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Fix the maturity horizon. Since $YB$ is a nonnegative local martingale and $r$ is bounded, $\mathbb{E}{Y}_T<\infty$: $B_T\ge B_0{e}^{-\Vert r{\Vert }_{\infty }T}$ and $\mathbb{E}(Y_T{B}_T)\le B_0$. The bounded nonnegative payoff thus makes $Y_T{\xi }_T$ integrable. SetThe Brownian martingale representation theorem yields $N_t=N_0+{\int }_0^t{\zeta }_{s}d{W}_s$, using its locally square-integrable version for an integrable terminal random variable. The Itô formula for $N/Y$ givesChoose the replicating strategyIts self-financing portfolio equation has exactly the displayed drift and diffusion, because $\mu -r=\sigma \lambda$. All coefficients are locally integrable after stopping; the holdings are predictable in the augmented natural Brownian filtration. This construction has $X_t\ge 0$ and $X_T={\xi }_T$, so it is an admissible replicating strategy under the question's nonnegative-wealth convention.

For any other nonnegative self-financing portfolio $\hat{X}$ replicating the same payoff, part (b) implies ${\hat{X}}_0\ge \mathbb{E}[Y_T{\xi }_T]$. The constructed strategy attains equality, since $Y_0=1$ and $N_0=\mathbb{E}[Y_T{\xi }_T]$ in the augmented natural Brownian filtration. HenceThis is deflator-based claim replication; it does not require upgrading the local martingale deflator to a true martingale density.

With constant coefficients, $\lambda =(\mu -r)/\sigma$ is constant, so the stochastic exponential $Z_t=e^{rt}{Y}_t$ is a true martingale. Under its equivalent martingale measure $\mathbb{Q}$, the stock follows the Black-Scholes model $d{S}_t=S_t(rdt+\sigma d{W}_t^{\mathbb{Q}})$. The Gaussian exponential moment gives the value of the power optionThe option delta is ${\partial }_{s}V=V/(2s)$, so the replicating strategy holdsHalf the wealth value is in the stock and half in the bank account, with continuous rebalancing. To check the self-financing portfolio property under the original measure, the Itô formula gives $d{V}_t=\frac{1}{2}(r+\mu )V_{t}dt+\frac{1}{2}\sigma V_{t}d{W}_t={\theta }_{t}d{S}_t+{\eta }_{t}d{B}_t$. Also $V_T=\sqrt{S_T}$ and $V_t>0$.

The payoff is unbounded, so the bounded-payoff restriction of part (c) is not invoked automatically; the Black-Scholes model has the necessary finite Gaussian moments, and the explicit strategy attains $V_0=\mathbb{E}[Y_T\sqrt{S_T}]$. The supermartingale bound from part (b) therefore proves its minimality. The cost is independent of the physical drift $\mu$.

The pair $(M_n,{\mathcal{F}}_n{)}_{n\ge 0}$ is a martingale when $({\mathcal{F}}_n)$ is a filtration, each $M_n$ is ${\mathcal{F}}_n$-measurable and integrable, and

The finite-horizon discrete-time fundamental theorem of asset pricing states that the market has no arbitrage if and only if there exists a probability measure $\mathbb{Q}$ equivalent to $\mathbb{P}$ under which every integrable discounted asset price $X^i$ is a martingale. Such a $\mathbb{Q}$ is an equivalent martingale measure.

Put $q=1-p$. Sincethe martingale condition isThe root in $(0,1)$ is therefore

Let $({\tau }_n)$ be a localizing sequence for $M$. For $s\le t$, the optional sampling theorem givesBoth sides converge almost surely to $M_t$ and $M_s$, and $|M_{u\wedge {\tau }_n}|\le Z$. Conditional dominated convergence therefore yields $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$, so $M$ is a martingale. Moreover, the family $(M_t{)}_{t\ge 0}$ is dominated by the integrable random variable $Z$, hence is uniformly integrable. Thus $M$ is a uniformly integrable martingale.

The pair $(M_n,{\mathcal{F}}_n{)}_{n\ge 0}$ is a martingale if $({\mathcal{F}}_n)$ is a filtration, $(M_n)$ is an adapted process, every $M_n$ is integrable, and

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.

Assume (ii). A martingale has constant expected value by the tower property of conditional expectation, so for every stopping time $\tau$ and every $n$,This is (iii).

In the one-period building block of the Cox--Ross--Rubinstein model, the money-market account grows by a deterministic factor $R>0$, while the stock grows independently at each time byThus $S_t^0=R^t{S}_0^0$ and $S_t^1=S_0^1{\prod }_{j=1}^t{Y}_j$. Equivalently, the discounted stock has multipliers $u=U/R$ and $d=D/R$.

The model has no arbitrage precisely when the risk-free return lies strictly between the stock returns:Indeed, failure of either strict inequality gives a one-period arbitrage, while under the strict inequalities the probabilitylies in $(0,1)$. Taking the moves independent under $Q$ with up probability $q$ makes the discounted stock a martingale, since $qu+(1-q)d=1$. This is the unique Risk-neutral probability in the Cox--Ross--Rubinstein model, hence the unique equivalent martingale measure.

The model has a unique equivalent martingale measure $\mathbb{Q}$ under whichand the discounted stock is a martingale. The time-zero value of an integrable time-$T$ contingent claim $C$ is therefore

A random time $T$ is a stopping time when the event $\{T\le n\}$ is determined by $X_0,\dots ,X_n$. The Strong Markov property says that, conditionally on $T<\infty$ and $X_T=x$, the process $(X_{T+k}{)}_{k\ge 0}$ is a fresh Markov chain started at $x$, independent of the history before $T$.

Use the state space $\{1,2,4\}\times {\mathbb{Z}}_{\ge 0}$, with $(2,0)$ absorbing. Observe the chain only when it is at square 2. From $(2,k)$, the change $Y$ in wealth by the next return to square 2 has distributionIndeed, heads lands on square 3 and returns to square 2 after losing £1. After tails reaches square 4, the remaining two or three moves give the other cases.

For $r=2/3$,Thus $r^{M_j}$ is a martingale for the embedded wealth random walk $M_j$. Stopping when it first reaches $m-1$ or a large upper level and then letting that level tend to infinity givesThe upper-bound contribution vanishes because $0<r<1$; equivalently, this is the smaller probability solution of the first-step equation.

If ${\xi }_k=S_k-S_{k-1}$, thenThe centered increments ${\xi }_k-\mu$ are independent of the past and have mean zero, sois a martingale.

The heat-semigroup form is $U(t,x)=P_{1-t}(f^2)(x)$. For $s<t$, independence and additivity of Brownian increments givewhere ${\tilde{B}}_{1-s}$ is an independent $N(0,1-s)$ increment. Thus $M_t=U(t,W_t)$ is a martingale.

A process $(X_n{)}_{n\ge 0}$ is a martingale with respect to a filtration $({\mathcal{F}}_n{)}_{n\ge 0}$ when:

Equivalently, every increment has conditional mean zero given the past.

For a bounded stopping time $T\le N$, writeThe indicator is measurable at time $n-1$. Taking conditional expectations term by term and using the supermartingale inequality gives $\mathbb{E}{\hat{M}}_T\le \mathbb{E}{\hat{M}}_0$. The same calculation gives equality for the martingale $M$. Since their initial values agree, $\mathbb{E}{M}_T\ge \mathbb{E}{\hat{M}}_T$.

Condition on the information through time $n-1$. Independence of ${\xi }_n$ and the definition of the Bellman equation for terminal-wealth utility implybecause the actual holding is one candidate in the supremum. Thus $V(n,X_n)$ is a supermartingale for every strategy.

For the stated strategy it is a martingale, soEvery competing strategy has expected terminal value at most $V(0,X_0)$ by the supermartingale inequality from part (a), evaluated at the bounded time $N$. Hence ${\theta }^{\ast }$ is optimal.

The partial sums $S_n$ are a martingale, as are $S_n^2-n$. Since the increment has mean zero and nonzero variance, there are ${\delta }_{+},{\delta }_{-}>0$ with $\mathbb{P}(X_1\ge {\delta }_{+})>0$ and $\mathbb{P}(X_1\le -{\delta }_{-})>0$. From any point in $(-a,b)$, a sufficiently long run of either kind exits the interval. Independence in consecutive blocks therefore bounds $\mathbb{P}(T>km)$ by a geometric sequence. In particular, $T<\infty$ almost surely and $\mathbb{E}T<\infty$.

Apply the optional stopping theorem first to $T\wedge n$. Since the increments are bounded and $\mathbb{E}T<\infty$, the stopped variables are uniformly integrable and passage to the limit givesApplying the same argument to $S_n^2-n$, using $|S_T|\le \max \{a,b\}+c$, gives

Under a risk-neutral measure, the discounted stock must be a martingale. If $q$ is the probability of the factor $1+b$, this condition issoThe hypothesis $a<r<b$ ensures $0<q<1$, making this equivalent to the original probability measure. Since the $N$ returns are independent under $Q$, the number of up moves has a binomial distribution, and hence

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.