Past exam of the martingales and financial mathematics course of the University of Oxford / 2009 / A (Martingales Through Measure Theory)

Almost sure convergence meansThus a single event of probability one supports ordinary pointwise convergence. Convergence in probability means that for every $ϵ>0$,The exceptional event is allowed to change with $n$ in the second definition.

Fix $ϵ>0$ and use the indicator random variables $Y_n=1_{\{|X_n-X|>ϵ\}}$. Almost sure convergence gives ${\mathrm{limsup}}_n{Y}_n=0$ almost surely, while $0\le Y_n\le 1$. The Reverse Fatou lemma states that for random variables bounded above by an integrable random variable,It follows thatSince $ϵ$ was arbitrary, this proves convergence in probability. The integrable upper bound is essential to the application of the Reverse Fatou lemma.

Independent random variables satisfy, for every finite selection of distinct indices $n_1,\dots ,n_m$ and Borel sets $C_1,\dots ,C_m$,The Second Borel-Cantelli lemma says that if events $E_n$ are independent and ${\sum }_n\mathbb{P}(E_n)=\infty$, thenHere ${\mathrm{limsup}}_n{E}_n={\bigcap }_N{\bigcup }_{n\ge N}{E}_n$ is the event that infinitely many of the $E_n$ occur.

By convergence in probability, choose recursively $n_k>n_{k-1}$ so thatThe corresponding events have a summable sequence of probabilities. The First Borel-Cantelli lemma, which requires no independence, therefore makes the probability of infinitely many of them zero. On the complementary event, $|X_{n_k}-X|\le 2^{-k}$ for all sufficiently large $k$, provingThe recursively chosen indices are strictly increasing and hence tend to infinity.

Suppose first that complete convergence holds. For each positive integer $j$, apply the First Borel-Cantelli lemma to $\{|X_n-X|>1/j\}$. Outside a countable union of null events, only finitely many of these exceedances occur for every $j$. This is almost sure convergence, with no independence assumption.

For the converse, an important point is that independence of $X_n$ does not immediately imply independence of $X_n-X$. Instead use the fact that an almost-sure limit of independent random variables is constant. Indeed, $Z={\mathrm{limsup}}_n{X}_n$ is measurable for the tail sigma-algebra. The Kolmogorov zero-one law makes each event $\{Z\le q\}$, $q\in \mathbb{Q}$, have probability zero or one. Since $Z=X$ almost surely and $X$ is finite, these probabilities form the distribution function of a point mass: there is a finite constant $c$ with $X=c$ almost surely.

Now $\{|X_n-c|>ϵ\}$ are independent events. Almost sure convergence says that their infinitely-often event has probability zero. The contrapositive of the Second Borel-Cantelli lemma givesReplacing $c$ by the almost surely equal $X$ leaves each probability unchanged. Thus complete convergence and almost sure convergence are equivalent for independent sequences.

Let $U$ have the uniform distribution on $(0,1)$, set $Y=1/U$, and take $X_n=Y/n$. The variable $Y$ is finite almost surely, so $X_n\to 0$ almost surely. For any $ϵ>0$ and all sufficiently large $n$,The harmonic series diverges, soThis dependent sequence fails complete convergence. All its members are scalar multiples of the same nonconstant random variable, which explains why the independence hypothesis cannot be dropped.

A real-valued random variable is a measurable function $Y:(\Omega ,\mathcal{F})\to (\mathbb{R},\mathcal{B}(\mathbb{R}))$: equivalently, $\{Y\le t\}\in \mathcal{F}$ for every real $t$.

For $p=1$ the asserted inequality is equality. For $p>1$, let $q=p/(p-1)$. If $\mathbb{E}{Y}^p=\infty$, the inequality is automatic. Otherwise Hölder's inequality on the probability space givesThus $\boxed{\Vert Y{\Vert }_1\le \Vert Y{\Vert }_p}$, using that the total measure is one.

Write ${\Delta }_0=X_0$ and ${\Delta }_j=X_j-X_{j-1}$ for $j\ge 1$. If $i<j$, then ${\Delta }_i$ is ${\mathcal{F}}_{j-1}$-measurable. The martingale property and linearity of conditional expectation giveThe product ${\Delta }_i{\Delta }_j$ is integrable by Cauchy-Schwarz inequality. The pull-out property of conditional expectation, followed by its tower property, therefore givesFor unbounded ${\Delta }_i$, the pull-out identity follows by truncation and Cauchy-Schwarz inequality. This proves martingale-difference orthogonality, including orthogonality to the initial value.

Expand $X_n={\sum }_{j=0}^n{\Delta }_j$ in the Hilbert space $L^2$. Orthogonality gives the exact identityThe terms are nonnegative and $\mathbb{E}{X}_0^2<\infty$, so boundedness of the left side over $n$ is equivalent to convergence of the series on the right.

Conditional on ${\mathcal{F}}_n$, put $x=X_n$. The two possible values have conditional meanThe process is adapted and bounded, hence integrable. Consequently it is a martingale. Since ${\sup }_n\mathbb{E}|X_n|\le 1$, the martingale convergence theorem gives an almost surely finite limit $X_{\infty }$, necessarily in $[0,1]$.

Moreover $|X_n-X_{\infty }{|}^2\le 1$ and these squared differences tend to zero almost surely. The dominated convergence theorem gives $\mathbb{E}|X_n-X_{\infty }{|}^2\to 0$. Thus the Bernoulli polarization martingale converges almost surely and in $L^2$.

Conditional on ${\mathcal{F}}_n$, the increment is $-X_n/2$ with probability $1-X_n$, and $(1-X_n)/2$ with probability $X_n$. ThereforeTaking expectations using the tower property yieldsThis is the conditional variance that drives the Bernoulli polarization martingale toward its endpoints.

The martingale is bounded in $L^2$, so the increment-square series from part (a)(ii) is finite. The identity just obtained implies ${\sum }_n\mathbb{E}[X_n(1-X_n)]<\infty$, hence $\mathbb{E}[X_n(1-X_n)]\to 0$. By almost sure convergence and the dominated convergence theorem,The integrand is nonnegative because $X_{\infty }\in [0,1]$. It must vanish almost surely, giving $X_{\infty }\in \{0,1\}$.

The martingale has constant expectation $a$. Its $L^2$ convergence implies $L^1$ convergence by Cauchy-Schwarz inequality, and hence $\mathbb{E}{X}_{\infty }=a$. ThereforeThe limiting law is the Bernoulli distribution with parameter $a$.

Independent and identically distributed random variables are independent as in question 1, and all have the same probability distribution. Their tail sigma-algebra isAn event is a tail event when it belongs to every sigma-algebra in this intersection. The Kolmogorov zero-one law says $\mathbb{P}(E)\in \{0,1\}$ for every $E\in \mathcal{T}$.

Use the intended simple symmetric random walk $S_n={\sum }_{j=1}^n{X}_j$ with $S_0=0$; the printed lower summation index zero refers to an undefined $X_0$. For each $m$, removing the finite initial sum changes all later positions by one finite constant. This cannot change whether the lower limit is $-\infty$ or the upper limit is $+\infty$. Thus $B_{-}$ and $B_{+}$ belong to $\mathcal{T}$, and

Each increment has the same law as its negative, and their independence makes the joint law of $(X_1,X_2,\dots )$ invariant under simultaneous sign reversal. Under this transformation, the simple symmetric random walk $S_n$ becomes $-S_n$, and the event $B_{-}$ becomes $B_{+}$. ThusThis uses symmetry of the entire increment sequence, rather than just the one-dimensional distributions of the positions.

Use ${\mathcal{F}}_0=\{\varnothing ,\Omega \}$; the printed $\{\varnothing ,\mathcal{F}\}$ does not define the intended trivial sigma-algebra. With the simple symmetric random walk convention from part (a), $M_n$ is adapted and integrable because $|S_n|\le n$. Independence givesBy linearity, the pull-out property for the ${\mathcal{F}}_n$-measurable $S_n$, and preservation of measurable variables by conditional expectation,This proves the square-minus-time martingale of a simple symmetric random walk.

A discrete stopping time is a random variable $T:\Omega \to {\mathbb{N}}_0\cup \{\infty \}$ such that $\{T\le n\}\in {\mathcal{F}}_n$ for every $n$. The convention is $\inf \varnothing =\infty$.

For the hitting time in question,Each event in the union is in ${\mathcal{F}}_j\subseteq {\mathcal{F}}_n$ because $S_j$ is adapted. This proves $T_k$ is a stopping time for every positive integer $k$.

The optional stopping theorem applies to $M$ at the bounded stopping time $n\wedge T_k$. As $M_0=0$, it yields the valid stopped second-moment identity for a simple symmetric random walkThe printed bound on $\mathbb{E}{T}_k$ is false. Before hitting the positive level $k$, the walk can travel arbitrarily far downwards, so $|S_{n\wedge T_k}|\le k$ is unavailable.

Here is a valid proof of finiteness of $T_k$. For a positive integer $L$, stop instead at $\tau =\inf \{n:S_n\in \{-L,k\}\}$. The same identity and the bound $|S_{n\wedge \tau }|\le \max (L,k)$ giveThe monotone convergence theorem makes $\mathbb{E}\tau <\infty$, so $\tau <\infty$ almost surely. Applying optional stopping to $S$ at $n\wedge \tau$ and then using dominated convergence gives $\mathbb{E}{S}_{\tau }=0$. Since $S_{\tau }\in \{-L,k\}$, this meansConsequently $\mathbb{P}(T_k<\infty )\ge L/(L+k)$ for every $L$, and letting $L\to \infty$ proves $T_k<\infty$ almost surely.

In fact this is the infinite mean first passage of a simple symmetric random walk. If $\mathbb{E}{T}_k<\infty$, the first displayed identity would bound the second moments of $S_{n\wedge T_k}$ uniformly. Their uniform integrability and their almost sure convergence to $k$ would imply $\mathbb{E}{S}_{n\wedge T_k}\to k$. But the optional stopping theorem applied to $S$ at every bounded $n\wedge T_k$ makes each expectation zero, a contradiction. Hence the correct conclusion is

Intersecting the probability-one hitting events over all positive integers shows that the simple symmetric random walk reaches arbitrarily large positive levels almost surely. The hitting times of levels tending to infinity also tend to infinity, since $|S_n|\le n$. Hence $\mathbb{P}(B_{+})=1$, and in particular $\mathbb{P}(B_{-}\cup B_{+})=1$.

Alternatively, the union conclusion, the equal probabilities from symmetry, and the Kolmogorov zero-one law force both probabilities to equal one. Thus almost every path has arbitrarily large positive and negative excursions. A nearest-neighbor integer path must pass through zero between a positive and a negative excursion. Such crossings occur after every finite time, givingThis is recurrence of the simple symmetric random walk.