Discrete test supermartingales and Ville — Examples

For each $i$, the uniform-law assumption gives $P_0(P_i>\alpha )=1-\alpha$.

Define $\sigma =\inf \{1\le i\le K:P_i\le \alpha \}$, with $\inf \mathrm{\varnothing }=\mathrm{\infty }$. For every $k\ge 0$, $\{\sigma \le k\}={\bigcup }_{i=1}^{\min (k,K)}\{P_i\le \alpha \}\in {\mathcal{F}}_k$, where the union is empty for $k=0$. Thus [F3] makes $\sigma$ a stopping time, and the rejection event by look $K$ is $\{\sigma \le K\}$.

By [F1]–[F2], the probability that every look misses is the product of the marginal probabilities, so $P_0({\bigcap }_{i=1}^K\{P_i>\alpha \})=(1-\alpha {)}^K$.

The event of at least one rejection is the complement of the event in step 2.1. Therefore its probability is $1-(1-\alpha {)}^K$.

Since $0<1-\alpha <1$ and $K\ge 2$, $(1-\alpha {)}^K<1-\alpha$, so step 3.1 is strictly greater than $\alpha$.

[F4] controls arbitrary threshold stopping only when its nonnegative test-supermartingale hypotheses hold. The calculation above concerns repeated fixed-look uniform p-values and does not assume such a process, so the excess rejection probability is not a violation of Ville's inequality.

Set $p=1\mathrm{/}4$, $c=2$, and $\mathrm{\Omega }=\{a\}\cup \{b_k:k\ge 1\}$ with $\mathcal{F}=2^{\mathrm{\Omega }}$, $P(\{a\})=p$, and $P(\{b_k\})=(1-p)2^{-k}$. The geometric series makes $P$ a probability measure. For $n\ge 0$, let $T_n=\{a\}\cup \{b_k:k>n\}$, let ${\mathcal{F}}_n$ be generated by the partition $\{\{b_1\},\dots ,\{b_n\},T_n\}$ (at $n=0$ this is the trivial partition), and put $m_n=P(T_n)=p+(1-p)2^{-n}$. These partitions refine as $n$ increases, so $({\mathcal{F}}_n)$ is a filtration.

Define $M_n=0$ on each singleton $\{b_j\}$ with $j\le n$, and set $M_n=cp\mathrm{/}{m}_n$ on $T_n$. This is ${\mathcal{F}}_n$-measurable and satisfies $0\le M_n<c$. By [F1], [F4], and [F5], ${\mathbb{E}}_P{M}_n\le cP(\mathrm{\Omega })=c<\mathrm{\infty }$, so it is integrable.

On every revealed singleton $\{b_j\}$, $j\le n$, both $M_n$ and $M_{n+1}$ are zero. The remaining atom $T_n$ splits into $\{b_{n+1}\}$, of mass $(1-p)2^{-(n+1)}$, and $T_{n+1}$, of mass $m_{n+1}$. The conditional mean of $M_{n+1}$ on $T_n$ is $\frac{m_{n+1}(cp\mathrm{/}{m}_{n+1})+(1-p)2^{-(n+1)}\cdot 0}{m_n}=\frac{cp}{m_n}=M_n,$ since the two parts' masses sum to $m_n$. These equalities on the finitely many atoms of ${\mathcal{F}}_n$ give a conditional-expectation version by [F2]; [F3] gives its almost-sure class, and [F6] makes $M$ a martingale.

$M_0=cp=1\mathrm{/}2$, so ${\mathbb{E}}_P{M}_0=cpP(\mathrm{\Omega })=1\mathrm{/}2\le 1$. With step 2.1 and step 3.1, [F7] shows that $M$ is a nonnegative test supermartingale for the singleton null family $\{P\}$.

Step 2.1 shows $M_n<c=2$ at every finite time, so ${\tau }_{1\mathrm{/}2}=\mathrm{\infty }$ everywhere. On $a$, $M_n=cp\mathrm{/}[p+(1-p)2^{-n}]\to c$, so ${\sup }_n{M}_n=c$; on each $b_k$, $M_n=0$ for $n\ge k$ and the finitely many earlier values are all below $c$. Hence $\{{\sup }_n{M}_n\ge 2\}=\{a\}$, with probability $1\mathrm{/}4$, whereas $P({\tau }_{1\mathrm{/}2}<\mathrm{\infty })=0$. Both are at most $\alpha =1\mathrm{/}2$, as [F8] requires, but they are not equal.