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Let $U_t=(1,0,0)+(B_t^1,B_t^2,B_t^3)$, with three independent standard Brownian motions, and set $R_t=|U_t|$ and $M_t=R_t^{-1}$. The radial stochastic process $R$ is a three-dimensional Bessel process, which never hits zero by the Hitting-zero classification for a Bessel process. In three dimensions $u\mapsto |u{|}^{-1}$ is a harmonic function away from zero: direct differentiation gives $\Delta (|u{|}^{-1})=0$.

For the stopping times $T_n=\inf \{t:R_t\notin (1/n,n)\}$, $n\ge 2$, the Itô formula givesThe coefficients are bounded by $n^2$, so each stopped Itô integral is a true martingale on every finite horizon. The stopping times form a localizing sequence; as permitted, its divergence need not be proved here. Thus $M$ is a nonnegative continuous local martingale.

To see that $M$ is not a true martingale, integrate the three-dimensional multivariate normal density of $U_t$ in spherical coordinates. The angular integral yieldswhere $\Phi$ is the standard normal distribution function. Therefore the reciprocal of a three-dimensional Bessel process is a strict local martingale.

Use the standard definition of a local martingale, including an integrable initial value, and let $T_n\uparrow \infty$ be a localizing sequence. If $X\ge 0$, the Fatou lemma gives $\mathbb{E}{X}_t\le {\mathrm{liminf}}_n\mathbb{E}{X}_{t\wedge T_n}=\mathbb{E}{X}_0<\infty$. For $s\le t$, the Conditional Fatou lemma givesFor clarity, the Conditional Fatou lemma itself follows by setting $U_m={\inf }_{n\ge m}{U}_n$ for nonnegative $U_n$, applying the conditional version of the monotone convergence theorem to $U_m\uparrow \mathrm{liminf}{U}_n$, and observing $\mathbb{E}(U_m\mid {\mathcal{F}}_s)\le {\inf }_{n\ge m}\mathbb{E}(U_n\mid {\mathcal{F}}_s)$. Hence a nonnegative local martingale is a supermartingale.

If $|X_t|\le C$ simultaneously for all $t$, with a deterministic $C<\infty$, the dominated convergence theorem applied to the stopped martingale identity gives, for each $A\in {\mathcal{F}}_s$,This is exactly the conditional expectation identity, so a bounded local martingale is a true martingale. The same proof works for deterministic bounds on every finite time interval; merely having a finite random pathwise bound supplies no dominating integrable random variable and is a different convention.

Take the stopping timesContinuity gives $|M_{t\wedge S_n}|\le n$, with no overshoot at the first crossing. Stopping preserves the local martingale property, and the bounded case proved in question 4(i) makes each $M^{S_n}$ a continuous true martingale.

Every continuous path is bounded on each compact time interval, so $S_n\uparrow \infty$ almost surely. Hence this is also a localizing sequence consisting of bounded stopped martingales.

Let $A=⟨M⟩$ be the quadratic variation and stop at $T_b=\inf \{u\ge 0:A_u>b\}$. Its continuity and monotonicity imply $A_{u\wedge T_b}\le b$ and $A_{T_b}=b$ whenever $T_b<\infty$. The standard martingale isometry after stopping by bounded quadratic variation gives $\mathbb{E}(M_{t\wedge T_b}^2)=\mathbb{E}{A}_{t\wedge T_b}\le b$; in particular the stopped local martingale is square-integrable. This criterion follows by localization and the Doob L2 maximal inequality, which bound the second moment of the stopped supremum by $4b$ and then permit removal of localization.

On $\{A_t\le b\}$ one has $T_b\ge t$, so the original and stopped paths agree through time $t$. Therefore the Doob L2 maximal inequality and the Markov inequality giveThus

The usual quadratic variation convention includes $⟨M{⟩}_0=0$. If $A,C$ are two continuous finite-variation processes with $A_0=C_0=0$ such that $M^2-A$ and $M^2-C$ are continuous local martingales, then $A-C$ is a continuous local martingale of finite variation. It must be constant, and its initial value is zero, so $A=C$ up to indistinguishability of stochastic processes.

Here is a proof of the fact used. Subtract the initial value of a continuous local martingale $N$ of finite variation, and stop so that both $|N|$ and its total-variation process are bounded by deterministic constants. The stopped local martingale is a bounded true martingale, so orthogonality of its increments gives, on any deterministic partition of an interval of $[0,t]$,Continuity and the dominated convergence theorem justify the last limit. Remove stopping and use continuity on a countable dense set of times to conclude $N\equiv 0$.

As literally printed, uniqueness is false without normalization. If $A$ works then $A+c$ also works for any deterministic constant $c$, since subtracting a constant preserves the local martingale property; this is a genuine omitted convention.

With normalization, existence always holds for a continuous local martingale. Its quadratic variation $[M]$, obtained as the limit in probability of squared increments, is continuous, increasing, starts at zero and satisfies $M^2-[M]$ is a continuous local martingale, so taking $⟨M⟩=[M]$ supplies the required finite-variation process.

The space-time Hermite polynomials obtained by scaling the Probabilists' Hermite polynomials are genuine polynomials in both variables, including at $t=0$:This expansion follows by differentiating the defining Gaussian exponential, and shows that $H_n(x,0)=x^n$ extends smoothly through $t=0$. Consequently the Itô formula is applicable from time zero. The given backward heat equation cancels the drift, and the derivative identity givesThe integrand is continuous and adapted, hence previsible, and stopping the Brownian motion and time makes it bounded. Thus $H_n(B_t,t)$ is a local martingale for every $n\ge 0$. The $n=0$ case is the constant $H_0=1$; in fact Gaussian moments make the displayed Itô integral square-integrable on every finite horizon, so these are true martingales as well.

The first three Probabilists' Hermite polynomials are $h_1(x)=x$, $h_2(x)=x^2-1$ and $h_3(x)=x^3-3x$. Hence

In the martingale residual$v_i$ is the observed event count up to the end of the individual's follow-up: 1 for a recorded event and 0 for right censoring. The second term is the fitted counting-process compensator, obtained by integrating that individual's fitted hazard function while they remain at risk:Thus the residual is observed events minus the fitted cumulative event intensity over follow-up. The fitted intensity is not an event probability and can exceed 1. Positive values indicate more events than predicted over that observed exposure; negative values indicate fewer. A right-censored individual has a nonpositive residual. With true model parameters and independent censoring, the underlying observed-minus-compensator process is a martingale; fitted terminal residuals are not independent identically distributed errors.

Let ${\tau }_n\uparrow \infty$ be a localizing sequence for the discrete-time local martingale $X$. For $t\ge 1$, the increment of the stopped martingale isand $\{{\tau }_n\ge t\}\in {\mathcal{F}}_{t-1}$. Thus, for any $A\in {\mathcal{F}}_{t-1}$,The assumed integrability gives the dominating random variable $|X_t|+|X_{t-1}|$. The dominated convergence theorem now yields $\mathbb{E}[1_A(X_t-X_{t-1})]=0$, or $\mathbb{E}[X_t\mid {\mathcal{F}}_{t-1}]=X_{t-1}$. Iterating the tower property of conditional expectation proves an integrable discrete-time local martingale is a true martingale. The proof uses the discrete-time increment indicator; the conclusion does not extend to continuous time.

Under the standard local martingale definition, $X_0$ is integrable. For a nonnegative $X$, the Fatou lemma and a localizing sequence giveHence every $X_t$ is integrable, and part (a) applies. Therefore a nonnegative discrete-time local martingale is a true martingale. This is stronger than the continuous-time supermartingale conclusion.

Take an increasing localizing sequence ${\tau }_n$ for $X$ and define the stopping timesThe predictable process condition $K_{u+1}\in {\mathcal{F}}_u$ makes $R_n$ a stopping time. For each fixed finite time range, the finitely many coefficients $K_1,\dots ,K_t$ are finite almost surely; hence $T_n\uparrow \infty$ almost surely.

The stopped martingale transform can be writtenIts coefficient is ${\mathcal{F}}_{s-1}$-measurable and bounded in absolute value by $n$. Each summand is consequently integrable with zero conditional expectation given ${\mathcal{F}}_{s-1}$, since $X^{{\tau }_n}$ is a true martingale. Thus $M^{T_n}$ is a true martingale andThe original PDF fixes $M_0=0$; the TeX extraction omits this equality.

For an integer $m$ and $d\in \{-1,0,1\}$, verify the single-step identityIf $m\ge 1$, the positive part is linear across the step and $f(m)=1$; if $m\le -1$, it remains zero and $f(m)=0$. At $m=0$, $d^{+}=(d+d^2)/2$, which is checked for the three permitted values of $d$.

Apply this with $m=S_{t-1}-K$ and $d=S_t-S_{t-1}$, and telescope over $t$. The result is the discrete Tanaka formulaThe first sum is a martingale transform; the second records the correction at the kink of the positive part. The algebraic identity itself needs only the integer values and permitted increments, not the martingale property.

Let $p_K=C(T,K+1)-2C(T,K)+C(T,K-1)>0$ and $D_K=C(T+1,K)-C(T,K)$. Given $S_T=K$, the next value is one of $K-1,K,K+1$. The martingale property makes its upward and downward conditional probabilities equal; their sum is the conditional variance of the increment, ${\sigma }^2(T,K)=2D_K/p_K$. ThusThe discrete Dupire equation therefore recovers all positive-probability one-step conditional transitions from the European call option price surface. No Markov property is required: these are conditional probabilities given the current value, not necessarily given the whole past.

Recovery is impossible at a zero-probability state. For example, $S_t\equiv S_0$ satisfies every stated assumption, but at $T\ge 1$ a different integer $K$ can satisfy $|K-S_0|\le T$ while $p_K=0$. Its conditional transitions can be specified arbitrarily without altering $C$. Thus the literal all-states claim needs the positive-probability qualification.

Assume the usual positive initial stock price and strike, so the logarithm is defined, and interpret $U$ as a classical solution of the displayed partial differential equation. For a fixed complex $z$, put $F_t(z)=S_t^{z}U(t,{\sigma }_t,z)$, defining $S_t^z=\exp (z\log S_t)$. The Itô formula for the two Brownian motions with correlation coefficient $\rho$ givesThe partial differential equation cancels the entire drift. By the permission to treat the resulting local martingales as true martingales, and the terminal condition, $F_t(z)=\mathbb{E}[S_T^z\mid {\mathcal{F}}_t]$.

The bounded spot volatility ensures $\mathbb{E}{S}_T^{x_0}<\infty$: the Itô formula for a real power and localization bound its moment by $S_0^{x_0}\exp (x_0(x_0-1)\Vert \sigma {\Vert }_{\infty }^{2}T/2)$. Rewrite the proposed integrand asNow $|F_t(x_0+iy)|\le \mathbb{E}[S_T^{x_0}\mid {\mathcal{F}}_t]$, an integrable bound independent of $y$. Conditional Fubini's theorem and part (b) therefore implyThis route justifies the contour exchange without assuming bounds on $U$ uniform in all complex $z$. Cash is constant, $S$ is a true martingale under the stated allowance, and the displayed $C$ is a true martingale. The original measure is thus an equivalent martingale measure for all three assets. The fundamental theorem of asset pricing gives the market has no arbitrage under the usual admissible trading convention.

Use positive initial asset prices and the usual augmentation of the natural Brownian filtration. Define the market price of riskContinuity and strict positivity of $\sigma$ imply ${\int }_0^T{\lambda }_t^{2}dt<\infty$ on every finite horizon almost surely, because each path has a positive minimum of $\sigma$ there. Hence the strictly positive stochastic exponentialis defined and has $Y_0=1$ and $d{Y}_t=-Y_t(r_{t}dt+{\lambda }_{t}d{W}_t)$. The Itô product rule givesso it is a local martingale deflator.

For uniqueness, let $\tilde{Y}$ be any normalized strictly positive local martingale deflator. The Brownian martingale representation theorem makes $B\tilde{Y}$ a continuous local martingale with a Brownian motion integral representation. Dividing by $B$ therefore gives $d\tilde{Y}=-r\tilde{Y}dt+\eta dW$. The vanishing drift of $S\tilde{Y}$ requires $\tilde{Y}(\mu -r)+\eta \sigma =0$, so $\eta =-\tilde{Y}\lambda$. This scalar linear stochastic differential equation has exactly the exponential solution above, proving the normalized local martingale deflator is unique.

The assumptions do not make $\lambda$ deterministically bounded: $\sigma$ need not be bounded away from zero uniformly over outcomes. Thus a true martingale density or Novikov condition is not inferred here; the required conclusion is local.