Past exam of the probability, measure and martingales course of the University of Oxford / 2021 / 3

For each integer $m\ge 1$, letIf $D_m$ had $m+1$ distinct members, finite additivity on their disjoint singleton Borel sets would give total measure at least $(m+1)/m>1$, contradicting $\mu (\mathbb{R})=1$. Thus $|D_m|\le m$.

Every positive singleton measure is at least $1/m$ for some sufficiently large $m$. Consequently the set of point atoms of a measure isA countable union of finite sets is at most countable. This proves the claim, including the possibility that the set is empty. The same argument gives the countability of point atoms of a finite measure, with $|D_m|\le m\mu (\mathbb{R})$ in place of $m$.

The product measure $\mu \otimes \nu$ is the unique probability measure on the sigma-algebra generated by measurable rectangles for whichHere the printed product notation $\mathcal{B}(\mathbb{R})\times \mathcal{B}(\mathbb{R})$ denotes the generated product sigma-algebra, conventionally written $\mathcal{B}(\mathbb{R})\otimes \mathcal{B}(\mathbb{R})$; it is not just the collection of rectangles. On ${\mathbb{R}}^2$ it equals $\mathcal{B}({\mathbb{R}}^2)$: coordinate projections show that rectangles are Borel sets, while open rectangles with rational endpoints give every open set as a countable union and hence generate the Borel sigma-algebra.

Existence is the product measure theorem for finite measures, which are in particular sigma-finite measures. Uniqueness follows because the rectangles form a pi-system generating the product sigma-algebra, and two finite measures agreeing there agree everywhere by the Dynkin lemma.^[ref]

Consider the Borel setIts indicator function is nonnegative and measurable, so the Tonelli theorem applies to the product measure, without needing any separate absolute-integrability argument. At a fixed $y\in (a,b]$, its section in the first coordinate is $(a,y]$; outside $(a,b]$ that section is empty. ThereforeThe last equality uses the definition of a cumulative distribution function on the half-closed interval. It retains all point atoms of a measure on the diagonal $x=y$, excludes a possible point atom of a measure at $x=a$, and includes $y=b$, exactly as required by the set $D$. If $a=b$, both sides vanish. This verifies the identity even when the probability measures are not non-atomic measures.^[ref]

The filtration is generated by the successive independent random variables $X_k$. Each $S_n$ is adapted, and $|S_n|\le n$ makes it an integrable random variable. Since independence gives $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=0$,Thus $(S_n)$ is a martingale.

Write the accumulated number of visits to zero asIt is adapted, and $0\le L_n\le n$. Thus $M_n=|S_n|-L_n$ is also adapted and an integrable random variable. For every integer $s$,Conditioning on $S_n$ and using independence of the next step therefore givesThe drift of the absolute value is cancelled exactly by the visit count, so $\mathbb{E}[M_{n+1}-M_n\mid {\mathcal{F}}_n]=0$. Hence both $(S_n)$ and $(M_n)$ are martingales. The latter is the absolute value martingale of a simple symmetric random walk, a discrete Tanaka formula in which visits to zero supply the correction at the corner of the absolute-value function.

For each $n$,since $S_k$ is adapted. This verifies directly that $\tau$ is a stopping time.

Set $d=b-a$. On $\{\tau >md\}$, the random walk is at some integer strictly between $a$ and $b$. If its next $d$ steps are all $+1$, it must hit $b$ within that block. These future steps are independent of ${\mathcal{F}}_{md}$ and have probability $2^{-d}$ of being all $+1$. HenceThe law of total expectation and mathematical induction give the geometric tail bound from a uniform escape probability:Therefore $\mathbb{P}(\tau <\infty )=1$. In fact this also proves $\mathbb{E}\tau \le d{2}^d<\infty$ by grouping the tail-sum formula into blocks of length $d$. No assertion of independence between successive survival events is needed.

The optional stopping theorem at a bounded stopping time states that an integrable martingale $Z_n$ satisfies $\mathbb{E}{Z}_T=\mathbb{E}{Z}_0$ whenever $T\le n$ for a fixed finite $n$. Apply this first to $S$ at $T=n\wedge \tau$. The simple symmetric random walk cannot cross an integer endpoint without hitting it, soThe preceding part gives $\tau <\infty$ almost surely, hence $S_{n\wedge \tau }\to S_{\tau }$ almost surely. The constant bound $C=\max \{-a,b\}$ permits the dominated convergence theorem, yielding $\mathbb{E}{S}_{\tau }=0$. Since $S_{\tau }$ is either $a$ or $b$, solving $ap+b(1-p)=0$ gives its full probability distribution:Now letThis counts the visit at time $0$, and excludes the exit time. Apply the same bounded-time optional stopping theorem to $M_n=|S_n|-L_n$. For each $n$ it givesThe visit counts increase to $L_{\tau }$, so monotone convergence applies to the left. Dominated convergence applies to the right using $|S_{n\wedge \tau }|\le C$. Consequently $L_{\tau }$ is an integrable random variable andThis is the origin case of expected occupation before interval exit for a simple symmetric random walk. The argument obtains integrability of the total occupation count before using any unbounded-time optional stopping for $M$, so it does not presuppose the uniform integrability asked about next. The endpoint probabilities also agree with the standard gambler's ruin calculation.^[ref]

Yes. The preceding calculation established $\mathbb{E}{L}_{\tau }=-2ab/(b-a)<\infty$. For every $n$, the stopped martingale satisfiesHere $H$ is a single nonnegative integrable random variable that bounds the whole family. Thus, for any $K>0$,The last limit follows from the dominated convergence theorem. This is precisely uniform integrability. It also gives $M_{n\wedge \tau }\to M_{\tau }=|S_{\tau }|-L_{\tau }$ with convergence in L1, and therefore $\mathbb{E}{M}_{\tau }=0$, consistently with the preceding answer. A common bound on the individual expected values alone would not have sufficed; the integrable dominating random variable is what supplies control of every tail.