Sequential Validity and Ville's Inequality

Fix any $\gamma \in (0,1)$ and define ${\tau }_{\gamma }=\inf \{t\ge 0:E_t\ge 1\mathrm{/}\gamma \}$, with $\inf \mathrm{\varnothing }=\mathrm{\infty }$. For every $n$, $\{{\tau }_{\gamma }\le n\}={\bigcup }_{k=0}^n\{E_k\ge 1\mathrm{/}\gamma \}\in {\mathcal{F}}_n$, so ${\tau }_{\gamma }$ is a stopping time. Also $\{{\tau }_{\gamma }\wedge N\le n\}=\{{\tau }_{\gamma }\le n\}$ for $n<N$ and is the whole space for $n\ge N$, so each ${\tau }_{\gamma }\wedge N$ is a bounded stopping time.

The constant time $0$ is a stopping time bounded by $0$, and step 1.1 shows ${\tau }_{\gamma }\wedge N$ is a bounded stopping time. Bounded optional sampling therefore gives ${\mathbb{E}}_P{E}_{{\tau }_{\gamma }\wedge N}\le {\mathbb{E}}_P{E}_0\le 1$.

On $\{{\tau }_{\gamma }\le N\}$, $E_{{\tau }_{\gamma }\wedge N}\ge 1\mathrm{/}\gamma$; on its complement the stopped value is $E_N\ge 0$ almost surely. Hence ${\mathbb{E}}_P{E}_{{\tau }_{\gamma }\wedge N}\ge {\gamma }^{-1}P({\tau }_{\gamma }\le N)$, and step 2.1 yields $P({\tau }_{\gamma }\le N)\le \gamma$.

The events $\{{\tau }_{\gamma }\le N\}$ increase with $N$ and their union is $\{{\tau }_{\gamma }<\mathrm{\infty }\}={\bigcup }_{t\ge 0}\{E_t\ge 1\mathrm{/}\gamma \}$. Continuity from below and step 3.1 therefore give $P(\mathrm{\exists }t\ge 0:E_t\ge 1\mathrm{/}\gamma )\le \gamma$.

Let $H=\{{\sup }_{t\ge 0}{E}_t\ge 1\mathrm{/}\alpha \}$ and suppose $q=P(H)>\alpha$. Since $q\le 1$, $\beta =(\alpha +q)\mathrm{/}2$ lies in $(\alpha ,1)$, so $1\mathrm{/}\beta <1\mathrm{/}\alpha$. If every $E_t\le 1\mathrm{/}\beta$, then ${\sup }_t{E}_t\le 1\mathrm{/}\beta <1\mathrm{/}\alpha$, contrary to $H$; hence $H\subseteq \{\mathrm{\exists }t:E_t\ge 1\mathrm{/}\beta \}$ even if the supremum is not attained. Step 4.1 with $\gamma =\beta$ gives $q\le \beta <q$, a contradiction. Therefore $P(H)\le \alpha$, and since $P$ was arbitrary, the bound holds under every null law.

Fix $P\in {\mathcal{P}}_0$ and $t\ge 0$. For every $\lambda$, $P(E_t^{\lambda }\ge 0)=1$, so $E_t^{\lambda ,+}=E_t^{\lambda }$ $P$-almost surely and has the same expectation. Apply F3 at the deterministic times $0$ and $t$ to get ${\mathbb{E}}_P{E}_t^{\lambda ,+}={\mathbb{E}}_P{E}_t^{\lambda }\le {\mathbb{E}}_P{E}_0^{\lambda }={\mathbb{E}}_P{E}_0^{\lambda ,+}\le 1$. The pointwise nonnegative map $(\lambda ,\omega )\mapsto E_t^{\lambda ,+}(\omega )$ is product-measurable; Tonelli F2 makes ${\bar{E}}_t$ ${\mathcal{F}}_t$-measurable and gives ${\mathbb{E}}_P{\bar{E}}_t={\int }_{\mathrm{\Lambda }}{\mathbb{E}}_P{E}_t^{\lambda ,+}\pi (d\lambda )\le 1$. Hence $N_t=\{{\bar{E}}_t=\mathrm{\infty }\}\in {\mathcal{F}}_t$ is null under every $P\in {\mathcal{P}}_0$. The common modification ${\tilde{E}}_t={\bar{E}}_t$ off $N_t$ and $0$ on $N_t$ is finite, adapted, nonnegative and integrable under every null law, with ${\mathbb{E}}_P{\tilde{E}}_0\le 1$.

Fix $P\in {\mathcal{P}}_0$, $t\ge 1$, and $A\in {\mathcal{F}}_{t-1}$. Since $E_s^{\lambda ,+}=E_s^{\lambda }$ $P$-almost surely for $s=t-1,t$, the component supermartingale condition and the event-integral identity F5 give ${\int }_A{E}_t^{\lambda ,+}dP={\int }_A{E}_t^{\lambda }dP\le {\int }_A{E}_{t-1}^{\lambda }dP={\int }_A{E}_{t-1}^{\lambda ,+}dP$ for every $\lambda$. Tonelli F2 integrates these inequalities over $\pi$; since $N_t$ and $N_{t-1}$ are $P$-null, replacing the mixtures by ${\tilde{E}}_t$ and ${\tilde{E}}_{t-1}$ does not change these integrals. Thus ${\int }_A{\tilde{E}}_{t}dP\le {\int }_A{\tilde{E}}_{t-1}dP$.

Let $U={\mathbb{E}}_P[{\tilde{E}}_t\mid {\mathcal{F}}_{t-1}]$, which exists by F4 and AC A1. For every $A\in {\mathcal{F}}_{t-1}$, the event-integral identity F5 and step 2.1 give ${\int }_A(U-{\tilde{E}}_{t-1})dP\le 0$. Take $A=\{U>{\tilde{E}}_{t-1}\}$. If $P(A)>0$, then for some integer $m\ge 1$ the subset where $U-{\tilde{E}}_{t-1}\ge 1\mathrm{/}m$ has positive measure, making that integral strictly positive, a contradiction. Therefore $U\le {\tilde{E}}_{t-1}$ almost surely. As $P$ and $t$ were arbitrary, $\tilde{E}$ is a test supermartingale for ${\mathcal{P}}_0$.

The map $g:\mathbb{R}\to [0,1]$ defined by $g(x)=1$ for $x\le 0$ and $g(x)=\min (1,1\mathrm{/}x)$ for $x>0$ is Borel measurable. Thus $p_t=g(E_t)$ is ${\mathcal{F}}_t$-measurable and $[0,1]$-valued. For $0<\alpha <1$, the threshold $1\mathrm{/}\alpha >1$, so $p_t\le \alpha$ holds exactly when $E_t\ge 1\mathrm{/}\alpha$.

Taking the union over all $t$ in the equivalence of step 1.1 and applying F3 gives $P(\mathrm{\exists }t\ge 0:p_t\le \alpha )=P(\mathrm{\exists }t\ge 0:E_t\ge 1\mathrm{/}\alpha )\le P({\sup }_{t\ge 0}{E}_t\ge 1\mathrm{/}\alpha )\le \alpha$ for every $P\in {\mathcal{P}}_0$. This is the defining anytime-valid condition.

For a fixed $t$ and $0<u<1$, $\{p_t\le u\}\subseteq \{\mathrm{\exists }s:p_s\le u\}$, so its probability is at most $u$ by step 2.1. Since $E_t$ is finite, $p_t>0$; hence the bound at $u=0$ is zero, and at $u=1$ it is trivial. For any stopping time $\tau$, $\{\tau <\mathrm{\infty },p_{\tau }\le u\}={\bigcup }_{t\ge 0}(\{\tau =t\}\cap \{p_t\le u\})$ is contained in the all-times crossing event for $0<u<1$; the endpoints follow as above. Thus fixed-time and stopping-time validity both hold.

If ${\mathcal{P}}_0$ is nonempty, form the fixed-time experiment with data space $(\mathrm{\Omega },{\mathcal{F}}_t)$, parameter set $\mathrm{\Theta }={\mathcal{P}}_0$, laws given by restriction to ${\mathcal{F}}_t$, and null set ${\mathrm{\Theta }}_0=\mathrm{\Theta }$. The observation is the identity map from the original sample space to this data space. Step 1.1 makes $p_t$ a statistic and step 3.1 gives its superuniform inequality under every law in ${\mathrm{\Theta }}_0$, so F4 makes it a valid p-value in this experiment. If ${\mathcal{P}}_0$ is empty, all lawwise assertions are vacuous.

Fix $\theta \in \mathrm{\Theta }$. Under $P_{\theta }$, define $p_t^{\theta }=1$ when $E_t^{\theta }\le 0$ and $p_t^{\theta }=\min (1,1\mathrm{/}{E}_t^{\theta })$ otherwise. By adaptedness in F1 and the Borel transformation used in F2, each $p_t^{\theta }$ is ${\mathcal{F}}_t$-measurable. Hence $\{\theta \in C_t\}=\{E_t^{\theta }<1\mathrm{/}\alpha \}=\{p_t^{\theta }>\alpha \}$ is ${\mathcal{F}}_t$-measurable.

For this fixed $\theta$, F2 gives $P_{\theta }(\mathrm{\exists }t\ge 0:p_t^{\theta }\le \alpha )\le \alpha$. The complement is exactly $\{\theta \in C_t\text{ for every }t\ge 0\}$, whose probability is therefore at least $1-\alpha$.

Steps 1.1 and 2.1 hold for every $\theta$. They give the membership measurability and simultaneous-coverage clauses in F3, so $(C_t)$ is a confidence sequence.