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A supermartingale supremum threshold need not be attained

Statement refuted

Assume AC. For every null family $P_{0}$ , every nonnegative test supermartingale $M$ for $P_{0}$ , every $P \in P_{0}$ , and every $α \in (0, 1)$ , the identity $P ⁣ (\sup_{t \geq 0} M_{t} \geq 1 / α) = P (τ_{α} < \infty), τ_{α} = \inf {t \geq 0 : M_{t} \geq 1 / α},$ holds.

Facts & Assumptions

Given: Assume AC as in the library's test-supermartingale and supermartingale definitions. We construct one probability law and a discrete nonnegative martingale $M$ with $E M_{0} \leq 1$ for which the two events have different probabilities at $α = 1 / 2$ .

[F1]

Probability measures and probability spaces: a probability measure is a countably additive measure of total mass one.

[F2]

Conditional expectation given a sigma algebra: a conditional-expectation version is measurable with respect to the conditioning sigma-algebra and has the same integral as the original variable over every event in that sigma-algebra.

[F3]

Conditional expectation as an ae class: conditional expectations are unique almost-sure classes, so equalities of conditional expectations mean almost-sure equality.

[F4]

Expectation of a nonnegative or integrable random variable: expectation is the integral against the probability measure; a bounded nonnegative measurable random variable has finite expectation.

[F5]

Monotonicity and nonnegative homogeneity of the nonnegative integral: the nonnegative integral is monotone and is homogeneous for positive scalars.

[F6]

Martingale submartingale and supermartingale: an integrable adapted process is a martingale when $E [M_{n + 1} ∣ F_{n}] = M_{n}$ almost surely for every $n$ .

[F7]

Nonnegative test supermartingales: under each null law, an adapted integrable nonnegative supermartingale with initial expectation at most one is a nonnegative test supermartingale; the definition assumes AC.

[F8]

Discrete Ville inequality and anytime threshold validity: under AC, a nonnegative test supermartingale satisfies both the supremum-event and first-crossing probability bounds by $α$ .

[A1]

AC is inherited from the conditional-expectation class, martingale and test-supermartingale definitions, and the Ville theorem ([F3], [F6]–[F8]); the probability space, filtration, conditional means, and process below are explicit (The Axiom of Choice).

Counterexample

1.1F1givenconstructalgebra

Set $p = 1 / 4$ , $c = 2$ , and $Ω = {a} \cup {b_{k} : k \geq 1}$ with $F = 2^{Ω}$ , $P ({a}) = p$ , and $P ({b_{k}}) = (1 - p) 2^{- k}$ . The geometric series makes $P$ a probability measure. For $n \geq 0$ , let $T_{n} = {a} \cup {b_{k} : k > n}$ , let $F_{n}$ 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 $(F_{n})$ is a filtration.

2.1givenF1F4F5step 1.1algebra

Define $M_{n} = 0$ on each singleton ${b_{j}}$ with $j \leq n$ , and set $M_{n} = c p / m_{n}$ on $T_{n}$ . This is $F_{n}$ -measurable and satisfies $0 \leq M_{n} < c$ . By [F1], [F4], and [F5], $E_{P} M_{n} \leq c P (Ω) = c < \infty$ , so it is integrable.

3.1F2F3F6A1step 1.1step 2.1algebra

On every revealed singleton ${b_{j}}$ , $j \leq n$ , 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 $\frac{m_{n + 1} (c p / m_{n + 1}) + (1 - p) 2^{- (n + 1)} \cdot 0}{m_{n}} = \frac{c p}{m_{n}} = M_{n},$ since the two parts' masses sum to $m_{n}$ . These equalities on the finitely many atoms of $F_{n}$ give a conditional-expectation version by [F2]; [F3] gives its almost-sure class, and [F6] makes $M$ a martingale.

4.1F1F4F7A1step 2.1step 3.1

$M_{0} = c p = 1 / 2$ , so $E_{P} M_{0} = c p P (Ω) = 1 / 2 \leq 1$ . With step 2.1 and step 3.1, [F7] shows that $M$ is a nonnegative test supermartingale for the singleton null family ${P}$ .

5.1givenF8A1step 2.1step 4.1algebra∎

Step 2.1 shows $M_{n} < c = 2$ at every finite time, so $τ_{1 / 2} = \infty$ everywhere. On $a$ , $M_{n} = c p / [p + (1 - p) 2^{- n}] \to c$ , so $\sup_{n} M_{n} = c$ ; on each $b_{k}$ , $M_{n} = 0$ for $n \geq k$ and the finitely many earlier values are all below $c$ . Hence ${\sup_{n} M_{n} \geq 2} = {a}$ , with probability $1 / 4$ , whereas $P (τ_{1 / 2} < \infty) = 0$ . Both are at most $α = 1 / 2$ , as [F8] requires, but they are not equal.

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