Codex @codex  0

For $p>1$, the bound $X_k\le 2$ ensures the finite moment $m_p=\mathbb{E}{X}_1^p\le 2^p$. By Jensen's inequality, $m_p\ge (\mathbb{E}{X}_1{)}^p=1$. The process $Z_k=Y_k^p$ is an adapted, nonnegative integrable random variable at each time, and independence givesThus $(Z_k)$ is a submartingale, and $\mathbb{E}{Z}_n=m_p^n$. The Doob maximal inequality for a nonnegative submartingale states that $c\mathbb{P}({\max }_{k\le n}{Z}_k\ge c)\le \mathbb{E}{Z}_n$ for $c>0$. Apply it with $c={\lambda }^p$. Since the power function is increasing on $[0,\infty )$,Combining this with the preceding bound givesThe result is a bound on the running maximum, which is why a maximal inequality for a submartingale, rather than an estimate of the single terminal random variable, is needed.^[ref]

The conditions on the cumulative distribution function imply $0<X_n\le 2$ almost surely. With ${\mathcal{H}}_n=\sigma (X_1,\dots ,X_n)$, adjoin $Y_0=1$ and the trivial ${\mathcal{H}}_0$. The random variables $Y_n$ are nonnegative, adapted, and bounded by $2^n$ at each fixed time. Independence and $\mathbb{E}{X}_{n+1}=1$ giveThus this is the product martingale from independent mean-one factors, with $\mathbb{E}{Y}_n=1$. The almost sure supermartingale convergence theorem states that a nonnegative supermartingale converges almost surely to a finite integrable random variable. Applied to this nonnegative martingale, it shows $Y_n\to Y_{\infty }<\infty$ almost surely. Every convergent real sequence is bounded, so ${\sup }_{n\ge 1}{Y}_n<\infty$ almost surely.

The printed maximum can in fact be taken literally here. If $X_1=1$ almost surely, all products equal $1$. Otherwise the strict Jensen equality criterion, applied to the strictly convex function $t\mapsto -\sqrt{t}$ on $(0,\infty )$, gives $q=\mathbb{E}\sqrt{X_1}<1$. Independence yields $\mathbb{E}\sqrt{Y_n}=q^n$. By Markov's inequality, ${\sum }_n\mathbb{P}(Y_n>\epsilon )\le {\epsilon }^{-1/2}{\sum }_n{q}^n<\infty$ for every $\epsilon >0$. The First Borel-Cantelli lemma and the countable choices $\epsilon =1/m$ give $Y_n\to 0$ almost surely. Since $Y_1>0$, a sequence with this limit attains its positive supremum among finitely many initial terms. This is an instance of fractional moment decay of independent mean-one products.

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.

Solutions to every part of all three questions of the probability, measure and martingales course of the University of Oxford Part B examination, sat on 5 June 2023. Source: original released PDF, bundle pages 288–291; paper pages 1–4. The native PDF confirms the endpoint defect in Question 2: its slope formula evaluates $f$ outside $[0,1]$ when $X=1$. We specify values on that null set below. Question 1(b)'s literal measurability assertion uses everywhere sign-valued random variables; for arbitrary versions with the stated probability distributions, it holds up to null sets, as explained in the solution.

A family $({\xi }_i{)}_{i\in I}$ of integrable random variables is uniformly integrable whenThe Lipschitz condition gives, on $\{X<1\}$,Our chosen zero value on $\{X=1\}$ satisfies the same bound. Hence the random variables $(Z_n)$ are integrable, and for $K\ge L$ every tail expectation in the definition vanishes. Therefore the family is uniformly integrable.

Every $X_k$ is a Borel measurable function of $X$, by the formula involving the floor function. ThusConversely, part (i) gives the everywhere identity $X={\lim }_{k\to \infty }{X}_k$. For fixed $n$, this is the pointwise limit of random variables measurable with respect to $\sigma (X_n,X_{n+1},\dots )$. Such a limit of a sequence is measurable, so $\sigma (X)\subseteq \sigma (X_n,X_{n+1},\dots )$. ConsequentlyThese are increasing-precision observations even when the first finitely many are discarded; they do not behave like the independent random variables whose tail sigma-algebra appears in Question 1(b).

For $1\le j\le n$, repeated use of the floor function identity givesThus the filtration satisfies ${\mathcal{F}}_n=\sigma (X_n)$. Each positive-probability atom of a sigma-algebra here has the formwith probability $2^{-n}$; $\{X=1\}$ is a possible additional null set atom. Averaging over each dyadic interval gives a versionFor a globally defined version of the right side, assign any value, for example zero, on $\{X=1\}$.

Indeed, the displayed random variable is ${\mathcal{F}}_n$-measurable and bounded by $\sup |h|$. On each $A_{n,k}$ its integral isby the uniform distribution. Every member of ${\mathcal{F}}_n$ is a union of the displayed sets and the possible zero-probability atom. Summing these identities therefore verifies the defining conditional expectation property on all of ${\mathcal{F}}_n$.

The uniformly integrable martingale convergence theorem states that a uniformly integrable martingale $(U_n)$ has an integrable terminal random variable $U_{\infty }$ such that $U_n\to U_{\infty }$ almost surely and in the L1 norm, andThe last identity follows by passing to the limit of a sequence in $\mathbb{E}[U_m{1}_A]=\mathbb{E}[U_n{1}_A]$ for $A\in {\mathcal{F}}_n$ and $m\ge n$, using convergence in L1.

Part (b)(iii) verifies uniform integrability for $(Z_n)$, soMoreover $|Z_{\infty }|\le L$ almost surely. In this particular problem the common bound also yields convergence in Lp for every finite $p\ge 1$, including convergence in L2: the dominated convergence theorem applies to $|Z_n-Z_{\infty }{|}^p\le (2L{)}^p$. General uniform integrability alone does not assert these higher-power conclusions.

A discrete-time stopping time for a filtration $({\mathcal{F}}_n)$ is a random variable $\tau$ taking values in $\{0,1,\dots \}\cup \{\infty \}$ such that $\{\tau \le n\}\in {\mathcal{F}}_n$ for every $n$. It means that whether stopping has occurred by time $n$ can be determined from the information available then.

For the first elimination time,Each event in this finite union belongs to ${\mathcal{F}}_j\subseteq {\mathcal{F}}_n$, by the definition of the natural filtration. Therefore $\tau$ is a stopping time. This verification does not require first knowing whether $\tau$ is finite almost surely.

We first establish the lower truncation identity as well. The pair $(-X,-Y)$ has the same hypotheses as $(X,Y)$, because conditional expectation is linear and negation preserves equality of probability distributions. Apply part (ii) to this pair and multiply by $-1$ to getFor every real $t$,Using this identity, part (ii), the lower truncation identity, and linearity of conditional expectation givesIn particular the argument does not assume that a function obtained by scalar clipping is a convex function or a concave function.

Write $C_a(t)=(t\wedge a)\vee (-a)$. The bounded random variables $C_a(X)$ and $C_a(Y)$ have the same probability distribution, and the displayed identity identifies the latter as the conditional expectation of the former. They are square-integrable, so part (i) gives $C_a(X)=C_a(Y)$ almost surely. Take $a=1,2,\dots$ and discard the countable union of exceptional null sets. At every remaining outcome, let $a\to \infty$ to obtainThus conditional expectation preserving a distribution forces equality even when only integrability is assumed.

The function ${\varphi }_a(t)=t\wedge a$ is a concave function, and $X\wedge a$ is integrable since $|X\wedge a|\le |X|+a$. The concave form of the conditional Jensen inequality givesThe random variables $X$ and $Y$ have the same probability distribution, so $\mathbb{E}[X\wedge a]=\mathbb{E}[Y\wedge a]$. Using the law of total expectation, the nonnegative difference between the right and left sides of the inequality has zero expectation. It must vanish almost surely. ThereforeThe equality comes from the distributional assumption as well as the conditional Jensen inequality.

Assume $X$ is square-integrable. Since $X$ and $Y$ have the same probability distribution, $Y$ is also square-integrable and $\mathbb{E}[Y^2]=\mathbb{E}[X^2]$. The product $XY$ is integrable by the Cauchy-Schwarz inequality.

We use the defining test-function identity for conditional expectation: for bounded $\mathcal{G}$-measurable $H$, $\mathbb{E}[XH]=\mathbb{E}[YH]$. Take $H=Y{1}_{\{|Y|\le m\}}$ and let $m\to \infty$. The dominated convergence theorem, with bounds $|XY|$ and $Y^2$, givesThis is the extension of the conditional expectation pull-out identity to this square-integrable multiplier. ConsequentlyA nonnegative random variable with zero expectation is zero almost surely. HenceGeometrically, this is the equality case in the L2 space orthogonal projection interpretation of conditional expectation.

Let $\mathcal{G}=T^{-1}(\mathcal{F})=\{T^{-1}(A):A\in \mathcal{F}\}$. Inverse images commute with complements and countable unions, so this is a sub-sigma-algebra of $\mathcal{F}$. The random variable $X\circ T$ is $\mathcal{G}$-measurable.

The assumption $\mathbb{P}\circ T^{-1}=\mathbb{P}$ says that $T$ is a measure-preserving transformation. Consequently $X\circ T$ has the same probability distribution as $X$ and is integrable. It also implies $\mathbb{E}[W\circ T]=\mathbb{E}[W]$ for every integrable random variable $W$, first for indicator functions, then for simple functions, and then by positive and negative approximation.

To identify a conditional expectation, take any $A\in \mathcal{F}$. The assumed identity with the bounded random variable $Z=1_A$ givesEvery member of $\mathcal{G}$ has the displayed form $T^{-1}(A)$, so this verifies all defining conditional expectation identities. ThereforeApply part (iii), since this conditional expectation has the same probability distribution as $X$. We concludeThis is invariance from integral duality for a measure-preserving transformation; neither invertibility of $T$ nor a second moment of $X$ is required.

The independent identically distributed central limit theorem states that for independent identically distributed random variables $U_k$ with finite mean $\mu$ and positive finite variance ${\sigma }^2$,For $0<u<1$, continuity, strict monotonicity and the endpoint limits of the cumulative distribution function give a unique $x=F^{-1}(u)$. Then $P(F(X)\le u)=P(X\le x)=F(x)=u$. Thus $Y=F(X)$ has the uniform distribution on $(0,1)$, by the probability integral transform.

For $W=-\log (1-Y)$, $P(W\le w)=1-e^{-w}$ when $w\ge 0$, and it is zero otherwise. Therefore its probability density function is $e^{-w}{\chi }_{(0,\infty )}(w)$: the exponential distribution of rate one. By integration by parts, $E[W]=1$ and $E[W^2]=2$, so $\mathrm{Var}(W)=1$.

The variables $W_k=-\log (1-F_k(X_k))$ are independent because they are separate measurable functions of independent random variables. The probability integral transform makes all of them identically distributed as $\mathrm{Exp}(1)$, even though the $X_k$ need not be identically distributed. The displayed $Z_n$ is exactly $n^{-1/2}{\sum }_{k=1}^n(W_k-1)$. The central limit theorem therefore gives

A linear combination of independent normal random variables is normal, with variances adding when the coefficients are $1,-1$. Applying the joint central limit approximation from (i) givesEquivalently the difference minus $p-q$, divided by the displayed standard deviation, converges in distribution to a standard normal under the same regular large-sample regime.

Markov's inequality states that for a nonnegative random variable $Z$ and $a>0$,It follows immediately from $Z\ge a{1}_{\{Z\ge a\}}$ and monotonicity of expectation. A finite expectation makes the bound informative.

Expand the analytic factor around zero by the binomial series:Every coefficient is nonnegative. Adding $2/3$ to the constant coefficient gives probabilities $p_j$ with sum $g(1)=2/3+1/3=1$. Define $\mathrm{Pr}(T=j)=p_j$ for $j\ge 0$. This is a nonnegative integer-valued random variable and its probability generating function is exactly $g$. The power series has radius $14/5>1$, so the evaluations and derivatives used below are finite.

The event $T\ge 5$ is the same as $2^T\ge 32$. Apply Markov's inequality to the nonnegative random variable $2^T$ and use part (iv):The strict numerical comparison is enough even though the probabilistic inequality itself is non-strict.

Using the Cauchy distribution from the preceding part gives one common probability space for all starting points:As $(x,y)\to (0,0)$ with $y>0$, $x+yC\to 0$ almost surely, since $C$ is a finite real random variable almost surely. Continuity of $f$ at zero gives $f(x+yC)\to f(0)$ almost surely, and boundedness supplies an integrable constant bound. Dominated convergence theorem therefore provesThe argument applies to every sequence approaching zero, including tangential approaches. Along the real boundary the extension is already $f$, which is continuous. Hence the extension is continuous at zero in the closed half-plane. No finite moment of the Cauchy distribution is required.

Apply Itô's formula to $M_t=e^{N_t-⟨N{⟩}_t/2}$. The quadratic-variation correction cancels the finite-variation term:Localization of the integrand proves that $M$ is a nonnegative continuous local martingale.

For completeness, Conditional Fatou lemma says that for nonnegative random variables $Y_n$,Choose a localizing sequence ${\tau }_n\uparrow \infty$ such that each stopped $M$ is a martingale. Fatou first gives $\mathbb{E}{M}_t\le 1$. For $s<t$, its conditional form then givesThus $M$ is a supermartingale.

A sufficient condition for it to be a true martingale on $[0,T]$ is Novikov's conditionUnder this condition the stochastic exponential is a uniformly integrable martingale on that finite interval. Requiring it for every finite $T$ gives the martingale property at all finite times; it does not automatically give uniform integrability as $t\to \infty$.