Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 201 (Advanced Probability)

For an integrable random variable $X$ and a sub-sigma-algebra $\mathcal{G}\subseteq \mathcal{F}$, the conditional expectation $Y=\mathbb{E}[X\mid \mathcal{G}]$ is an integrable, $\mathcal{G}$-measurable random variable satisfyingfor every $G\in \mathcal{G}$.

Conditional expectation is unique up to almost sure equality. If $Y$ and $Z$ both satisfy the definition, then for every $G\in \mathcal{G}$,The events $\{Y>Z\}$ and $\{Z>Y\}$ belong to $\mathcal{G}$. Testing on them, or first on $\{Y-Z\ge 1/k\}$ and $\{Z-Y\ge 1/k\}$, shows that both have probability zero. Hence $Y=Z$ almost surely.

If $G\in \mathcal{G}\cap \mathcal{H}$, independence makes $G$ independent of itself, soThus every event in $\mathcal{G}\cap \mathcal{H}$ has probability zero or one. The intersection is trivial modulo null sets, and the independent sigma-algebras have trivial intersection result givesalmost surely.

Put $S=X_1+X_2$. By symmetry,Their sum is $S$, which is measurable with respect to $\mathcal{G}=\sigma (S)$, soThereforeThis also follows from the Gaussian conditional expectation formula because $\mathrm{Cov}(X_1,S)/\mathrm{Var}(S)=1/2$.

If $\mathcal{G}\subseteq \mathcal{H}$, then $\mathbb{E}[X\mid \mathcal{G}]$ is already $\mathcal{H}$-measurable, soIf $\mathcal{H}\subseteq \mathcal{G}$, the tower property of conditional expectation givesFinally, if $\mathcal{G}$ and $\mathcal{H}$ are independent, the $\mathcal{G}$-measurable variable $\mathbb{E}[X\mid \mathcal{G}]$ is independent of $\mathcal{H}$. Its conditional expectation given $\mathcal{H}$ is its mean $\mathbb{E}[X]$. Part c shows that the right side is also $\mathbb{E}[X]$.

The equation fails for general nonnested sigma-algebras. On the four-point space $\Omega =\{1,2,3,4\}$ with uniform probability, letand take $X=1_A$. The intersection $\mathcal{G}\cap \mathcal{H}$ is trivial, soBut $X$ is $\mathcal{G}$-measurable andwhich is zero on $B^c$ and is not almost surely $1/2$. This exhibits the failure of iterated conditional expectation over nonnested sigma-algebras.

For a finite horizon $N$, let $U_N[a,b]$ be the number of completed upcrossings by time $N$. Use the predictable strategy that holds one unit of the process after a visit below $a$ until the next visit above $b$. For a supermartingale, the expected gain of this nonnegative predictable martingale transform is nonpositive. Pathwise, the completed trades earn at least $(b-a)U_N[a,b]$, while an unfinished final trade can lose at most $(X_N-a{)}^{-}$. HenceTaking expectations and using $X_N\ge 0$ gives the Doob upcrossing inequalityAs $N\to \infty$, monotone convergence yields

For every rational $0\le a<b$, part a implies $U[a,b]<\infty$ almost surely. The intersection of these probability-one events over the countable collection of rational pairs still has probability one. On this event, ifsome rational interval $[a,b]$ lies strictly between them, forcing infinitely many upcrossings, a contradiction. Thus $X_n$ has an extended limit almost surely.

The limit cannot be $+\infty$ on a set of positive probability: Fatou lemma and the supermartingale property giveNonnegativity excludes $-\infty$. Therefore $X_n$ converges almost surely to a finite random variable, proving the almost sure supermartingale convergence theorem in this case.

If ${\xi }_k=S_k-S_{k-1}$, thenThe centered increments ${\xi }_k-\mu$ are independent of the past and have mean zero, sois a martingale.

The increment variance isso independence givesLinear interpolation makes the supremum of the absolute centered process occur at an integer time. The Doob L2 maximal inequality therefore gives

The moment-generating function of one increment isThus withindependence givesThis is the exponential martingale of a biased simple random walk.

LetOn $A$, one has $T\le n$, $S_T\ge \mu T+\epsilon n$, and convexity of $\psi$ gives $\psi (\theta )-\mu \theta \ge 0$. Hence for $\theta >0$,The optional stopping theorem applies because $T$ is bounded, so $\mathbb{E}{Z}_T=1$. Therefore

Optimize over $\theta >0$ for the upper deviation and apply the same argument with $\theta <0$ to the lower deviation. Since the supremum of the linearly interpolated centered walk is attained at grid points, the Legendre transform of a cumulant-generating functionand the union bound give

A process $(X_t{)}_{t\ge 0}$ is Brownian motion in ${\mathbb{R}}^d$ when $X_0=0$, its paths are almost surely continuous, and for $0\le t_0<\cdots <t_k$ the increments $X_{t_j}-X_{t_{j-1}}$ are independent centered Gaussian vectors with covariance $(t_j-t_{j-1})I_d$.

The paths of $U{X}_t$ are continuous and start at zero. Its increments are independent because they are deterministic functions of the independent increments of $X$. They are centered Gaussian, and orthogonality givesThus $UX$ is Brownian motion. This is the orthogonal invariance of Brownian motion.

Apply the orthogonal transformationThe processes $D$ and $C$ are independent one-dimensional Brownian motions, with $D_0=\sqrt{2}a$ and $C_0=0$. The meeting time is the first time $D$ hits zero, which is almost surely finite by one-dimensional Brownian recurrence.

The Brownian reflection principle gives the first-passage density from $x>0$ to zero asSubstituting $x=\sqrt{2}a$ gives the meeting time of two independent Brownian motions density

At the meeting time,The process $C$ is independent of $T$, so conditional on $T=u$ the meeting position is $N(0,u/2)$. Independently, $B_t-B_u$ is $N(0,t-u)$. ThereforeFor $0<s\le t$, integrate this conditional Gaussian distribution against the density from part c:

The random-walk form of the Skorokhod embedding theorem says the following. If $S_n=Y_1+\cdots +Y_n$, where the $Y_i$ are independent and identically distributed withthen on a space carrying a Brownian motion $B$ there are stopping timessuch that $(B_{T_n}{)}_{n\ge 0}$ has the same law as $(S_n{)}_{n\ge 0}$, and the increments $T_n-T_{n-1}$ are independent and identically distributed with mean ${\sigma }^2$.

To prove the one-step statement, first note that every centered distribution is a mixture of centered two-point distributions. Indeed, match the equal-mass size-biased measures $x\mathbb{P}(Y\in dx)$ on $(0,\infty )$ and $|x|\mathbb{P}(Y\in dx)$ on $(-\infty ,0)$. This produces a random pair $(L,R)$ of positive numbers such that, conditionally on $(L,R)$, $Y$ has values $-L,R$ with probabilitiesand $\mathbb{E}[LR]=\mathbb{E}{Y}^2={\sigma }^2$. Include the atom at zero by taking the stopping time zero.

Choose $(L,R)$ independently of $B_t$ and stop Brownian motion on first leaving $(-L,R)$. The Brownian exit from an interval formulas give the displayed two-point probabilities and conditional mean stopping time $LR$. Thus $B_T$ has the law of $Y$ and $\mathbb{E}T={\sigma }^2$.

Starting from $T_0=0$, repeat this construction after each $T_{n-1}$. The Strong Markov property makes the new Brownian increments independent copies of the first embedding, proving the Skorokhod embedding of a centered random walk.

By the strong law of large numbers,almost surely. Brownian scaling and a maximal inequality show that changing Brownian time by $o(n)$ changes its value by $o_{\mathbb{P}}(\sqrt{n})$; explicitly, first restrict to $|T_n-n{\sigma }^2|\le \delta n$, bound the Brownian maximum over a time interval of length $2\delta n$, and then let $\delta \downarrow 0$. Consequentlyin probability.

Butfor every $n$. Since $B_{T_n}$ has the law of $S_n$, Slutsky theorem proves the Central limit theorem from the Skorokhod embedding:

A real Lévy process satisfies $X_0=0$ almost surely, has stationary independent increments, is stochastically continuous, and is taken in its almost surely càdlàg version.

The Lévy–Khintchine theorem states that there is a unique triplet $(a,b,K)$, where $a\in \mathbb{R}$, $b\ge 0$, and $K$ is a measure on $\mathbb{R}\setminus \{0\}$ satisfyingsuch thatConversely every such triplet is the characteristic triplet of a Lévy process.

Let $B$ be Brownian motion and let $N(ds,dx)$ be an independent Poisson random measure with intensity $dsK(dx)$. Writing $\tilde{N}=N-dsK(dx)$ for its compensated version, the Lévy–Itô decomposition constructsThe four terms are independent drift, Gaussian, compensated small-jump and compound-Poisson large-jump components.

Almost surely differentiable paths must be continuous, so the jump measure must vanish: $K=0$. A nonzero Brownian component has almost surely nowhere-differentiable paths, so also $b=0$. Conversely, if $b=0$ and $K=0$, then $X_t=at$ is differentiable. Thus

The Brownian and drift components are continuous. Any nonzero Lévy measure produces jumps: a set bounded away from zero with positive finite $K$-measure gives a nontrivial compound Poisson component, and increasing such sets detects every nonzero $K$. Hence paths are almost surely continuous exactly when

The compensated small-jump integral is integrable after localization and has mean zero, while the number of large jumps on a compact time interval is finite. Its absolute first moment is finite exactly when the large-jump sizes have finite first moment. Thus $(X_t)$ is integrable exactly when

The Brownian component has finite variance $bt$, and the compensated jump integral has variancewhen this integral is finite. Conversely, a finite second moment forces the jump measure to have a finite second moment. Therefore $(X_t^2)$ is integrable exactly whenThese four equivalences are the Path and moment criteria from a Lévy triplet.