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Use the independent increments of Brownian motion, rather than merely checking that an Itô formula drift vanishes. For $0\le s<t$, put $h=t-s$ and write $B_t=B_s+Z$, where $Z$ is independent of ${\mathcal{F}}_s$ and has normal distribution $N(0,h)$. Its first four moments are $0,h,0,3h^2$. ThereforeFor the cubic expression, the coefficient of $B_s$ after conditioning is $3(t-s)+\alpha (t)$, so $\alpha (t)=-3t$ makes it $\alpha (s)$. For the quartic expression, choose $\beta (t)=-6t$. Its conditioned coefficient of $B_s^2$ is then $6(t-s)-6t=-6s$. The constant term becomeswhich equals $3s^2$ when $\gamma (t)=3t^2$. Thus a standard choice isThe resulting stochastic processes are Hermite polynomial martingales $H_3(B_t,t)$ and $H_4(B_t,t)$. They are genuine integrable martingales, since Gaussian moments are finite at every finite time and the displayed conditional identities establish the martingale property directly.

The choice is not unique. Constants $c,d,e\in \mathbb{R}$ give the valid family $\alpha (t)=c-3t$, $\beta (t)=d-6t$, and $\gamma (t)=3t^2-dt+e$: these add $c{B}_t$ to the cubic martingale and $d(B_t^2-t)+e$ to the quartic one. The boxed choice sets these harmless additions to zero.

Write $\tau ={\tau }_x$ and $T_t=t\wedge \tau$. Path continuity gives $|B_{T_t}|\le x$. Stop the martingale $B_s^2-s$ at the bounded stopping time $T_t$ and apply the optional stopping theorem:By the monotone convergence theorem, $\mathbb{E}\tau \le x^2$, so $\tau <\infty$ almost surely. Path continuity then gives $B_{\tau }\in \{-x,x\}$. The dominated convergence theorem for the bounded variables $B_{T_t}^2$ shows

Next stop the quartic Hermite polynomial martingale from part (a), again only at $T_t$. Its expectation is zero, soMonotone convergence proves $\mathbb{E}{\tau }^2\le 2x^4$, establishing the needed second-moment integrability before the final passage to the limit. Since $T_t{B}_{T_t}^2\le x^2\tau$ and $B_{T_t}^4\le x^4$, dominated convergence givesConsequently the Brownian symmetric interval-exit moments are

To obtain the Laplace transform of symmetric Brownian interval-exit time, put $a=\sqrt{2\lambda }$. The Exponential martingale for Brownian motion shows thatis a martingale with $H_0=1$. Bounded-time stopping gives $\mathbb{E}{H}_{T_t}=1$. Its stopped values are bounded by $\cosh (ax)$, so dominated convergence applies as $t\to \infty$. Since $\cosh (a{B}_{\tau })=\cosh (ax)$, it yieldsEvery use of stopping at $\tau$ has thus been justified through bounded stopping and an explicit integrability or domination argument.

Fix $t$ and put $\sigma =t\wedge \tau$. The stopping time property implies that $\sigma$ is an ${\mathcal{F}}_t$-measurable random variable with values in $[0,t]$: for $s<t$,and for $s\ge t$ the event is the whole space. By part (c), the restriction of $X$ to $[0,t]\times \Omega$ is $\mathcal{B}([0,t])\otimes {\mathcal{F}}_t$-measurable.

The evaluation map $\omega \mapsto (\sigma (\omega ),\omega )$ is measurable from $(\Omega ,{\mathcal{F}}_t)$ to this product space: the inverse image of a measurable rectangle $A\times C$ is $\{\sigma \in A\}\cap C$. Composing it with the jointly measurable stochastic process gives an ${\mathcal{F}}_t$-measurable random variableSince this holds for every fixed $t$, the stopped process is adapted. If path regularity holds only almost surely, first apply the proof to its pathwise regular representative; with a completed filtration, the original stopped variable differs only on a null event and is also ${\mathcal{F}}_t$-measurable. The adaptedness of a stopped right-continuous process requires neither boundedness of $\tau$ nor a martingale assumption; $\tau =\infty$ is harmless because $t\wedge \tau \le t$.

Put $t_j=j{2}^{-n}$ and ${\Delta }_j=X_{t_{j+1}}-X_{t_j}$. Telescoping $X_{t_{j+1}}^2-X_{t_j}^2=2X_{t_j}{\Delta }_j+{\Delta }_j^2$ givesThe martingale transform summands are orthogonal in $L^2$: for an earlier summand, conditioning on the sigma-algebra at the start of the later increment makes the cross expectation zero. HenceThe martingale increments themselves are also orthogonal. Since $X_0=0$, their variance sum equals $\mathbb{E}{X}_1^2\le C^2$. ThereforeOnly discrete martingale orthogonality is used here; no pre-existing quadratic variation calculation is needed.

Use the Chordal Loewner equation with Loewner driving function ${\xi }_t=\sqrt{\kappa }{W}_t$, and set $Z_t=g_t(z)-{\xi }_t=X_t+i{Y}_t$. Before the Loewner swallowing time,The upper-half-plane branch of the logarithm has imaginary part $h_t$. The Itô formula givesand therefore the SLE angle process satisfiesAt $\kappa =4$ the drift vanishes. Since $0<h_t<\pi$, the stopped local martingale is a true bounded martingale, and the SLE4 angle martingale is global for each fixed point almost surely, using fixed-interior-point avoidance of SLE4 for the simple parameter-four Loewner trace.

Conversely, take $\kappa >0$. If this semimartingale were a local martingale, uniqueness of its finite-variation decomposition would force $(\kappa -4)X_t{Y}_t=0$ throughout every compact interval before swallowing. Because $Y_t>0$, for $\kappa \ne 4$ this would force $X_t$ to be identically zero on such an interval. Its Brownian component has quadratic variation $\kappa t$, which makes that impossible. Thus for positive $\kappa$, the martingale parameter is exactly four.

If the degenerate value $\kappa =0$ is admitted, there is one exception to a fixed-point reading of the assertion: for $z$ on the positive imaginary axis the deterministic flow stays on that axis until swallowing, and $h_t=\pi /2$ is constant. For all starting points simultaneously, the unique parameter giving the martingale property is still four.

The multidimensional Itô formula includes a mixed second derivative multiplied by the quadratic covariation, with no extra factor $1/2$. HereConsequently the Itô formula for $M_t=U(t,{\sigma }_t,X_t)$ has drift equal to the left side of the stated backward partial differential equation. That drift vanishes, leavingA stochastic integral against Brownian motion with locally square-integrable predictable integrand is a continuous local martingale. The smoothness of $U$ and localization of the diffusion and its coefficients give this integrability on the model's lifetime. Thus $M$ is a local martingale, as required. The partial differential equation cancellation alone does not establish a true martingale or justify replacing $U$ by a terminal-payoff expectation without an additional integrability argument.

The first-contact time ${\tau }_{\ast }\le T$ is a stopping time: the event $\{{\tau }_{\ast }\le t\}$ is the finite union of events $\{U_s=Y_s\}$ for $s\le t$. It is finite because $U_T=Y_T$. On $\{{\tau }_{\ast }>t\}$ we have $U_t>Y_t$, so the maximum defining the Snell envelope selects continuation andThe stopped process $U_{t\wedge {\tau }_{\ast }}$ is therefore a martingale: before stopping its conditional increment is zero and after stopping its increment vanishes. Applying the stopped-sum argument with zero conditional increments givesThe upper bound from part (a) now proves that first contact is an optimal stopping time.

Construct the strictly positive rolling one-period bond account using successive one-period zero-coupon bonds:At time $t-1$, invest the entire account value in the bond maturing at $t$. This gives a positive self-financing portfolio usable as a numéraire.

We use the finite-discrete-time fundamental theorem of asset pricing: in a frictionless market with finitely many adapted assets and trading dates, no arbitrage is equivalent to existence of an equivalent martingale measure for asset prices, including dividends, expressed in a positive traded numéraire. Maturing bond payoffs are reinvested in that numéraire. Let $\mathbb{Q}$ be such a measure and $L_t=\mathbb{E}[d\mathbb{Q}/d\mathbb{P}\mid {\mathcal{F}}_t]$ its positive density process. The martingale pricing relation for a unit zero-coupon bond isDefine $Z_t=L_t/{\beta }_t$. The Bayes formula for conditional expectation then givesThe positive expectations are finite because these are the traded finite bond prices; in particular $\mathbb{E}{Z}_T=Z_0{P}_0(T)$ when initial information is trivial. Normalize $Z_0=1$ without changing any ratio. With nontrivial initial information the same identities are conditional on that information. The state-price density need not be unique when the bond market is incomplete; existence is sufficient.

Assume the usual positive initial asset values. The coefficients determine a unique normalized local martingale deflator. The state-price density and local deflator distinction matters here: calling it a state-price density uses the local convention, while true expectation pricing needs an additional qualification, addressed below.

Let $ZB$ and $ZS$ be local martingales, with $Z_0=1$. The Brownian martingale representation theorem says that every local martingale in the usual natural Brownian filtration is continuous and is a stochastic integral against $W$. Applying it to $ZB$, and dividing by the positive finite-variation account, forcesfor a locally square-integrable predictable $\eta$. The Itô product rule for $ZS$ has driftIt must vanish. Since $S,Z,\sigma$ are positive, this yieldsConversely this drift choice makes both $ZB$ and $ZS$ local martingales. Continuity and strict positivity of $\sigma$ make $\lambda$ bounded along each path on every finite time interval, so its pathwise square integral is finite. The unique linear SDE solution isUniqueness follows from the forced drift and diffusion coefficients and uniqueness of this linear stochastic differential equation.

Continuity alone does not make the density a true martingale. For a true equivalent martingale measure, the stochastic exponential $D_t=Z_t{B}_t/B_0$ must have expectation one, for example under the Novikov condition on each horizon. True martingale pricing of all desired deflated payoffs also requires the relevant integrability. These stronger conclusions do not follow just from pathwise continuity.

An explicit counterexample to the stronger reading uses a three-dimensional Bessel process $R$ with $R_0=1$ and $dR=dW+R^{-1}dt$, which is a positive strong solution in the Brownian filtration. Take $B=1$ and $S=R$. Then $r=0$, $\mu =R^{-2}$ and $\sigma =R^{-1}$ are continuous and $\sigma >0$. The unique local candidate is $Z=R^{-1}$, with $ZS=1$. The reciprocal three-dimensional Bessel strict local martingale is not a true martingale. To verify the loss of expectation, use the standard Bessel transition densityIntegrating $p_t(1,y)/y$ gives $\mathbb{E}[R_t^{-1}]=2\Phi (1/\sqrt{t})-1<1$ for $t>0$, where $\Phi$ is the standard normal cumulative distribution function. A true pricing density for the constant bank account would have expectation one. Hence the printed hypotheses establish the local deflator statement, while a true-density reading needs an additional condition and is false as stated.

Let $h_t$ and $k_t$ be predictable holdings in the stock and account, with the stochastic integrability needed for a self-financing portfolio. Then $V=hS+kB$ and $dV=hdS+kdB$. The Itô product rule and self-financing identity giveSince $V=hS+kB$ and $\sigma \lambda =\mu -r$, the finite-variation term is zero. Thereforeand the deflated wealth is a local martingale.

For zero-capital nonnegative wealth under a local deflator, a nonnegative local martingale is a supermartingale, by localization and the Conditional Fatou lemma. Under the required nonnegative-wealth condition, $ZV\ge 0$ and starts at zero, so $0\le \mathbb{E}(Z_t{V}_t)\le 0$. Thus $Z_t{V}_t=0$ almost surely at every fixed $t$. Strict positivity of $Z$ gives $V_t=0$ almost surely. Taking a countable intersection over rational times and then using continuous wealth paths strengthens this toThis argument only needs the local deflator, so it remains valid without promoting $ZB$ to a true martingale.

The random walk $S_n$ is a martingale, since its integrable increments are independent of the past and have mean zero. For the bounded stopping time $n\wedge \eta$, the bounded optional stopping theorem, proved in the next question, gives $\mathbb{E}{S}_{n\wedge \eta }=x$.

If $0<x<r$, the value immediately before exit lies in $(0,r)$, soBefore exit the stopped value has absolute value less than $r$, and after exit it equals $S_{\eta }$. Consequently $|S_{n\wedge \eta }|\le r+|X_{\eta }|$ for every $n$. This is an integrable dominating random variable by the preceding part. Since $\eta <\infty$ with probability one, the dominated convergence theorem givesFor $r\le x$, $S_{\eta }=S_0=x$ directly, without any convention about $X_0$. Thus the expected stopped position is well-defined and has the stated value for every $r>0$.

Let $T\le N$ for a deterministic integer $N$. The stopped martingale has the finite-sum representationEach summand is integrable, and $\{T\ge k\}=\{T>k-1\}\in {\mathcal{F}}_{k-1}$ by the stopping time property. Therefore the conditional expectation identity for a martingale givesTaking expectations in the finite sum proves the bounded optional stopping theorem:No limiting argument or uniform-integrability assumption is needed for a bounded stopping time.

Put $V_t=⟨X{⟩}_t$ and $S_t={\sup }_{s\le t}|X_s|$. Suppose first that $|X|\le K$. The Itô formula for $|x{|}^p$, which is twice continuously differentiable for $p\ge 2$, givesFor $p=2$, the second derivative is interpreted as the constant $2$. The stochastic term has mean zero: its integrand is bounded, and $\mathbb{E}{V}_t<\infty$ makes it a square-integrable martingale. ConsequentlyThis is the first required estimate.

A bounded local martingale is a true martingale. Apply the allowed Doob Lp maximal inequality, and then Hölder's inequality with conjugate exponents $p/(p-2)$ and $p/2$ for $p>2$. With $D_p=(p/(p-1){)}^p$, this yieldsIf the left side is zero there is nothing to prove; otherwise divide by its indicated power and raise to $p/2$. For $p=2$, the same conclusion follows directly from $\mathbb{E}{X}_t^2=\mathbb{E}{V}_t$ and Doob's inequality. A usable constant is thereforeIn particular $C_2=4$. This is the upper maximal moment bound for a continuous local martingale.

For an unbounded continuous local martingale, stop at ${\tau }_n=\inf \{s:|X_s|\ge n\}$. Its stopped path is bounded, and its bracket at $t$ is $V_{t\wedge {\tau }_n}\le V_t$. The proved inequality gives a bound by $C_p\mathbb{E}{V}_t^{p/2}$, independent of $n$. Continuity makes ${\tau }_n\uparrow \infty$, and the stopped maxima increase to $S_t$. Monotone convergence proves the same inequality for the original unbounded process, with the same constant. This localization step also shows that the assumed bracket moments supply all the needed maximal moments.

Use $V=⟨X⟩$. Applying the Itô formula and the Itô product rule givesThe finite-variation terms cancel in the prescribed combination, leavingThus $Y$ is initially a continuous local martingale. It is a true martingale, not merely local. On each fixed horizon $T$,Part (a) with $p=4$, the assumed bracket moments and Cauchy-Schwarz inequality make this bound integrable. The integrable-supremum martingale criterion now applies, by localization and dominated conditional expectations. The same reasoning makes $X^2-V$ a martingale, so $\mathbb{E}{V}_t=\mathbb{E}{X}_t^2=t$.

The zero covariance now means $\mathbb{E}(X_t^2{V}_t)=t^2$. Since $Y_0=0$ and $\mathbb{E}{Y}_t=0$,This is the fourth-moment deficit and bracket variance identity. If equality holds for every $t$, then $V_t=t$ almost surely for each $t$. Take a single probability-one event for all rational times and use continuity to obtain $V_t=t$ simultaneously for all times. The Lévy characterization of Brownian motion then proves $X$ is Brownian motion in its given filtration. A proof of that characterization by conditional characteristic functions is included in Question 2(a).