Tests, Neyman–Pearson and Valid P-values

By [F1] and [F8], $\int pd\mu =P(X)=1$. The functions $f^{+},f^{-}$ are measurable by [F3] and satisfy $0\le f^{+},f^{-}\le M$. Products $f^{+}p$ and $f^{-}p$ are measurable by [F4].

Monotonicity [F5], [F1], and [F8] give $\int f^{\pm }dP\le MP(X)=M$ and $\int f^{\pm }pd\mu \le M\int pd\mu =M$. Hence $f^{+},f^{-}$ are integrable under $P$ and $f^{+}p,f^{-}p$ are integrable under $\mu$ by [F6].

Apply [F2] separately to the nonnegative measurable functions $f^{+}$ and $f^{-}$. It gives $\int f^{+}dP=\int f^{+}pd\mu$ and $\int f^{-}dP=\int f^{-}pd\mu$.

Subtract the second equality in step 2.2 from the first. By [F6] and [F7], both sides are the integrals of the corresponding signed functions, so $\int fdP=\int fpd\mu$.

Put $q(t)=Q(\{x:T(x)>t\})$ for $t\ge 0$ and $S=\{t\ge 0:q(t)\le \alpha \}$. The sets $\{T>n\}$ decrease to the empty set because $T$ is finite-valued and the natural numbers are unbounded by [F7]. Their first member has measure at most one, so [F4] gives $q(n)\to 0$. Thus $S\mathrm{\ne }\mathrm{\varnothing }$. If $s\in S$ and $t\ge s$, then $\{T>t\}\subseteq \{T>s\}$, so [F3] gives $q(t)\le q(s)\le \alpha$. Hence $S$ is upward closed, is bounded below by zero, and has a finite infimum $c$ by [F6].

If $t>c$, then $t\in S$: otherwise no member of $S$ lies below $t$, making $t$ a lower bound larger than $c$. Therefore $q(c+1\mathrm{/}n)\le \alpha$ for every positive integer $n$. By [F7], for every $T(x)>c$ some $n$ satisfies $1\mathrm{/}n<T(x)-c$, so the sets $\{T>c+1\mathrm{/}n\}$ increase to $\{T>c\}$. Thus [F5] gives $Q(T>c)\le \alpha$.

If $c>0$, set $s_n=c(1-1\mathrm{/}(n+1))$ for $n\ge 1$. Then $s_n<c$, and [F7] gives $s_n\uparrow c$; also $s_n\mathrm{\notin }S$, so $q(s_n)>\alpha$. The sets $\{T>s_n\}$ decrease to $\{T\ge c\}$, since every $T(x)<c$ is eventually below some $s_n$ by [F7]. Their first member has finite measure since $Q(X)=1$. Thus [F4] gives $Q(T\ge c)\ge \alpha$. If $c=0$, the same lower bound follows from $Q(T\ge 0)=1$.

Since $\{T\ge c\}$ is the disjoint union of $\{T>c\}$ and $\{T=c\}$, finite additivity gives $Q(T=c)=Q(T\ge c)-Q(T>c)$. If this mass is positive, define $\gamma =(\alpha -Q(T>c))\mathrm{/}Q(T=c)$; the bounds in steps 2.1 and 2.2 give $0\le \gamma \le 1$ and the required equality. If the mass is zero, those same bounds force $Q(T>c)=\alpha$, and $\gamma =0$ works.

Define $h(x)=1\mathrm{/}{p}_0(x)$ when $p_0(x)>0$ and $h(x)=0$ when $p_0(x)=0$. It is finite and nonnegative. For $t<0$, $\{h>t\}=X$; for $t=0$, $\{h>t\}=\{p_0>0\}$; for $t>0$, $\{h>t\}=\{0<p_0<1\mathrm{/}t\}$. Each set is measurable by [F1] and [F2], so [F3] makes $h$ measurable.

By [F1] and step 1.1, $p_{1}h$ is a finite nonnegative measurable function by [F4]. On $\{p_0>0\}$ it equals $p_1\mathrm{/}{p}_0$, and on $\{p_0=0\}$ it is zero. Hence $R=p_{1}h$ and has all the asserted properties.

Fix $x\in X$. The two weights $\phi (x)$ and $1-\phi (x)$ are nonnegative and sum to one. Thus $K(x,\mathrm{\varnothing })=0$ and $K(x,\{0,1\})=1$. If $(B_j{)}_{j\ge 0}$ are pairwise disjoint subsets of $\{0,1\}$, each action belongs to at most one $B_j$, so ${\mathbf{1}}_{{\cup }_j{B}_j}(a)={\sum }_j{\mathbf{1}}_{B_j}(a)$ for $a=0,1$. Substitution into the formula for $K$ gives $K(x,{\cup }_j{B}_j)={\sum }_{j}K(x,B_j)$; hence $K(x,\cdot )$ is a probability measure on the binary action sigma-algebra.

For the four action events, the evaluation functions are $K(x,\mathrm{\varnothing })=0$, $K(x,\{0\})=1-\phi (x)$, $K(x,\{1\})=\phi (x)$, and $K(x,\{0,1\})=1$. They are measurable by [F1] and [F2]. Therefore [F3] makes $K$ a probability kernel; [F5] makes it a randomized procedure, and [F6] identifies its test function as $K(x,\{1\})=\phi (x)$.

For fixed $\theta$, the action section $a\mapsto L(\theta ,a)$ is one of $a$, $1-a$, or $0$. Each is nonnegative and measurable on the full sigma-algebra of $\{0,1\}$, so [F2] defines its risk.

On ${\mathrm{\Theta }}_0$, the section is $a={\mathbf{1}}_{\{1\}}(a)$, whose integral under $K(x,\cdot )$ is $K(x,\{1\})=\phi (x)$ by [F1] and [F3]. On ${\mathrm{\Theta }}_1$, the section is $1-a={\mathbf{1}}_{\{0\}}(a)$, whose integral is $K(x,\{0\})=1-\phi (x)$ because $K(x,\cdot )$ is a probability measure.

Substituting these inner integrals into [F2] and using [F4] gives $R(\theta ,K)=E_{\theta }\phi (X)$ on ${\mathrm{\Theta }}_0$ and $R(\theta ,K)=E_{\theta }[1-\phi (X)]$ on ${\mathrm{\Theta }}_1$.

Since $0\le \phi \le 1$ and $0\le 1-\phi \le 1$, [F5] and [F8] show that each has integral at most $1$. Thus their absolute-value integrals are finite, so they are integrable by [F6]. Linearity [F7] and [F8] give $E_{\theta }[1-\phi ]=1-E_{\theta }[\phi ]$. The two risk identities now have the asserted form.

Let $A_{\alpha }=\{x:p(x)\le \alpha \}$. This event is measurable because $p$ is a statistic and $[0,\alpha ]$ is Borel in $[0,1]$. Thus ${\phi }_{\alpha }={\mathbf{1}}_{A_{\alpha }}$ is a measurable map into $[0,1]$, and [F3] supplies a randomized test with this rejection function. Since ${\phi }_{\alpha }$ takes only the values $0$ and $1$, that kernel assigns probability one to the corresponding action, so the test is deterministic.

For every $\theta \in {\mathrm{\Theta }}_0$, [F4] and validity [F1] give $E_{\theta }[{\phi }_{\alpha }(X)]=P_{\theta }(A_{\alpha })\le \alpha$, proving the pointwise bound. If a nonempty ${\mathrm{\Theta }}_1\subseteq \mathrm{\Theta }\setminus {\mathrm{\Theta }}_0$ is specified, taking the supremum over the nonempty set ${\mathrm{\Theta }}_0$ gives size at most $\alpha$ by [F5], and therefore level $\alpha$.

The sets $U,A,B$ in [F1] are measurable by [F2] and [F3]. For $\alpha >0$ they are pairwise disjoint, so [F4] makes ${\phi }_{\alpha }^{\ast }={\mathbf{1}}_U+{\mathbf{1}}_A+\gamma {\mathbf{1}}_B$ measurable; for $\alpha =0$, [F4] makes ${\mathbf{1}}_U$ measurable.

Let $N=\{p_0=0\}$. By [F1], [F6], and the zero-function case of [F7], $P_0(N)={\int }_N{p}_{0}d\mu =0$, since $p_0$ vanishes on $N$. Thus $P_0(U)=0$ by [F8].

The disjointness and $0\le \gamma \le 1$ show that every value of ${\phi }_{\alpha }^{\ast }$ lies in $[0,1]$. Hence [F5] makes it a randomized test.

If $\alpha =0$, [F1] gives ${\phi }_0^{\ast }={\mathbf{1}}_U$. The simple integral identity [F7] and [F9] give $E_{P_0}[{\phi }_0^{\ast }]=P_0(U)=0$, so [F10] makes its size $0=\alpha$.

Suppose $\alpha >0$. Since $R=0$ on $N$ and $c\ge 0$, $A=\{R>c\}$. Also $B=\{R=c\}$ when $c>0$; when $c=0$, the two sets differ only by $N$. Step 1.2 makes that difference $P_0$-null. The three disjoint indicator terms in [F1] therefore have expectation $P_0(R>c)+\gamma P_0(R=c)=\alpha$ by [F7], [F9], and the threshold equation in [F1]. By [F10], the test has size $\alpha$.

First suppose $\alpha >0$, so the cutoff $c$ in [F5] is finite. Put $D={\phi }_{\alpha }^{\ast }-\phi$. Then $D$ is measurable and $\mathrm{\mid }D\mathrm{\mid }\le 1$. By [F2] and [F6], $\int p_{i}d\mu =P_i(\mathcal{X})=1$, so $\mathrm{\mid }D{p}_i\mathrm{\mid }\le p_i$ makes $D{p}_i$ integrable for $i=0,1$. Since $c$ is finite, $D(p_1-c{p}_0)$ is also integrable.

On $\{p_0=0\}$, if $p_1>0$ then [F5] gives ${\phi }_{\alpha }^{\ast }=1$, so $D(p_1-c{p}_0)\ge 0$; if $p_1=0$, the product is zero. On $\{p_0>0\}$, write $R=p_1\mathrm{/}{p}_0$. If $R>c$, then ${\phi }_{\alpha }^{\ast }=1$ and both $D$ and $p_1-c{p}_0$ are nonnegative. If $R<c$, then ${\phi }_{\alpha }^{\ast }=0$ and both are nonpositive. If $R=c$, the second factor is zero. Thus $D(p_1-c{p}_0)\ge 0$ everywhere.

The signed density identity [F15] and linearity [F8] give $\int D(p_1-c{p}_0)d\mu =(E_{P_1}{\phi }_{\alpha }^{\ast }-E_{P_1}\phi )-c(E_{P_0}{\phi }_{\alpha }^{\ast }-E_{P_0}\phi )$.

By monotonicity of the nonnegative integral, the integral in step 2.2 is nonnegative. Hence $E_{P_1}{\phi }_{\alpha }^{\ast }-E_{P_1}\phi \ge c(E_{P_0}{\phi }_{\alpha }^{\ast }-E_{P_0}\phi )$. The candidate has null rejection probability $\alpha$ by [F5], while the competitor has null rejection probability at most $\alpha$ by [F4]. Since $c\ge 0$, the right-hand side is nonnegative, proving the desired power inequality for $\alpha >0$.

Now suppose $\alpha =0$. Then $E_{P_0}\phi =0$, so [F6] gives $\int \phi p_{0}d\mu =0$. For positive integers $n,m$, set $B_{n,m}=\{\phi \ge 1\mathrm{/}n\}\cap \{p_0\ge 1\mathrm{/}m\}$. On this set, $\phi p_0\ge 1\mathrm{/}(nm)$; [F7] and [F11] therefore give $0\ge {\int }_{B_{n,m}}\phi p_{0}d\mu \ge \mu (B_{n,m})\mathrm{/}(nm)$. Thus $\mu (B_{n,m})=0$, and absolute continuity [F1] gives $P_1(B_{n,m})=0$. For each positive-integer pair $(n,m)$ put $C_{n,m}=\{\phi \ge 1\mathrm{/}n\}\cap \{p_0\ge 1\mathrm{/}m\}$, enumerate these pairs by increasing $n+m$ and then by increasing $n$, and write the resulting sequence of sets as $(C_k{)}_{k\ge 0}$. If $\phi (x)>0$ and $p_0(x)>0$, [F16] supplies positive integers $n,m$ with $1\mathrm{/}n<\phi (x)$ and $1\mathrm{/}m<p_0(x)$, so $x\in {\bigcup }_k{C}_k$. Conversely every $C_k$ is contained in $\{\phi >0,p_0>0\}$; hence this union is exactly that set. Put $D_0=C_0$ and $D_k=C_k\setminus {\bigcup }_{j<k}{C}_j$ for $k\ge 1$. The $D_k$ are measurable by [F13], pairwise disjoint, and each $D_k\subseteq C_k$, so the preceding null-set calculation and [F12] give $P_1(D_k)=0$. Their union equals ${\bigcup }_k{C}_k$: if $x$ lies in that union, the nonempty set of indices $k$ with $x\in C_k$ has a least member by [F14], which puts $x$ in the corresponding $D_k$. Countable additivity [F9] makes the union $P_1$-null. On its complement, $\phi =0$ wherever $p_0>0$; where $p_0=0<p_1$, [F5] gives ${\phi }_0^{\ast }=1$, and the remaining set $\{p_0=p_1=0\}$ has $P_1$ measure zero because $p_1=0$ there and [F2], [F6] give $P_1(A)={\int }_A{p}_{1}d\mu$. Therefore $\phi \le {\phi }_0^{\ast }$ $P_1$-almost everywhere, and monotonicity of expectation proves the power inequality at $\alpha =0$.