Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 201 (Advanced Probability)

Put $M_n=\mathbb{E}[Z\mid {\mathcal{F}}_n]$. Each $M_n$ is ${\mathcal{F}}_n$-measurable and integrable, with $|M_n|\le \mathbb{E}[|Z|\mid {\mathcal{F}}_n]\le 1$. Since a filtration is increasing, the tower property of conditional expectation givesThus $(M_n)$ is the conditional-expectation martingale associated with $Z$.

The Martingale convergence theorem gives an almost-sure limit $M_{\infty }$, because $|M_n|\le 1$. The dominated convergence theorem also gives $M_n\to M_{\infty }$ in $L^1$.

To identify the limit, take $A\in {\bigcup }_n{\mathcal{F}}_n$. For some $N$, $A\in {\mathcal{F}}_N$, and for every $n\ge N$,Passing to the $L^1$ limit preserves this equality. The sets for which $\mathbb{E}[M_{\infty }{1}_A]=\mathbb{E}[Z{1}_A]$ form a monotone class containing the algebra ${\bigcup }_n{\mathcal{F}}_n$, so the equality holds throughout ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. Since $M_{\infty }$ is ${\mathcal{F}}_{\infty }$-measurable, it is $\mathbb{E}[Z\mid {\mathcal{F}}_{\infty }]$. This proves the conditional-expectation convergence along a filtration both almost surely and in $L^1$.

Define $Z_n={\sup }_{m\ge n}|X_m-X|$. Then $0\le Z_n\le 2$ and $Z_n\downarrow 0$ almost surely. The dominated convergence theorem gives $\mathbb{E}{Z}_n\to 0$.

Set $W_n=\mathbb{E}[Z_n\mid {\mathcal{F}}_n]$. Since $Z_{n+1}\le Z_n$, the tower property of conditional expectation givesso $(W_n)$ is a nonnegative supermartingale. The almost sure supermartingale convergence theorem gives $W_n\to W_{\infty }$ almost surely, and Fatou lemma yields $\mathbb{E}{W}_{\infty }\le {\mathrm{liminf}}_n\mathbb{E}{W}_n=0$. Hence $W_n\to 0$ almost surely; because $\mathbb{E}{W}_n=\mathbb{E}{Z}_n\to 0$, convergence also holds in $L^1$.

Finally,The second term tends to zero almost surely and in $L^1$ by part b. The first does so by the preceding argument, proving the moving-variable conditional-expectation convergence.

The almost sure supermartingale convergence theorem says that a supermartingale whose negative parts have uniformly bounded expectations converges almost surely to a finite integrable limit. In particular, every nonnegative supermartingale converges almost surely.

Here the process is uniformly bounded, say $0\le X_n\le C$. Let $X_{\infty }={\lim }_n{X}_n$, whose existence follows from the theorem. The dominated convergence theorem then givesThe limit on the right also exists directly because the expectations of a supermartingale form a decreasing sequence.

For $x\ne 0$, the Strong Markov property at the first step gives the discrete mean-value identityAt $x=0$, $v(0)=1$ while the same average is at most one. Thus $v$ is a bounded superharmonic function on ${\mathbb{Z}}^3$. Conditioning on the natural filtration and using the one-step Markov property givesTherefore $(v(X_n))$ is a nonnegative supermartingale.

Let $A_n=\{X_m=0\text{ for some }m\ge n\}$. By the Markov property,The events $A_n$ decrease to the event that the walk visits zero infinitely often, which has probability zero by the stated transience assumption. HencePart a and the almost sure supermartingale convergence theorem give an almost-sure limit $V\ge 0$. By Fatou lemma, $\mathbb{E}V\le {\mathrm{liminf}}_n\mathbb{E}[v(X_n)]=0$, so $V=0$ almost surely.

The probability generating function of each $X_i$ is $\exp (z-1)$. Independence makes the generating function of $S_n$ equal to $\exp (n(z-1))$, so the addition of independent Poisson random variables gives $S_n\sim \mathrm{Pois}(n)$.

The Poisson distribution has mean and variance $n$, hence $\mathbb{E}{Y}_n=0$ and $\mathbb{E}[Y_n^2]=1$. Since $Y_n^{-}\le |Y_n|$, Chebyshev inequality gives

The Poisson central limit theorem, or the ordinary central limit theorem applied to the $X_i$, gives $Y_n\stackrel{d}{\to }Y$ for a standard normal random variable $Y$. The negative-part map $x\mapsto x^{-}:=\max (-x,0)$ is continuous, so the continuous mapping theorem gives

For $K>0$, the Cauchy-Schwarz inequality and part a giveThus $(Y_n^{-})$ is uniformly integrable. Combining this with the weak convergence of random variables from part b yields convergence of the first moments:By symmetry of the standard normal density,

For $S_n\sim \mathrm{Pois}(n)$,ConsequentlyPart c says that this tends to $1/\sqrt{2\pi }$. Rearranging gives the Stirling formula

For $0\le s\le t$, write $B_t=B_s+(B_t-B_s)$. The increment is independent of ${\mathcal{F}}_s$ and is normally distributed with variance $t-s$. Its moment generating function givesThe process is integrable for every real $\lambda$, so it is the exponential Brownian martingale.

Differentiate the conditional identity from part a. To justify doing so, fix a compact parameter interval $|\lambda |\le L$. Every $n$th derivative of $M_{\lambda }(t)$ is a polynomial in $B_t,t,$ and $\lambda$ times $M_{\lambda }(t)$, and its absolute value is bounded byThis bound is integrable because a Gaussian random variable has every polynomially weighted exponential moment. Dominated differentiation of conditional expectation therefore givesThus every parameter derivative of the exponential Brownian martingale is itself a martingale.

Let $T={\tau }_b\wedge {\tau }_{-a}$ and $A=\{{\tau }_b<{\tau }_{-a}\}$. The Brownian exit time $T$ is finite almost surely. Optional stopping of the bounded martingale $B_{t\wedge T}$ givesFor $T_n=T\wedge n$, optional stopping of $B_t^2-t$ gives $\mathbb{E}{T}_n=\mathbb{E}[B_{T_n}^2]\le \max (a^2,b^2)$. Letting $n\to \infty$ by monotone and bounded convergence proves $\mathbb{E}T=\mathbb{E}[B_T^2]=ab$.

The third derivative in part b at $\lambda =0$ is the cubic martingale $B_t^3-3t{B}_t$. Optional stopping at $T_n$ is valid because $T_n$ is bounded. Since $B_{T_n}$ is bounded and $T_n\to T$ in $L^1$, its stopped identity passes to the limit and givesPut $x=\mathbb{E}[T{1}_A]$ and $y=\mathbb{E}[T{1}_{A^c}]$. Then $x+y=ab$, whileSolving gives $x=ab(2a+b)/(3(a+b))$. Dividing by $\mathbb{P}(A)=a/(a+b)$ proves the conditional Brownian interval-exit time formula

The Brownian reflection principle reflects a path after its first hit of $x>0$ and givesSymmetry of the centered normal distribution gives $2\mathbb{P}(B_t\ge x)=\mathbb{P}(|B_t|\ge x)$. Both random variables are nonnegative, so their tail distributions agree for every $x$, proving the Brownian running maximum identity $M_t\stackrel{d}{=}|B_t|$.

Continuity on the compact interval ensures that a maximum time exists. Fix a rational $r\in (0,1)$ and defineBy Brownian time reversal and the Brownian running maximum law, $U_r\stackrel{d}{=}|N(0,r)|$. Independent increments give $V_r\stackrel{d}{=}|N(0,1-r)|$ independently of $U_r$. Their continuous distributions imply $\mathbb{P}(U_r=V_r)=0$.

If the maximum were attained at two distinct times, a rational $r$ strictly between them would make the maxima on $[0,r]$ and $[r,1]$ equal, hence $U_r=V_r$. A countable union over rational $r$ still has probability zero. Therefore the time of the Brownian maximum is almost surely unique.

For fixed $s\in (0,1)$, uniqueness givesThe two sides of the comparison are independent and distributed as $\sqrt{s}|Z_1|$ and $\sqrt{1-s}|Z_2|$ for independent standard normal random variables. Rotational invariance of $(Z_1,Z_2)$ makes its angle uniform, soThe endpoint values follow by continuity. Thus the time of the Brownian maximum has the arcsine distribution.

For $m>0$, the Brownian reflection principle givesas $t\to \infty$. Hence Brownian motion hits every positive integer almost surely. Applying the same argument to $-B$ shows that it hits every negative integer almost surely. Taking the countable intersection of these probability-one events provesArbitrarily late positive and negative values occur, and path continuity forces a zero between successive values of opposite sign. Thus recurrence of one-dimensional Brownian motion gives infinitely many visits to zero.

The pair $(B,\tilde{B})$ is a Brownian motion in ${\mathbb{R}}^{2d}$. Assume inductively that $T_{i-1}<\infty$. By the Strong Markov property at $T_{i-1}$, the differenceis a one-dimensional Brownian motion with variance rate two, started from its current value. By recurrence of one-dimensional Brownian motion, it hits zero in finite time almost surely. Hence $T_i<\infty$. Induction through $i=1,\dots ,d$ proves $T_d<\infty$ almost surely.

Start Brownian motions at arbitrary $x,y\in {\mathbb{R}}^d$. Use the successive meeting times from part b, but after coordinate $i$ meets, drive that coordinate of the second process with the first process's increments forever. The Strong Markov property shows that each marginal remains a $d$-dimensional Brownian motion. By part b every coordinate is eventually locked, so the resulting coordinatewise coalescing coupling of Brownian motions has an almost surely finite coalescence time $T$.

Because $f$ is bounded and harmonic, Dynkin formula for Brownian motion shows that $f(B_t^x)$ and $f(B_t^y)$ are bounded martingales. ThereforeSince $T<\infty$ almost surely, the right side tends to zero. Thus $f(x)=f(y)$ for all $x,y$, proving the Brownian coupling proof of the harmonic Liouville theorem.