step 2.1.

The full power set of $\{0,\dots ,n\}$ is an algebra closed under countable unions, hence a sigma-algebra by [F5]. The binomial theorem [F1] gives ${\sum }_{j=0}^n\left(\genfrac{}{}{0px}{}{n}{j}\right)p^j(1-p{)}^{n-j}=(p+1-p{)}^n=1$; all terms are nonnegative for $p\in [0,1]$, so [F6] makes the displayed masses a probability measure, including $p=0,1$. These laws form an experiment as in [F7].

For $k<n$, count pairs consisting of a $j$-element subset of an $n$-element set and a distinguished element inside it. Counting in the two orders gives $j\left(\genfrac{}{}{0px}{}{n}{j}\right)=n\left(\genfrac{}{}{0px}{}{n-1}{j-1}\right)$ for $1\le j\le k$. Counting instead a distinguished element outside the subset gives $(n-j)\left(\genfrac{}{}{0px}{}{n}{j}\right)=n\left(\genfrac{}{}{0px}{}{n-1}{j}\right)$ for $0\le j\le k$, by [F4]. The first identity is not used at $j=0$: in differentiating $p^j$, that boundary term is zero, so the first derivative sum starts at $j=1$, while the second identity includes $j=0$. Since $F_p(k)$ is a polynomial, it is continuous on $[0,1]$, and for $0<p<1$ termwise differentiation and the two identities give $\frac{d}{dp}{F}_p(k)={\sum }_{j=1}^{k}j\left(\genfrac{}{}{0px}{}{n}{j}\right)p^{j-1}(1-p{)}^{n-j}-{\sum }_{j=0}^k(n-j)\left(\genfrac{}{}{0px}{}{n}{j}\right)p^j(1-p{)}^{n-j-1}=-n\left(\genfrac{}{}{0px}{}{n-1}{k}\right)p^k(1-p{)}^{n-k-1}<0.$ When $k=0$, the first sum is empty and the $j=0$ term of the second sum gives the same formula. By [F9], $F_p(k)$ is strictly decreasing on $[0,1]$. Also $F_0(k)=1$ and $F_1(k)=0$ because $k<n$. Applying [F10] to this continuous polynomial on $[0,1]$ gives a point $U_k\in (0,1)$ with $F_{U_k}(k)=\alpha$; strict decrease makes it unique.

For each fixed candidate $p_0$, $F_{p_0}(k)$ is nondecreasing in integer $k$. Let $k_0$ be the least $k\in \{0,\dots ,n\}$ with $F_{p_0}(k)\ge \alpha$, which exists because $F_{p_0}(n)=1$. The deterministic test function is ${\phi }_{p_0}(x)={\mathbf{1}}_{\{F_{p_0}(x)<\alpha \}}$; it is measurable on this finite space. Its rejection event is $\{X<k_0\}$, whose probability is $0$ if $k_0=0$, and otherwise is $F_{p_0}(k_0-1)<\alpha$. The null is $\{p_0\}$, and the alternative $[0,1]\setminus \{p_0\}$ is nonempty; by [F8] the test size equals this rejection probability and is at most $\alpha$.

For an observed $x<n$, strict decrease of $F_p(x)$ in $p$ makes $\{p_0:F_{p_0}(x)\ge \alpha \}=[0,U_x]$; for $x=n$, the set is $[0,1]=[0,U_n]$. Thus $C(X)$ is precisely the inversion of those acceptance events. The level calculation in step 2.1 and [F3] imply its coverage is at least $1-\alpha$, in the sense of [F2].