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For a discrete-time martingale relative to the filtration $({\mathcal{F}}_n)$, each $M_n$ is ${\mathcal{F}}_n$-measurable, $\mathbb{E}|M_n|<\infty$, and $\mathbb{E}[M_n\mid {\mathcal{F}}_{n-1}]=M_{n-1}$ for $n\ge 1$. The tower property then gives the corresponding identity conditional on any earlier time.

A bounded previsible process has ${\varphi }_i$ measurable at time $i-1$ and $|{\varphi }_i|\le C$. Its martingale transform $V$ is adapted and integrable, since each increment has absolute expectation at most $C(\mathbb{E}|M_i|+\mathbb{E}|M_{i-1}|)$. Moreover,Thus $V$ is a martingale, with $V_0=0$.

For the Itô integral, begin with a bounded adapted step integrand $H_s=H_i$ on $(t_i,t_{i+1}]$. Its integral is $I={\sum }_i{H}_i(W_{t_{i+1}}-W_{t_i})$. Each summand has zero conditional mean. For $i<j$, conditioning the cross product on ${\mathcal{F}}_{t_j}$ makes its expectation zero. The conditional second moment of a Brownian increment is $t_{i+1}-t_i$, henceApproximate the predictable integrand $g(s,W_s)$ in $L^2(dsd\mathbb{P})$ by such step processes. The same identity makes their integrals Cauchy in $L^2$ and defines the limit; both its first and second moments converge. This gives the requested mean-zero result and the Itô isometry:

For the Ornstein-Uhlenbeck process, the deterministic integrating factor has no quadratic covariation with $X$. The Itô formula therefore gives $d(e^{\kappa t}{X}_t)=\kappa \theta e^{\kappa t}dt+\sigma e^{\kappa t}d{W}_t$. Integrating and dividing by $e^{\kappa t}$ yieldsThe mean-zero integral and Itô isometry giveAt $\kappa =0$ these formulas are read by continuity: $X_t=x+\sigma W_t$, with mean $x$ and variance ${\sigma }^{2}t$.

For the CIR model, let $C_0=c_0\ge 0$ and use the usual positive parameters. Before a possible visit to zero, the derivatives of $\sqrt{c}$ are $1/(2\sqrt{c})$ and $-1/(4c^{3/2})$. Applying the Itô formula gives the Square root of a CIR diffusion:The PDF's claim under $2ab={\sigma }^2$ is false: that choice leaves the nonzero drift ${\sigma }^2/(8Z_t)-a{Z}_t/2$, which is not affine in $Z_t$. The actual interior cancellation condition is $4ab={\sigma }^2$. Even then the principal square root is nonnegative and cannot be an unrestricted Gaussian OU process.

To give the intended corrected construction, let a signed process solve $d{Y}_t=-a{Y}_{t}dt/2+(\sigma /2)d{B}_t$, with $Y_0=\sqrt{c_0}$. The Itô formula givesWriting ${\tilde{W}}_t={\int }_0^t\mathrm{sgn}(Y_s)d{B}_s$, with either sign chosen at zero, produces a Brownian motion; the diffusion term is $\sigma |Y_t|d{\tilde{W}}_t$. Thus $Y^2$ is a CIR model with $ab={\sigma }^2/4$, while $Z=|Y|$ is reflected at zero. The reflection contributes local time of a semimartingale, so the interior SDE alone is not a global unrestricted OU equation. Its second moment is $c_0{e}^{-at}+{\sigma }^2(1-e^{-at})/(4a)$, agreeing with the CIR mean for that corrected parameter choice.

The requested mean under the parameters actually printed is still available without the false OU assertion. Taking expectation in the original CIR equation gives $m^{\prime }=a(b-m)$; the stochastic integral has zero mean after the usual moment/localization justification. Consequently, for the printed choice as well as general admissible parameters,

Assume the forward field is sufficiently differentiable in maturity to differentiate its deterministic and stochastic integrals. Moving along the diagonal adds the maturity derivative to the fixed-maturity dynamics:The Heath-Jarrow-Morton model no-arbitrage drift is ${\alpha }_Q(t,T)=\sigma (t,T){\int }_t^T\sigma (t,u)du$, so ${\alpha }_Q(t,t)=0$. Equivalently, from the integrated forward equation,That drift can depend on the entire previous forward curve or driving path, rather than on the current scalar rate.

The short rate need not be Markov. For a concrete One Brownian factor does not imply a Markov short rate example, take the deterministic volatility $\sigma (t,T)=T-t$. Its admissible risk-neutral drift is $(T-t{)}^3/2$, andSubtracting the deterministic part leaves $I_t={\int }_0^t{W}_{s}ds$. The past of $I$ reveals its derivative $W_t$, and $\mathbb{E}[I_{t+h}-I_t\mid {\mathcal{F}}_t^I]=h{W}_t$. Yet $W_t$ is not determined by $I_t$: the Gaussian covariance calculation givesConsequently the conditional law of the future depends on more than $I_t$, proving failure of the Markov property for the scalar short rate. The enlarged pair $(I_t,W_t)$ is Markov. Special maturity-volatility structures, such as those producing a Vasicek-type short-rate diffusion, can instead admit a scalar Markov realization.

For a simple predictable process, define its stochastic integral byIt is constant after $t_n$, so the symbol $(H\cdot M{)}_{\infty }$ means its value there. Put ${\Delta }_{k}M=M_{t_{k+1}}-M_{t_k}$. If $k<\ell$, then $Z_k{\Delta }_{k}M{Z}_{\ell }$ is ${\mathcal{F}}_{t_{\ell }}$-measurable, and the conditional expectation of ${\Delta }_{\ell }M$ given that sigma-algebra is zero. Boundedness of the coefficients and square-integrability of $M$ justify the expectations, so all cross terms vanish:For each diagonal term, the martingale increment identity also givesIndeed, the omitted term $2Z_k^2{M}_{t_k}{\Delta }_{k}M$ has expectation zero. Apply the assumed martingale property of $M^2-[M]$ to replace this difference of squares by the corresponding quadratic variation increment. ThusThe intervals $(t_k,t_{k+1}]$ in the Lebesgue–Stieltjes integral exactly match these increments, including possible jumps of $M$.

Let $X,Y$ be two solutions with the same Brownian driver and the same initial value. Write $D=X-Y$, and take a common Lipschitz constant $L$ for both coefficients. Stop when $|X|+|Y|$ reaches $n$. Before this stopping time ${\tau }_n$, all difference integrands below are bounded. The Itô formula givesThe stopped stochastic integral is a true martingale on every finite horizon. Lipschitz continuity therefore yieldsSince $D_0=0$, the Gronwall inequality makes the left side zero. Let $n\to \infty$; the continuous global solutions are bounded on each compact time interval, so their stopping times exhaust it. Equality first at every rational time and then by continuity gives indistinguishability. This proves pathwise uniqueness for globally Lipschitz drift and diffusion coefficients. The argument does not require the initial value to have a global second moment: after stopping, the difference is bounded and initially zero.

At exit, continuity gives $X_T^2=1$. The preceding density identity becomesSince $|X_s|\le 1$ before exit, ${\int }_0^T{X}_s^{2}ds\le T$. On $\{T\le t\}$ it follows that $Z_T\ge e^{1/2-T}\ge e^{1/2-t}$. The definition of the changed measure now yieldsThe positive sign of the stochastic integral in the density is what introduces the positive drift $X_{t}dt$; reversing that sign would give a different measure and would not prove this inequality.

The strain is an Ornstein-Uhlenbeck process. Multiplying its equation by the integrating factor $e^{t/\tau }$ and using the zero initial condition givesThe second expression is a stochastic integral against Brownian motion. A deterministic linear functional of Gaussian white noise is Gaussian, and its expected value is zero. For $t\ge t^{\prime }$, the noise covariance givesFor $t^{\prime }\gg \tau$, the initial-condition term is negligible. The stationary autocorrelation is thereforeIts normalized exponential decay has correlation time $\tau$. The zero-start Ornstein-Uhlenbeck covariance also shows how stationarity is approached.

Fix $T<\infty$ and a standard $d$-dimensional Brownian motion $B$ starting at zero. Let $D_n$ be deterministic nested finite partitions of $[0,T]$ with mesh tending to zero; dyadic partitions are the basic example. Let $B^n$ be the polygonal interpolation through the partition values and put ${\mathbf{B}}^n=S_2(B^n)$. The limit is enhanced Brownian motion, with incrementsHere the integral is a Stratonovich integral. In terms of Itô integrals,Its symmetric part is $\frac{1}{2}{B}_{s,t}^{\otimes 2}$, while its antisymmetric part is Lévy area,These identities and splitting the integral at an intermediate time give the Chen identity, so $\mathbf{B}$ has values in $G^2({\mathbb{R}}^d)$.

A precise nested polygonal approximation of Brownian rough paths statement is: for every $0<\alpha <1/2$,where the rough path metric used here isIn particular, for each $2<p<3$, choosing $1/p<\alpha <1/2$ gives convergence in the inhomogeneous $p$-variation rough path metric. The limiting signed area is the Lévy area, and the limiting full second level is Stratonovich, not Itô. This distinction is necessary: a polygonal lift has the prescribed symmetric second level, whereas an Itô lift would lack half the quadratic-variation correction.

Here is a proof sketch emphasizing the nested martingale structure. SetA Brownian bridge between two prescribed endpoints has the affine interpolation as conditional mean. ConsequentlyFor $i\ne j$ the coordinate processes remain independent when their respective partition values are conditioned on. Approximating the Itô integral by its left sums, conditioning each product, and then taking its $L^2$ limit givesIndeed each conditioned factor is its polygonal interpolation, and their sums converge to the ordinary integral of those piecewise linear functions. For a fixed pair $s,t$, the right side is therefore a martingale in the refinement index $n$. It is not being asserted to be a martingale in physical time; polygonal interpolation uses the right endpoint of its current cell.

The union of the partitions is dense. Continuity of $B$ implies that ${\mathcal{G}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{G}}_n)$ contains the full Brownian path and its stochastic integrals, up to completion. The martingale convergence theorem for conditional expectations therefore yields convergence of $B_t^n$ and all off-diagonal ${\mathbb{B}}_{s,t}^{n,ij}$ at every fixed pair, almost surely and in $L^q$ for finite $q$. Simultaneously it holds for a countable dense collection of pairs. The diagonal terms must be handled separately:They are not conditional expectations of the diagonal Brownian second level: the latter would also contain a conditional variance.

Pointwise convergence alone is insufficient for the rough path metric. The Burkholder-Davis-Gundy inequalities, Brownian scaling and Gaussian increment moments give, for arbitrarily large $q$,Equivalently, using the homogeneous norm on $G^2$, the $q$-th moment of the Carnot-Carathéodory distance of an increment is bounded by $C_q|t-s{|}^{q/2}$. The metric form of the Kolmogorov continuity theorem, with $q$ chosen large, gives for every $\beta <1/2$ a random $K_{\beta }$ having all finite moments such thatConditioning these bounds givesThe diagonal bounds follow by squaring the first bound. The two conditional-expectation processes are nonnegative martingales; the Doob Lp maximal inequality makes their suprema over $n$ finite almost surely, with all needed moments. Thus all the polygonal lifts have a common random bound at exponent $\beta$.

These bounds and the Chen identity give joint equicontinuity of the increments as functions of $(s,t)$. Convergence on the dense collection of pairs consequently gives uniform convergence of both levels on the time triangle. Take $\alpha <\beta <1/2$. For either level $k=1,2$, with $M$ a common bound and ${\epsilon }_n$ its uniform difference, interpolation givesThis proves the stated Hölder rough path metric convergence. Finally, for a partition $D$, the $k$-th level difference satisfiesbecause $\alpha p>1$. Taking the appropriate powers and suprema proves $p$-variation convergence too. Nesting is what permits the conditional-expectation martingale proof; without nesting one instead needs direct approximation estimates.

Interpret the otherwise undefined $b_k(t)$ here as the moment ${\beta }_k(t)$ defined above. For $k=1$, the process is $B_t$, and for $k=2$ it is $B_t^2-t$; both are martingales.

For $k\ge 3$, the Itô formula and the derivative of the Brownian moment recursion yieldThe stochastic integral is a true martingale on finite horizons, since the required $2k-2$ moment is integrable in time. The continuous finite-variation term is not identically zero. If it were, its continuous derivative would vanish at every time, whereas at a fixed $t>0$, the nondegenerate Gaussian $B_t$ does not almost surely satisfy the polynomial equation $B_t^{k-2}={\beta }_{k-2}(t)$. Uniqueness of continuous semimartingale decomposition therefore rules out even the local martingale property. HenceFor example $\mathbb{E}(B_t^3\mid {\mathcal{F}}_s)=B_s^3+3(t-s)B_s$, so merely subtracting its zero expectation cannot make the cubic power a martingale. The centered Brownian powers need not be martingales distinction is resolved by subtracting the random drift compensator $k(k-1){\int }_0^t{B}_s^{k-2}ds/2$ instead of the deterministic mean.

Write $A=\sigma {\sigma }^T$ and let $V(x,t)$ denote the optimal remaining expected reward from positive portfolio wealth $x$ at time $t$. The Hamilton-Jacobi-Bellman equation and terminal condition areThe controls are progressively measurable, the investment integrals exist, consumption is nonnegative and locally integrable, and the resulting portfolio wealth stays in the utility domain. At zero consumption use the right limit of the utility function if needed. This is the usual admissible trading strategy formulation; without restrictions preserving the domain the reward need not be defined.

For the verification bound, assume $V$ is a nonnegative classical solution, continuous through the terminal time, with one time derivative and two wealth derivatives in the interior. Fix any admissible controls and putThe Itô formula givesThe drift is nonpositive because the actual controls cannot exceed the supremum in the Hamilton-Jacobi-Bellman equation. Localize by stopping times on which the stochastic integral is a true martingale, the derivatives and wealth stay in their interior domain, and the accumulated reward is finite. For each such stop ${\tau }_n$,Both utilities and $V$ are nonnegative. Thus the Fatou lemma, continuity, and the terminal condition giveFor admissible paths strictly positive on the closed horizon, the domain-localizing stops eventually reach $T$. If zero wealth is allowed, the same proof uses the continuous boundary extension of $V$ and the utilities there. Nonnegativity is what permits removing the stops without assuming the uncontrolled stochastic term already has zero expectation. This is verification by a nonnegative control supermartingale.

The square-root case needs an additional no-arbitrage qualification because the printed volatility matrix need not be invertible. If there is $v\in \ker {\sigma }^T$ with $v^T\mu >0$, investment $\theta =Kv$ has zero volatility and arbitrarily large positive drift. With any fixed positive consumption, sufficiently large $K$ preserves positive wealth and makes terminal square-root utility tend to infinity. For example, in one dimension $\sigma =0$, $\mu =1$, $\theta =K$, $C=1$ gives $X_s=x+(K-1)(s-t)$, so the value is infinite when $T>t$. A finite separated solution does not exist for all matrices allowed by the printed hypotheses.

The needed condition is $\mu \in \mathrm{Range}A$, equivalent to $\mu$ being orthogonal to $\ker {\sigma }^T$, because $A$ is a symmetric matrix and $\ker A=\ker {\sigma }^T$. Assume this condition and let $B=A^{+}$ be its Moore-Penrose inverse; if $A$ is invertible, simply use $B=A^{-1}$. Put $q={\mu }^{T}B\mu \ge 0$. Seek the square-root investment-consumption value in the formThen $V_x=g/\sqrt{x}$ and $V_{xx}=-g/(2x^{3/2})$. Completing the quadratic investment expression, using $AB\mu =\mu$, shows that its maximum is attained at ${\theta }^{\ast }=2xB\mu$ and has value $qg\sqrt{x}$:Any investment component in $\ker A$ has no effect. For consumption, set $z=\sqrt{c}$. Its contribution is the concave quadratic $2z-(g/\sqrt{x})z^2$, whose maximum is $\sqrt{x}/g$ at $c^{\ast }=x/g^2$. Substituting both maxima into the Hamilton-Jacobi-Bellman equation givesWith $h=g^2$, this becomes $h^{\prime }+qh+1=0$, $h(T)=1$. Solving this linear ordinary differential equation givesIn particular $h\ge 1$, so $g=\sqrt{h}$ stays positive, and the two controls areThese controls are admissible and actually attain the value. Indeed their portfolio wealth satisfiesSince $\Vert 2{\mu }^{T}B\sigma {\Vert }^2=4q$, the explicit positive solution starting at $(x,t)$ isIts finite-horizon moments of every positive order are finite. More directly,Differentiating $g(s)\mathbb{E}\sqrt{X_s}$ and using the equation for $g$ gives $\frac{d}{ds}(g(s)\mathbb{E}\sqrt{X_s})=-\mathbb{E}\sqrt{X_s}/g(s)$. Integrating from $t$ to $T$ therefore yieldsThis proves attainment without relying on an unjustified expectation of a local martingale. When $\mu$ is outside $\mathrm{Range}A$, the singular-volatility drift arbitrage counterexample applies instead.

Use the usual stochastic-integrability assumptions implicit in the price stochastic differential equation: the coefficients have progressively measurable versions, the volatility is locally square integrable, and initial prices are finite deterministic positive numbers. A no-consumption admissible trading strategy consists of predictable risky share holdings $\pi$ for which the price stochastic integral exists and the portfolio wealth of the self-financing portfolio isThe remainder $X_t-{\pi }_t^T{S}_t$ is held in the unit cash asset. Admissibility requires $X_t\ge -a$ throughout the horizon for some deterministic finite $a$; the more restrictive nonnegative-wealth convention also works. A finite-horizon arbitrage is such a strategy with $x=0$, $X_T\ge 0$ almost surely and $P(X_T>0)>0$. The lower bound is essential to exclude doubling strategies.

Interpret the stated uniform ellipticity as ${\sigma }_t{\sigma }_t^T\ge \epsilon I$ for a deterministic $\epsilon >0$. If $|{\mu }_t|\le K$, the market price of risk is bounded:The authoritative PDF puts both integrals inside the exponent. With that interpretation,On every finite horizon $T$, boundedness of $\lambda$ gives the Novikov condition:Hence the positive stochastic exponential $Z$ is a true martingale with $\mathbb{E}{Z}_T=1$. Define the equivalent probability measure $Q_T$ by $d{Q}_T=Z_{T}dP$. The Girsanov theorem says that $W_t^Q=W_t+{\int }_0^t{\lambda }_{s}ds$ is a Brownian motion under $Q_T$, up to time $T$. Consequentlyso the prices are local martingales under $Q_T$.

Every admissible zero-initial-capital self-financing portfolio has portfolio wealth which is therefore a $Q_T$-local martingale bounded below by $-a$. The shifted process $X+a$ is a nonnegative local martingale, hence a supermartingale by localization and the Conditional Fatou lemma. In particular ${\mathbb{E}}_{Q_T}{X}_T\le 0$. If the strategy were an arbitrage, nonnegativity of $X_T$ would force $X_T=0$ $Q_T$-almost surely. Equivalence of $P$ and $Q_T$ would then force $P(X_T>0)=0$, a contradiction. The market has no admissible arbitrage on any finite horizon. This proves the exclusion argument itself, rather than just asserting the fundamental theorem of asset pricing.

For the deflated price calculation, let ${\sigma }_{t,i}$ denote the column vector whose transpose is the $i$th row of ${\sigma }_t$. The Itô product rule includes the quadratic covariation termSince ${\sigma }_{t,i}^T{\lambda }_t={\mu }_t^i$, the full product equation isLocalization makes the last stochastic integral a true martingale on each stopped interval, proving each component of the deflated price vector is a local martingale under the original measure. This is the bounded-coefficient asset deflator identity; bounded volatility has not yet been used.

If $\sigma$ is bounded, then ${\sigma }_{t,i}-{\lambda }_t$ is bounded as well. Solving the last linear stochastic differential equation givesThe Novikov condition now holds for this stochastic exponential on every finite horizon, because its bracket is bounded by a deterministic constant times the horizon. ThereforeThe entire deflated price vector is a true martingale when volatility is bounded. This finite-horizon argument does not assert uniform integrability as time tends to infinity, which is not required. Integrable random initial prices can also be handled by the conditional martingale argument; without any integrability assumption on initial prices, a true-martingale assertion would not be meaningful.

Use the PDF's starred image hull $K_t^{\ast }$ and write $h_t={\Phi }_t=g_{K_t^{\ast }}\circ \Phi \circ g_{K_t}^{-1}$. For $t<T$, this map is analytic across a real neighborhood of the driver. SetLet $u(t)=\mathrm{hcap}(K_t^{\ast })/2$. The classical conformal change of half-plane capacity gives $\dot{u}(t)=a_t^2$, and the image driver, indexed by original time, is $h_t({\xi }_t)$. Differentiating the conjugacy relation givesThese are precisely the standard classical identities allowed in this question. The singularities cancel at $z={\xi }_t$. To calculate the remaining derivative, put $\delta =z-{\xi }_t$ and expand at fixed $t$:Thereforewhere each ${\partial }_t$ keeps the spatial variable fixed before evaluation. Applying the Itô formula to $h_t^{\prime }({\xi }_t)$ gives the boundary derivative diffusion under conformal Loewner conjugacyAt $\kappa =8/3$, the third-derivative drift cancels. A second application of the Itô formula to $a_t^p$ givesChoosing $p=1-1/\kappa =5/8$ cancels the remaining drift. Hence the SLE eight-thirds restriction martingale isLocalizing away from $T$ and from vanishing derivatives makes the displayed stochastic integral a martingale on each localized interval, so it defines a continuous local martingale up to $T$.

Set $N_t={\int }_0^t{Z}_s^{-1}d{Z}_s$, the stochastic logarithm of $Z$. On each finite horizon, a strictly positive continuous path has its reciprocal bounded; stopping at levels of $Z$, $Z^{-1}$ and $[Z]$ rigorously localizes this stochastic integral. Thus $N$ is a zero-start continuous local martingale, withThe Itô formula for the ordinary logarithm givesAccordingly, with the exponential convention printed in this question,The initial constant in $M$ contributes no quadratic variation. Under the standard normalized Doléans-Dade exponential convention, which subtracts the driver's initial value in the exponent, the same formula is written $Z=Z_0\mathcal{E}(N)$.