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The martingale convergence theorem says that a discrete-time martingale $(M_n,{\mathcal{F}}_n)$ satisfyinghas an almost sure convergence limit $M_{\infty }$, finite almost surely and in $L^1$. The theorem asserts that the limit is integrable; it does not assert convergence in L1. More generally, the almost sure submartingale convergence theorem applies to a submartingale with ${\sup }_n\mathbb{E}{M}_n^{+}<\infty$. Uniform integrability is the additional condition that upgrades a martingale's convergence to convergence in L1.

For the requested distinction, let $({\xi }_k)$ be independent fair Bernoulli variables and use their natural filtration. The coin-doubling martingaleis a nonnegative martingale: conditionally on ${\mathcal{F}}_n$, the next factor is $2{\xi }_{n+1}$ with mean one, so $\mathbb{E}(M_{n+1}\mid {\mathcal{F}}_n)=M_n$. Also $\mathbb{E}|M_n|=\mathbb{E}{M}_n=1$ for every $n$, giving the required uniform $L^1$ bound. The probability that all the Bernoulli variables equal one is ${\lim }_n{2}^{-n}=0$. Therefore a zero is eventually encountered almost surely, after which $M_n$ stays zero. ThusThere can be no other $L^1$ limit, since convergence in L1 implies convergence in probability, whose limit must agree with the almost sure limit. This martingale satisfies the almost sure theorem but does not converge in $L^1$.

Use the natural filtration ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$. The first exit time is a stopping time. If $r\le x$, it is zero, so its expectation is already finite. Now assume $0<x<r$.

The positive-increment hypothesis gives a number $h>0$ with $\mathbb{P}(X_1>h)>0$. Choose an integer $m$ so large that $r/m<h$, and put $p=\mathbb{P}(X_1>r/m)>0$. A block of $m$ increments all exceeding $r/m$ has probability $p^m$. From any point still in $(0,r)$, that block forces an upper exit before the block ends.

By independence, conditional on ${\mathcal{F}}_{km}$ and survival to time $km$, the next block has this same probability. Thus the geometric tail bound from a uniform escape probability givesUsing the tail-sum formula for the expected value of a nonnegative integer-valued random variable,This is the random-walk exit bound from a positive-increment block. In particular, the exit occurs with probability one. The mean-zero assumption is not needed for this first bound.

Let independent Bernoulli random variables $Y_n$ have success probabilities $2^{-n}$, for $n\ge 1$, and use their natural filtration. DefineEach increment is integrable, finite, and independent of the previous filtration, with expectationThus $X$ is a martingale. The increments have no common finite bound, since a success produces a jump of size $2^n-1$.

Since ${\sum }_n\mathbb{P}(Y_n=1)<\infty$, the First Borel-Cantelli lemma says that only finitely many successes occur almost surely. Consequentlyfor all sufficiently large $n$, with the random sum finite almost surely. Therefore $X_n\to +\infty$ almost surely, so neither finite convergence nor two-sided oscillation occurs. This rare-jump martingale divergence example gives probability zero, rather than one, for the union in (i).

Work on a filtered probability space $(\Omega ,\mathcal{F},({\mathcal{F}}_n{)}_{n\ge 0},\mathbb{P})$. A submartingale is a real-valued stochastic process satisfying the following three conditions: $Y_n$ is ${\mathcal{F}}_n$-measurable, $\mathbb{E}|Y_n|<\infty$, andThus it is adapted, has integrable random variables at every time, and its conditional expected next value is at least its current value. Repeated conditioning gives $\mathbb{E}[Y_m\mid {\mathcal{F}}_n]\ge Y_n$ for every $m\ge n$. The inequality concerns conditional expectation; individual sample paths need not increase. If no filtration is specified, the definition can be applied to the natural filtration generated by the process.

Take independent and identically distributed random variables ${\xi }_k$ with the Bernoulli distribution of parameter $1/2$, and their natural filtration. SetThe event at $n=0$ is the whole sample space. Independence givesso this is a nonnegative martingale; it is integrable with $\mathbb{E}{M}_n=1$. The first zero among the ${\xi }_k$ is almost surely finite, since the probability of having only ones through time $n$ is $2^{-n}\to 0$. After that first zero, every $M_n$ is zero. Hence the almost sure limit is zero:This fair-coin doubling martingale also exhibits failure of uniform integrability: convergence in L1 would force the expectations to converge to zero.

Let $({\mathcal{F}}_n)$ be the natural filtration of the random walk on a graph. For a bounded harmonic function on a graph, the transition rule and the Markov property giveThus $h(X_n)$ is a bounded martingale. The martingale convergence theorem gives a finite almost sure limit $L$.

On the probability-one recurrence event, every vertex $v$ is visited along a sequence $n_k\to \infty$. Along that sequence $h(X_{n_k})=h(v)$, while convergence of the whole sequence forces $h(v)=L$. Therefore every pair of vertices has the same value. This is the bounded harmonic function theorem on a recurrent graph:Recurrence makes each vertex value a subsequential limit of the same convergent martingale. A connected locally finite graph is countable, since it is the union of its finite distance balls, so these almost sure assertions can also be combined by a countable intersection. The one-vertex case is immediate.

A standard real Brownian motion relative to a filtration $({\mathcal{F}}_t)$ is an adapted stochastic process with $B_0=0$ almost surely, continuous paths almost surely, andEquivalently, it has independent increments and stationary increments, each with a normal distribution of mean zero and variance equal to the elapsed time, together with continuous paths and the stated initial value. The filtration may be its natural filtration. A Brownian motion started at $z$ is $z+W_t$ for a standard process $W$; multidimensional standard Brownian motion has independent standard real coordinate processes. The normalization is variance $t$ at time $t$.

A weak solution of a stochastic differential equation consists of a filtered probability space, an adapted continuous process $X$, and a Brownian motion $B$ relative to that filtration, with specified initial distribution, such that for every finite $t$ the coefficient integrals are finite andIn one dimension the integrability conditions are ${\int }_0^t\sigma (X_s{)}^{2}ds<\infty$ and ${\int }_0^t|b(X_s)|ds<\infty$. The noise integral is the Itô integral. A solution on a prescribed stochastic basis with prescribed $B$ is distinguished from this weak construction.

A strong solution of a stochastic differential equation uses the prescribed $B$ and initial variable, and is adapted to their augmented natural filtration; thus no additional randomness is needed. For deterministic $X_0=x$, each $X_t$ is determined by the driving Brownian motion through time $t$. Merely being adapted to a larger filtration is insufficient for this definition.

Uniqueness in law means that any two weak stochastic solutions with the same initial distribution have the same law of $X$ on continuous path space. Pathwise uniqueness means that two solutions on one stochastic basis, with the same $B$ and almost surely equal initial values, are indistinguishable stochastic processes. The latter compares paths under the same noise, rather than just comparing their probability distributions.

The Girsanov theorem construction proves a weak stochastic solution, but does not by itself establish a strong stochastic solution. It constructs $X=x+W$ first, and then constructs a new driving Brownian motion $B=W-\int b(X_s)ds$. Adaptation of $X$ to the original filtration of $W$ does not by itself prove adaptation to the smaller natural filtration of $B$. Inverting this relation requires an additional result; uniqueness in law alone is insufficient in general.

For the particular coefficients in part (b), however, the literal claim that a strong solution may fail is incorrect. The strong existence theorem for additive-noise SDEs with bounded measurable drift gives strong existence and pathwise uniqueness for bounded Borel state-dependent $b$ and unit diffusion. Consequently every weak realization here is determined by its driving noise and initial value. This theorem is deeper than the Girsanov theorem argument and no proof is needed here. Examples with path-dependent drift or with a nonconstant diffusion coefficient do not contradict it.

The valid distinction is therefore: the argument in part (b) has established weak existence and uniqueness in law; strong existence requires an additional theorem, and in this class that theorem is available. A primary reference for the correction is On strong solutions and explicit formulas for solutions of stochastic integral equations.

The natural filtration of $S$ is the natural Brownian filtration: since $\sigma >0$, the equation $W_t=(\log (S_t/S_0)-\mu t)/\sigma$ recovers the entire Brownian motion history from the stock history. For $s<t$, independent increments and the moment-generating function of a normal distribution giveAll these expectations are finite. Since $S_s>0$, the martingale identity holds for every $s<t$ exactly when the last exponential equals one. The required logarithmic drift isThis is the distinction between the drift of $\log S$ and the drift of $S$: here the latter is zero.

First work with the filtration ${\mathcal{G}}_t={\mathcal{F}}_t^W\vee \sigma (1_{\{u\le \tau \}}:0\le u\le t)$, which records the Brownian motion and whether default has occurred. On $\{s>\tau \}$ all subsequent defaultable stock prices vanish. On $\{s\le \tau \}$, the memoryless property of the exponential distribution gives survival from $s$ to $t$ with conditional probability $e^{-\lambda (t-s)}$. The residual survival event is independent of the future independent increments of the Brownian motion. Thus, for deterministic $s<t$,Also $\mathbb{E}{\hat{S}}_t=e^{\lambda t}{e}^{-\lambda t}{S}_0=S_0$, so this is a true martingale, with no localization needed. Its natural filtration is contained in $\mathcal{G}$; applying the tower property of conditional expectation to the displayed identity yields the requested natural-filtration martingale property:The printed convention keeps the stock alive at $t=\tau$. It is a left-continuous convention at that one random time. Replacing it by $1_{\{t<\tau \}}$ gives the usual càdlàg zero-recovery default model; since $\mathbb{P}(\tau =t)=0$ at each fixed $t$, the fixed-time martingale calculations above are unchanged.

A martingale with the specified terminal value must equal its conditional expectation. If ${\hat{S}}_t=0$, default has already occurred and the terminal European call option payoff is zero. If ${\hat{S}}_t>0$, put $h=T-t$. Conditional on survival to $T$,where $Y$ has the standard normal distribution; the conditional probability of this survival is $e^{-\lambda h}$. The conditional expectation of the payoff in the enlarged filtration used in part (c) is consequentlyThis expression is already measurable with respect to the natural filtration of $\hat{S}$, so the tower property of conditional expectation gives the same value there. The survival factor cancels the compensating growth factor:The separate zero branch avoids division by zero. At $t=T$ it gives the required payoff, using $F(0,m)=(1-m{)}^{+}$; integrability follows from $({\hat{S}}_T-K{)}^{+}\le {\hat{S}}_T$.

A standard one-dimensional Brownian motion is a real-valued stochastic process $(B_t{)}_{t\ge 0}$ satisfying $B_0=0$ almost surely, having continuous paths almost surely, and having independent increments withHere $N(0,t-s)$ is the normal distribution with zero expected value and variance $t-s$. Thus the increments are also stationary increments.

If Brownian motion is defined relative to a specified filtration $({\mathcal{F}}_t)$, additionally require that it is an adapted process and that $B_t-B_s$ is independent of ${\mathcal{F}}_s$. For its natural filtration, this follows from independent increments. These are the defining properties of standard Brownian motion.

Let ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$ be the natural filtration. An integer-valued random variable $M$ is a stopping time with respect to $(X_i)$ whenThus whether one has stopped by time $n$ can be decided from the first $n$ observations, without seeing future variables. Equivalently, $\{M\ge i\}\in {\mathcal{F}}_{i-1}$.

Write the stopped sum asThe event $\{M\ge i\}$ depends only on $X_1,\dots ,X_{i-1}$, so its indicator and $X_i$ are independent random variables. Since $\mathbb{E}|X_1|<\infty$ and $\mathbb{E}M<\infty$, Tonelli theorem givesWe may consequently exchange expectation and summation. Using the tail-sum formula for $M$ proves Wald's equation:

For the renewal process, put $S_n=X_1+\cdots +X_n$ and $N_t=\max \{n:S_n\le t\}$. The random variable $N_t+1$ is a stopping time. Since $S_{N_t+1}>t$, Wald's equation givesand hence

For $c>0$, set $Y_i=X_i\wedge c$ and ${\mu }_c=\mathbb{E}{Y}_i$. Wald's equation still applies to the stopping time $N_t+1$ and the independent variables $Y_i$. Because $S_{N_t}\le t$,ThereforeBy the monotone convergence theorem, ${\mu }_c\uparrow \mu$ as $c\to \infty$. Combining the bounds proves the elementary renewal theorem

For a fixed word over $q$ equally likely symbols, the waiting time for a word in independent uniform symbols has mean $\sum q^k$, where $k$ runs through the lengths of the borders of a word, including the full word. The word lava' has no proper nonempty border, so $$ \boxed{\mathbb E\tau_{\mathtt{lava}}=100^4=100\,000\,000}. $$ The word aa' has borders of lengths $1$ and $2$, soFor a direct check, let $E_0$ be the expected remaining time with no trailing a' and $E_1$ with one trailing a'. Thenwhich gives $E_0=10100$.

The process is adapted to its natural filtration. At each fixed time it has finite support and is therefore integrable. Conditional on $X_{n-1}=0$,Conditional on $X_{n-1}\ne 0$,The Markov property now givesso $(X_n)$ is a martingale.

A random time $\tau$ is a stopping time for the natural filtration $({\mathcal{F}}_n)$ of a Markov chain when $\{\tau \le n\}\in {\mathcal{F}}_n$ for every $n$. The Strong Markov property says that, conditionally on $\tau <\infty$ and $X_{\tau }=i$, the process after $\tau$ is a fresh copy of the chain started from $i$, independent of the pre-$\tau$ history.

Before capture define the half-separationFor $D_n=k\ge 2$, independence of the two moves gives the transition probabilitiesIgnore the holding steps. The resulting embedded chain is a biased random walk that moves left with probability $q$ and right with probability $1-q$. Its expected number of moves needed to descend one level is $1/(2q-1)$; this follows either from first-step analysis or from its drift $1-2q$. A non-holding move occurs with probability $1/2$ at each time, so its mean waiting time is two. Hence, for every $m\ge 2$, the expected time to go from separation $2m$ to $2(m-1)$ is

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.

False. Let $Z\sim N(0,1)$ and define $X_t=tZ$. This is a centered continuous Gaussian process. Its natural filtration satisfies $Z=X_s/s\in {\mathcal{F}}_s^X$ for every $s>0$, and hence, for $0<s<t$,with positive probability. Thus $X$ is not a martingale and does not belong to the stated martingale class.

For $s<t$, write $X_t=X_s+Y$, where the increment $Y=X_t-X_s$ is independent of the natural filtration at time $s$, has mean zero, and has variance $(t-s){\sigma }^2$. HenceIt follows that $M_t=X_t^2-t{\sigma }^2$ is a martingale, as asserted by the centered square-integrable Lévy martingale identity.

Let $X_1,X_2$ be independent fair signs, each taking $-1$ and $1$ with equal probability. Set $N=1$ if $X_2=1$, and $N=2$ if $X_2=-1$. Deciding whether $N\le 1$ requires the second sign, so $N$ is not a stopping time for the natural filtration. The four equally likely possibilities giveBoth the increments and the count are bounded, but the identity fails because the second contribution is retained only when it is negative. This is selection using an unobserved outcome, rather than a legitimate sequential decision. Boundedness alone cannot replace the stopping time hypothesis.^[ref]