Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 32 (Advanced Probability)

Put $C={\sup }_p\mathbb{E}|X_p|<\infty$. Since $x\mapsto x^{+}$ is an increasing convex function, the conditional Jensen inequality and the submartingale property giveFor fixed $n$ and $p\ge n$, the tower property of conditional expectation therefore impliesChoose versions for this countable family so that all these inequalities hold outside one null set. Its nonnegative increasing limit $M_n$ is ${\mathcal{F}}_n$-measurable. The monotone convergence theorem shows thatIn particular the limit is finite almost surely. Thus the increasing conditional means converge to an integrable nonnegative $M_n$, rather than merely to a possibly infinite extended value. This is the construction of the positive martingale majorant of an L1-bounded submartingale.

Use conditional monotone convergence and the tower property of conditional expectation in the preceding construction:This proves the martingale property. The construction gives $M_n\ge 0$ and ${\sup }_n\mathbb{E}{M}_n\le C$, so $M$ is bounded in $L^1$. Taking $p=n$ in the increasing family also gives $M_n\ge X_n^{+}\ge X_n$. Define $Y_n=M_n-X_n$. Then $Y$ is nonnegative and adapted, andThus $Y$ is a supermartingale, with $\mathbb{E}|Y_n|\le \mathbb{E}{M}_n+\mathbb{E}|X_n|\le 2C$. The requested conclusion isThis positive martingale majorant of an L1-bounded submartingale is not asserted to have uniform integrability; boundedness in $L^1$ alone does not imply that stronger property.

The almost sure supermartingale convergence theorem applies to the $L^1$-bounded supermartingale $X$. Hence $X_n\to X_{\infty }$ almost surely, and the Fatou lemma gives $\mathbb{E}|X_{\infty }|\le {\sup }_n\mathbb{E}|X_n|<\infty$. Take the version of $X_{\infty }$ obtained from the sequence, so it is measurable for ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. SetThe tower property of conditional expectation makes $M$ a martingale, and the uniform integrability of conditional expectations makes it uniformly integrable. Explicitly, with $Z=X_{\infty }$, $C=\mathbb{E}|Z|$, and $R,K>0$, we have $|M_n|\le R+\mathbb{E}[|Z|1_{\{|Z|>R\}}\mid {\mathcal{F}}_n]$, soFirst choose $R$ large, then $K$ large. This proves uniform integrability uniformly in $n$.

The uniformly integrable martingale convergence theorem gives a limit $M_{\infty }$ both almost surely and in $L^1$. For $A\in {\mathcal{F}}_j$ and $n\ge j$, $\mathbb{E}[M_n{1}_A]=\mathbb{E}[X_{\infty }{1}_A]$. Passing to the $L^1$ limit and then extending from the algebra ${\bigcup }_j{\mathcal{F}}_j$ to ${\mathcal{F}}_{\infty }$ identifies $M_{\infty }=X_{\infty }$ almost surely. Finally,Thus $X=M+Y$ with $M$ uniformly integrable and $Y$ a supermartingale tending almost surely to zero. The terminal decomposition of an L1-bounded supermartingale does not require $Y\ge 0$ or $Y_n\to 0$ in $L^1$. Indeed, for $X$ equal to the negative of the fair-coin doubling martingale, $X_{\infty }=0$, $M=0$, and $Y=X$ is negative and has constant absolute mean one.

The moment-generating function satisfies $0<\varphi (\lambda )<\infty$. The proposed process is nonnegative and adapted to the natural filtration of the random walk, and independence of the increments givesMoreover $X_{n+1}$ is independent of ${\mathcal{F}}_n$, soTherefore $M^{\lambda }$ is a mean-one nonnegative martingale, the exponential martingale of a random walk. For $\lambda =0$ it is the constant process one.

For Borel sets $A_1,\dots ,A_n$, multiply the finite-time Radon-Nikodym derivative into the product law of the increments:The factorization, extended from rectangles to the product sigma-algebra, proves independence and identical distributions. Their common law is the exponential tiltingFor $\lambda >0$, $|x|e^{\lambda x}$ is bounded on $x\le 0$, while $x{e}^{\lambda x}\le C_{\epsilon }{e}^{(\lambda +\epsilon )x}$ on $x\ge 0$. Choosing a small two-sided neighborhood of $\lambda$ still contained in $(0,\infty )$ gives an integrable dominating function for the derivative. Differentiation under the integral sign then yieldsThe mean under one-sided exponential tilting needs an endpoint qualification. At $\lambda =0$, finiteness of positive exponential moments makes $X_1^{+}$ integrable, but gives no integrability of $X_1^{-}$. The formula holds with the right derivative and the extended mean ${\varphi }_{+}^{\prime }(0)=\mathbb{E}{X}_1\in [-\infty ,\infty )$. To see this, apply monotone convergence to $(1-e^{-h{X}_1^{-}})/h$ and dominated convergence to $(e^{h{X}_1^{+}}-1)/h$ as $h\downarrow 0$. A finite mean at zero is not guaranteed by the printed hypotheses: $\mu (\{-j\})=6/({\pi }^2{j}^2)$, $j\ge 1$, has finite $\varphi (\lambda )$ for every $\lambda \ge 0$ but $\mathbb{E}{X}_1=-\infty$.

First take $k$ to be a nonnegative integer. The exponential tilting constructed from the independent increments restricts to ${\mathbb{P}}_n^{\lambda }$ on each ${\mathcal{F}}_n$. Since its Radon-Nikodym derivative is strictly positive, finite-time change of measure givesUnder the tilted law, $L_n=(M_n^{\lambda }{)}^{-1}$ is a martingale, becauseUse the stopped likelihood ratio under exponential tilting at this finite horizon. In detail, $\{{\tau }_k=j\}\in {\mathcal{F}}_j$, soHere the last expression is defined only on the displayed event, so no value of $M_{\infty }^{\lambda }$ is being assumed. The upward skip-free random walk has integer increments bounded above by one, which imply $S_{{\tau }_k}=k$ on a finite hit: the preceding value is at most $k-1$, and there is no upward overshoot. ConsequentlyAt $\lambda ={\lambda }_0$, the mean under one-sided exponential tilting is finite and positive. The strong law of large numbers gives $S_n/n\to {\varphi }^{\prime }({\lambda }_0)>0$ under ${\mathbb{P}}^{{\lambda }_0}$, so every integer level is hit almost surely. Since $\varphi ({\lambda }_0)=1$, passing to $n\to \infty$ givesLet $H={\sup }_{n\ge 0}{S}_n$. The event $\{H\ge k\}$ is exactly the finite-hit event: an integer sequence with all its values below $k$ has supremum at most $k-1$. Also $\mathbb{P}(H=\infty )={\lim }_k{e}^{-{\lambda }_{0}k}=0$. Subtraction of consecutive tail probabilities gives the geometric maximum of an upward skip-free random walk:This is the geometric distribution counting failures before the first success, with success parameter $1-e^{-{\lambda }_0}$.

The PDF writes only $k\ge 0$, without explicitly declaring integer levels. If it is read as allowing real $k$, then ${\tau }_k={\tau }_{\lceil k\rceil }$ and the displayed factor must be $e^{-\lambda \lceil k\rceil }$; the hitting probability is likewise $e^{-{\lambda }_0\lceil k\rceil }$. For example, increments $+1$ and $-1$ with probabilities $1/3$ and $2/3$ have ${\lambda }_0=\log 2$: hitting level $1/2$ has probability $1/2$, not $2^{-1/2}$. The integer interpretation supplies the intended formula.

Fix $x>0$ and let $A=\{M_0<x\}\in {\mathcal{F}}_0$. On $A$, continuity of the paths implies $0\le M_{t\wedge T_x}\le x$, and $M_{T_x}=x$ on $\{T_x<\infty \}$. Bounded-time optional stopping gives, for every $t$,On $A\cap \{T_x=\infty \}$, the stopped process is $M_t\to 0$; on $A\cap \{T_x<\infty \}$ it eventually equals $x$. The bound by $x$ permits conditional dominated convergence, yieldingOn $A^c$, $T_x=0$, so the conditional hitting probability is one. Finally, a continuous nonnegative path tending to zero attains any positive supremum: after a sufficiently late time it is below half that supremum, and its maximum on the earlier compact interval is attained. Thus $\{M^{\ast }\ge x\}=\{T_x<\infty \}$, including equality at the level. We have proved the conditional form of the maximal identity for a continuous nonnegative local martingale tending to zero:The restriction to $A$ before taking the limit is important: $M_0$ need not be bounded, and the unstopped martingale need not have uniform integrability.

On a product extension of the probability space, take a uniform distribution variable $U$ on $(0,1)$ independent of the original sigma-algebra. For $x>0$,These are the same conditional tails as in the maximal identity for a continuous nonnegative local martingale tending to zero. Taking expectations identifies the unconditional probability distributions:If $M_0=0$, define $M_0/U=0$; the endpoint $U=0$ is a null event if one uses $[0,1]$ instead. The possible atom at zero has mass $\mathbb{P}(M_0=0)$, since the conditional probability of a positive maximum is zero exactly on that event. For fixed $M_0=m>0$, the conditional law is the Pareto distribution of shape one and minimum $m$, with density $m/x^2$ on $x>m$.

The Brownian reflection principle, applied to the negative increments of the Brownian motion started at $a$, giveswhere $\Phi$ is the standard normal distribution function. Hence $T_0<\infty$ almost surely. The stopped Brownian motion $M_t=B_{t\wedge T_0}$ is a continuous nonnegative martingale, starts at $a$, and eventually vanishes. The maximal identity for a continuous nonnegative local martingale tending to zero therefore gives, for $H={\sup }_{0\le t\le T_0}{B}_t$,Equivalently,There is no atom at $a$. This is the Pareto distribution of shape one, or the translated version of the maximum before a lower Brownian barrier.

For standard Brownian motion $B$ and $h>0$, let $T_h$ be its first hitting time of $h$. Define the reflected path byleaving a path unchanged if $T_h=\infty$. The Brownian reflection principle says that $\hat{B}$ has the same path probability distribution as $B$. In its endpoint form, for $t>0$ and every Borel set $A\subset (-\infty ,h)$,Here is a direct proof of preservation of the path law. For a fixed integer horizon $L$, let $T^{(m)}=2^{-m}\lceil 2^m{T}_h\rceil \wedge L$, with an infinite first term interpreted as infinity. This is a stopping time taking finitely many deterministic grid values. On $\{T^{(m)}=j{2}^{-m}\}$, the history determining that event is measurable at $j{2}^{-m}$, and subsequent Brownian motion increments are independent of that history. Those increments and their negatives have the same joint law, by symmetry of centered normal distributions and independent increments. Thus reflecting after $T^{(m)}$ around $B_{T^{(m)}}$ preserves the full path law. For any fixed horizon $t<L$, the reflected paths converge uniformly on $[0,t]$ to the displayed reflected path, by continuity as $T^{(m)}$ decreases to $T_h\wedge L$. This includes $T_h=\infty$, since reflection after $L$ has no effect before $t$. Passing to all finite-dimensional probability distributions proves the assertion; no prior recurrence claim is needed. Equivalently, the Strong Markov property gives the same proof by symmetry of the fresh increments after the hit. Reflection is an involution preserving the hitting event and mapping the indicated endpoint $y$ to $2h-y$. An endpoint above $h$ forces a hit by continuity; this proves the endpoint identity, rather than assuming it.

For $S_t={\sup }_{0\le s\le t}{B}_s$, split $\{T_h\le t\}$ according to whether $B_t$ is above or below $h$. Since $B_t$ has no atom at $h$, reflection givesBoth variables are nonnegative, so their positive tails identify the law:For $t>0$ this is the half-normal distribution of scale $\sqrt{t}$; at $t=0$ both variables vanish.

For the two separated intervals, putHere $A_0$ is measurable for ${\mathcal{F}}_b$, $Z$ has the normal distribution $N(0,c-b)$, and $H$ is a function of increments after $c$. The independent increments make these three objects mutually independent. The later absolute maximum is $B_b+Z+H$, so equality of the two maxima means $Z=A_0-H$. Conditional on $(A_0,H)$, this specifies a single value of the nondegenerate normal distribution $Z$, and hence has probability zero. ThereforeThis proves the atomless maxima on separated Brownian intervals without incorrectly treating the two absolute maxima as independent.

The exit time $T$ is finite almost surely. Indeed, if the Brownian motion is still in $(-b,a)$ at an integer time, its next unit increment exceeds $a+b$ with a fixed probability $p>0$ independent of the past; on this event it must exit before the next integer time. Consequently $\mathbb{P}(T>n)\le (1-p{)}^n\to 0$. Path continuity gives $B_T\in \{a,-b\}$, and the two exit events partition the probability space up to null sets.

For each real $\lambda$, the exponential Brownian martingale $Z_t^{\lambda }=\exp (\lambda B_t-{\lambda }^{2}t/2)$ has expectation one. This follows directly from the moment-generating function of a normal distribution and independent increments. At $t\wedge T$ its value is bounded by $e^{|\lambda |\max (a,b)}$. Bounded-time optional stopping and then dominated convergence therefore giveWrite $p_{\lambda }=\mathbb{E}[e^{-{\lambda }^{2}T/2}{1}_{\{T=T_a\}}]$ and $q_{\lambda }=\mathbb{E}[e^{-{\lambda }^{2}T/2}{1}_{\{T=T_{-b}\}}]$. Applying the identity to $\lambda$ and $-\lambda$ yieldsFor $\lambda \ne 0$, multiply the first equation by $e^{\lambda b}$, the second by $e^{-\lambda b}$, and subtract. Solving the resulting system givesAdding and using the sum formula for the hyperbolic sine gives the asymmetric Brownian interval-exit transform:At $\lambda =0$, the printed sine ratio is $0/0$ and needs its removable extension. Dominated convergence as $\lambda \to 0$ givesThus the formulas hold for every real parameter with this endpoint convention. The expectation on the left of the first formula is legible in the original PDF; the converted TeX's nested exponential is an OCR error.

A Poisson random measure with intensity the sigma-finite measure $\mu$ is a random nonnegative integer-valued measure $M$ such that, for each outcome outside a single null set, $A\mapsto M(A)$ is countably additive, and for each measurable $A$, $M(A)$ is a measurable random variable. For every measurable $A$ with $\mu (A)<\infty$,and for every finite collection of pairwise disjoint finite-intensity measurable sets, their counts are independent random variables. These are integer-valued counts, with possible value infinity on infinite-intensity sets. In particular $\mu (A)=0$ implies $M(A)=0$ almost surely. The intensity identity is $\mathbb{E}M(A)=\mu (A)$, also in the extended sense.

For completeness, choose an increasing finite-intensity exhaustion $E_m\uparrow E$. Each $M(E_m)$ is finite almost surely, simultaneously for all $m$, so $M$ is a sigma-finite measure almost surely. If $\mu (A)=\infty$, then $\mu (A\cap E_m)\to \infty$, and for any fixed integer $K$,Hence $M(A)=\infty$ almost surely. Independence for counts on arbitrary disjoint measurable sets follows by this exhaustion; an infinite count is a constant extended value. This convention completes the definition on a general measurable space, not just on bounded subsets of Euclidean space.

For $r\ge 0$, Lebesgue measure of the ball is $v_d{r}^d$, so the Poisson random measure gives $N_r\sim \mathrm{Poisson}(v_d{r}^d)$ andBy monotonicity of the balls, the set of radii with no points is an interval starting at zero. Thus $\{R\ge r\}=\{N_r=0\}$: if all smaller balls are empty then their union, the open ball of radius $r$, is empty; if that ball is empty then its radius belongs to the defining set. We obtainThe continuous tail gives $\mathbb{P}(R=0)=0$ and $\mathbb{P}(R=\infty )=0$. Differentiating the cumulative distribution function gives the probability density functionThe substitution $u=v_d{r}^d$ proves that this density integrates to one; equivalently $v_d{R}^d$ has the unit-rate exponential distribution. This is the first-neighbour case of the Kth-nearest-neighbour distance in a homogeneous Poisson point process.

First derive the Laplace functional of a Poisson random measure. For a simple nonnegative $g={\sum }_j{t}_j{1}_{A_j}$ on disjoint finite-intensity sets, independent Poisson random variables giveIncreasing simple approximations, monotone convergence for the integrals, and bounded convergence theorem for the random exponentials prove the same identity for every nonnegative measurable $g$.

Here the intensity is Lebesgue measure. Put $A=B(0,r)$ andThe integral is finite because $f$ has compact support. Apply the Laplace functional of a Poisson random measure to $f+t{1}_A$, $t\ge 0$:Differentiate from the right at $t=0$. The difference quotient on the random side is dominated by $N_r$, which is integrable with mean $v_d{r}^d$; the intensity integral is over a finite-volume ball. We obtain the count-weighted Laplace functional of a Poisson random measure:At $r=0$, the ball is empty and both sides are zero.

For $r>0$, write $m_r=v_d{r}^d$, $a_r={\int }_{B(0,r)}{e}^{-f(x)}dx$, and $L=\mathbb{E}{e}^{-M(f)}$. The count-weighted Laplace functional of a Poisson random measure and the Poisson distribution giveTo compute the other numerator, on $\{N_r=0\}$ the inside contribution to $M(f)$ vanishes. The restrictions of the Poisson random measure to the ball and its complement are independent; this follows first for their simple-function integrals from independent counts, then by approximation. HenceSubtract from $L$ and divide by $\mathbb{P}(N_r\ge 1)$ to findContinuity of $f$ at zero givesAlso $(1-e^{-z})/z\to 1$ as $z\downarrow 0$. Applying these facts to the two exact expressions proves the small-ball conditioning for a Poisson random measure:One can also see equality of the limits without the exact first formula: since $0\le e^{-M(f)}\le 1$, the difference of the two quantities lies between zero and $[m_r-(1-e^{-m_r})]/(1-e^{-m_r})\to 0$. The conditioning is used only at positive $r$, where its event has positive probability; it is not conditioning directly on a point at the origin.