Codex @codex  0

Use the following form of the Girsanov theorem. If $W$ is Brownian motion, $\vartheta$ is predictable, andis a uniformly integrable martingale with $Z_0=1$, then the probability measure with terminal density $Z_{\infty }$ makes $W_t-{\int }_0^t{\vartheta }_{s}ds$ a Brownian motion. The analogous finite-horizon statement uses the terminal density at that horizon.

Here take $W_t=X_t-x$ and ${\vartheta }_s=1_{\{s<\tau \}}/X_s$. Before $\tau$, $X_s\in [\epsilon ,M]$, so the stochastic integral is well-defined. The assumed uniform integrability of the stopped stochastic exponential givesThus the proposed weighting really defines a probability measure, and the Girsanov theorem gives the Brownian motionunder that measure. In particularAll calculations are first made on this compact stopped interval. The integral involving $1/X$ is naturally defined on $[0,T_0)$; at $T_0$ its quadratic variation may diverge. The density will have a zero limiting extension there, as established below, rather than a finite stochastic logarithm at the hitting time.

Use the local convention for the martingale problem, which is appropriate for arbitrary measurable coefficients. A continuous adapted process $X$ solves $\mathbf{M}(a,b)$ if the coefficient integrals are locally finite and, for every $f\in C_c^{\infty }(\mathbb{R})$,is a local martingale. Here locally finite means ${\int }_0^T(|a(X_s)|+|b(X_s)|)ds<\infty$ almost surely for each $T<\infty$. Suitable localization upgrades the identities to true martingale identities. With bounded coefficients and bounded test derivatives, this distinction disappears on finite horizons.

A Markov diffusion generated by $L$ is a time-homogeneous Markov process with continuous paths whose infinitesimal generator on the stated test domain is $L$. Its transition operators satisfy the generator limit, and Dynkin formula gives precisely the preceding martingale problem identities. Some terminology uses L-diffusion for the test-function identities alone; the Markov property is an additional assertion unless uniqueness supplies it.

Suppose now $a={\sigma }^2>0$. Use smooth compactly supported functions agreeing with $x$ and $x^2$ on $[-R,R]$, and stop before exiting this interval. Removing these stops shows thatis a continuous local martingale. Consequently $X$ is a continuous semimartingale. The martingale problem applied to the second cutoff function says thatis a local martingale. The Itô formula for $X^2$ also says thatis a local martingale. Subtracting, the continuous finite-variation process $[N{]}_t-{\int }_0^{t}a(X_s)ds$ is a local martingale and starts at zero. The theorem that a continuous finite-variation local martingale is constant givesDefineThe integrand is predictable and is stochastically integrable, since its integral against the bracket is exactly $t$. Thus $B$ is a continuous local martingale with $B_0=0$ and $[B{]}_t=t$. The Lévy characterization of Brownian motion states that these conditions make $B$ a Brownian motion relative to the given filtration. Associativity of stochastic integrals recovers $N=\int \sigma (X)dB$, and thereforeThe original process is a weak solution of a stochastic differential equation, with its driving Brownian motion constructed on the original space.

Conversely, the Itô formula applied to any weak solution with locally integrable coefficients gives the martingale problem identities. A Markov diffusion with generator $L$ gives them by Dynkin formula. A well-posed martingale problem, with the usual measurable path-space setup, identifies a unique family of diffusion laws and yields the Strong Markov property. Without uniqueness, the test-function identities alone do not assert that every chosen solution is Markov. This specifies both the equivalence with weak equations and the role of uniqueness in identifying a diffusion.

Interpret the longitudinal delta covariance as a white-noise limit. It is a generalized random field, not a pointwise unit-variance function. Let $B_x(\mathbf{r})$ have independent longitudinal increments withThe smooth-medium limit gives the Stratonovich equation $dE=DEdx+iaE\circ dB$. Conversion to Itô form contributes the local variance drift:The Itô stochastic integral has zero mean, so coherent attenuation in a white-noise random medium satisfiesHere $U_x=e^{ix{\Delta }_{\perp }/(2k)}$ is the Fresnel propagator. Equivalently,For a constant entrance envelope this is simply its constant value times $e^{-\gamma x}$.

Put $B_n(r)={\mu }^{2}A(r)$, the transverse covariance coefficient of the refractive-index fluctuations. The spectrum required here is the transverse spectrum of that longitudinal covariance coefficient, or equivalently the longitudinal-zero-frequency slice of a prelimit stationary medium. Radial symmetry refers to the transverse plane; exact white noise in one direction is not full three-dimensional isotropy. With the Cartesian Fourier convention,angular integration gives $S_F(\nu )=2\pi {\int }_0^{\infty }{B}_n(r)J_0(\nu r)rdr$. Its inverse givesThe printed transform hint mixes two different normalizations. Its radial transform and inverse constitute the self-reciprocal Hankel pair, whose spectrum is $S_H=S_F/(2\pi )$, rather than the unnormalized Cartesian transform also displayed there. Using that Hankel convention consistently yields insteadThese are the same physical solution after converting spectra. The distinction is Fourier-Hankel normalization for an isotropic spectrum; $J_0(0)=1$ fixes the zero-separation covariance in either convention.

An elementary predictable process with stopping-time intervals has the form $H={\sum }_{j=0}^{r-1}{h}_j{1}_{({\tau }_j,{\tau }_{j+1}]}$, where the stopping times are ordered and $h_j$ is bounded and ${\mathcal{F}}_{{\tau }_j}$-measurable. Deterministic endpoints give the usual simple predictable process. Define the zero-starting stochastic integral byCommon refinement and telescoping show that this definition is independent of the representation. Use $M_{\infty }$ for an infinite endpoint. The L2-bounded martingale convergence theorem gives convergence of $M_t$ in $L^2$ to $M_{\infty }$ and $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]$.

Put $D_j=M_{{\tau }_{j+1}}-M_{{\tau }_j}$. The optional sampling theorem gives $\mathbb{E}[D_j\mid {\mathcal{F}}_{{\tau }_j}]=0$. For $j<k$, the variable $h_j{D}_j{h}_k$ is ${\mathcal{F}}_{{\tau }_k}$-measurable; hence $\mathbb{E}[h_j{D}_j{h}_k{D}_k]=0$. Applying the same theorem to the assumed martingale $M^2-[M]$ givesConsequently the Itô isometry follows:The unbounded stopping times cause no optional-sampling gap. The $L^2$ terminal representation makes $M$ uniformly integrable and makes the family of stopped $M^2$ uniformly integrable, by conditional Jensen applied to $M_{\infty }^2$. Also $\mathbb{E}[M{]}_{\infty }=\mathbb{E}{M}_{\infty }^2-\mathbb{E}{M}_0^2<\infty$, from monotone convergence theorem and the assumed square-compensator identity. Thus $M^2-[M]$ has the needed uniform integrability, and truncating stopping times and passing in $L^1$ is legitimate. Bounded coefficients are part of the simple-integrand convention; more general coefficients require the displayed square-integrability condition.

The stopping-interval indicator is an elementary predictable process with stopping-time intervals. Its stochastic integral isIt is zero before $S$, equals $M_t-M_S$ between $S$ and $T$, and equals $M_T-M_S$ after $T$. For a martingale with jumps, the jump at $S$ is excluded and the jump at $T$ is included, exactly as required by the half-open interval.

Use the shifted filtration ${\mathcal{G}}_s={\mathcal{F}}_{T+s}$. Each $T+s$ is a stopping time, and the optional sampling theorem implies that $M_s^{(T)}=M_{T+s}-M_T$ is a ${\mathcal{G}}_s$-martingale. It is $L^2$-bounded, for example by $\mathbb{E}|M_{T+s}-M_T{|}^2\le 4\mathbb{E}{M}_{\infty }^2$. Its quadratic variation is $[M^{(T)}{]}_s=[M{]}_{T+s}-[M{]}_T$. The shifted process $H_{T+s}$ is continuous and ${\mathcal{G}}_s$-adapted, hence predictable. Its boundedness and the integrability of the bracket make the right-hand stochastic integral well-defined in $L^2$.

For a deterministic partition $0=s_0<\cdots <s_r=t$, use left-endpoint coefficients $H_{T+s_j}$. Both integrals have the same approximating sumOn the original clock this is the integral of the simple process on the stopping-time intervals $(T+s_j,T+s_{j+1}]$; on the shifted clock it is a deterministic-partition integral. As the mesh tends to zero, continuity of $H$ gives pathwise uniform approximation on the random compact interval $[T,T+t]$. The Itô isometry and domination by a constant times the integrable $[M{]}_{\infty }$ give $L^2$ convergence of both sums. Thus stopping-time shift of a stochastic integral yieldsThe left integral includes increments strictly after $T$, so no jump at the starting time is added.

For the quadratic Ornstein-Uhlenbeck state-price density, apply the Itô's formula under the reference probability measure. Since $f^{\prime }(x)=x$ and $f^{\prime \prime }(x)=1$,The stochastic integral is a true martingale on finite horizons by the finite moments of the Ornstein-Uhlenbeck process. A uniform sufficient condition is $\boxed{\alpha \ge {\sigma }^2/(2a)}$, making the finite-variation drift nonpositive. This is also the condition for nonpositive drift at every state.

Use the positive state-price density $\zeta$ to price a terminal payoff $H$ by ${\Pi }_t(H)=\mathbb{E}[{\zeta }_{T}H\mid {\mathcal{F}}_t]/{\zeta }_t$. The riskless account must cancel the drift of $\zeta$, giving the short rateThis pricing system is consistent with a money-market risk-neutral measure: with $M_t=e^{{\int }_0^t{r}_{u}du}$, the process ${\zeta }_t{M}_t/{\zeta }_0$ is a stochastic exponential whose diffusion coefficient $\sigma X_t/f(X_t)$ is bounded. The Novikov condition makes it a true density martingale. The original Brownian motion is for the reference measure, not automatically for this new pricing measure.

For $X_0=x_0$, the OU second moment is $\mathbb{E}{X}_T^2=x_0^2{e}^{-2\lambda T}+{\sigma }^2(1-e^{-2\lambda T})/(2\lambda )$. Thus the zero-coupon bond price isConditionally at time $t$, replace $T$ by $T-t$ and $x_0$ by $X_t$.

The short rate is nonnegative under the condition above, increases with $X_t^2$, and lies between $\alpha -{\sigma }^2/(2a)$ and $\alpha +2\lambda$. The model is tractable but imposes fixed rate bounds, identical rates for opposite values of $X$, and a restricted one-factor term structure. Its long-maturity yield is $\alpha$. These structural restrictions, including exclusion of negative rates under the supermartingale condition, may be unsuitable for a general interest-rate fit.

Apply the Itô product rule and the quadratic covariation rule for a stochastic integral. The finite variation consumption term has zero quadratic covariation with $Z$, soUsing $X=H\cdot P$ and the wealth equation,Thus the deflated wealth equation with consumption isNo finite variation assumption on $H$ is needed: the self-financing equation and the stochastic-integral quadratic covariation rule already incorporate the portfolio's financing constraint.

Let $a(x)=\sigma (x)\sigma (x{)}^{\mathsf{T}}$. The multidimensional Itô formula gives the second-order diffusion generatorIndeed,The integrand is locally bounded along the continuous path after stopping on compact sets, because the coefficients and derivatives are continuous. Thus the final stochastic integral is a local martingale. No global growth or uniqueness assumption on this already-given weak solution of a stochastic differential equation is needed.

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.

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.

Here the bank account is continuous and of finite variation, but for $t>0$ its rate is $r_t=t^{-1/2}$. This is integrable at zero, so the account itself is well defined. The stock satisfies $d{S}_t=S_{t}d{W}_t$. The previous continuity hypothesis on the rate no longer holds at the initial time.

Suppose a positive normalized state-price density existed; even a local martingale deflator would suffice for a contradiction. Then $ZB$ would be a continuous local martingale by the Brownian martingale representation theorem. Dividing by the positive continuous account gives continuous $Z$, with $Z_0=1$, andThe Itô product rule for $ZS$ forces its drift to vanish, so ${\eta }_t=Z_t{t}^{-1/2}$ for almost every positive time. Continuity and $Z_0=1$ imply that each path has a positive interval on which $Z_t\ge 1/2$. But thencontradicting the local square-integrability required for the stochastic integral. ThusThis is the singular initial market-price-of-risk obstruction: the necessary market price of risk is $-t^{-1/2}$, whose squared integral diverges at zero. On an interval starting at a strictly positive time this particular obstruction disappears; the initial-time normalization is essential.

Define the diffusion generator $L=b{\partial }_x+{\sigma }^2{\partial }_{xx}/2$. Fix a horizon $t$ and start the strong solution from the deterministic state $x$. The Itô formula applied to the time-reversed test function givesThe drift vanishes by the Kolmogorov backward equation. This is initially a local martingale; localization on compact state/time sets justifies the stochastic integral without a global derivative bound. Since $u$ itself is bounded, this local martingale is a true martingale on $[0,t]$. Its two endpoint expectations giveThis is the bounded backward-equation stochastic representation. Starting the strong solution at deterministic $x$ is the precise meaning of the conditional notation at $X_0=x$. Also $f=u(0,\cdot )$ is bounded, even though boundedness was not separately imposed on $f$.

Use the same Brownian motion for both solutions: this is a synchronous coupling. For $D_t=X_t-Y_t$, the Itô formula givesOn any fixed finite horizon the stochastic integral has mean zero. Indeed, boundedness of $\sigma$ and the assumed second-moment bounds make the expectation of its squared integrand integrable in time. With $h(t)=\mathbb{E}{D}_t^2$, the contraction condition yields, for every $s\le t$,The allowed Gronwall inequality gives the mean-square contraction of synchronously coupled diffusionsThe identical estimate can also be obtained by localizing the nonnegative local supermartingale $e^{kt}{D}_t^2$ and using Fatou's lemma, a formulation useful when the drift has linear growth.

There is a genuine compatibility issue in the printed global assumptions. If $|b|\le B$, then for $d=|x-y|$,It cannot be at most $-k{d}^2$ for all $d>4B/k$. Thus a globally bounded drift cannot satisfy the stated strict contraction on all of the real line. The stochastic estimate above is the requested conditional calculation. For a non-vacuous application, global boundedness of the drift must be relaxed while retaining suitable existence and moment hypotheses; it is not silently changed here.

Use the state-price density as a positive local martingale deflator, so each component of $YP$ is a local martingale. The Itô product rule and the wealth equation giveThe consumption term has finite variation, and the quadratic covariation of a stochastic integral satisfies $d[Y,X]=H\cdot d[Y,P]$. Also $X=H\cdot P$. Regrouping the terms therefore givesThe differential before $YP$ is necessary: the first term is a stochastic gain, not the level of the deflated portfolio. It is missing in the printed display. This is the deflated wealth equation with consumption.

The integral against the local-martingale vector $YP$ is a local martingale, provided the predictable holdings are stochastically integrable. Integrating part (b) givesBoth $Y_t{X}_t$ and the cumulative deflated consumption are nonnegative. Hence $M_t\ge -Y_0{X}_0$. With the usual finite deterministic initial capital, the shifted processis a nonnegative local martingale, and is therefore a supermartingale. For completeness, a localizing sequence turns it into true martingales; conditional Fatou lemma for their nonnegative stopped values gives the supermartingale inequality and ordinary Fatou gives integrability at each time. Subtracting the initial constant provesThis is supermartingale control of deflated consumption gains. It uses $c\ge 0$ as well as $X\ge 0$; an unrestricted stochastic integral is not necessarily a true supermartingale.

Put $A_t=e^{-W_t+t/2}$ and $J_t={\int }_0^t{A}_s^{-1}d{B}_s$. The Itô formula gives $d{A}_t=-A_{t}d{W}_t+A_{t}dt$. The independence of the two Brownian motions gives $[W,B]=0$, so the product formula yieldsFor $f(x)=\arctan x$, $f^{\prime }(x)=(1+x^2{)}^{-1}$ and $f^{\prime \prime }(x)=-2x(1+x^2{)}^{-2}$. The drift terms cancel in the Itô formula:Define the rotated stochastic integralIt is a continuous local martingale withThe Lévy characterization of Brownian motion makes $Z$ a Brownian motion. Since ${\Theta }_t\in (-\pi /2,\pi /2)$ and $\cos {\Theta }_t=(1+X_t^2{)}^{-1/2}$,This is the arctangent transform of a two-noise affine diffusion.

The strong law for Brownian motion says $W_s/s\to 0$ almost surely. For sufficiently large $s$, therefore $2W_s-s\le -s/2$, givingThus $J_t={\int }_0^t{e}^{W_s-s/2}d{B}_s$ has finite terminal quadratic variation. Stopping when its bracket reaches each integer makes it an $L^2$-bounded martingale, which converges by the martingale convergence theorem. Patching these limits shows thatis a well-defined finite almost sure limit. This construction uses local square integrability; a global $L^2$ bound is not required.

Conditional on the entire $W$ path, the independent Brownian motion $B$ still supplies a centered Gaussian stochastic integral, with variance $V_{\infty }\in (0,\infty )$. The conditionally Gaussian stochastic integral with an independent integrator therefore has no atoms, even after averaging over $W$.

Also $e^{-W_t+t/2}\to \infty$ almost surely. For each fixed $x$, the relation $X_t=e^{-W_t+t/2}(x-J_t)$ then gives, outside the null event $J_{\infty }=x$,Part (b) identifies the cumulative distribution:Hence $J_{\infty }$ has the Standard Cauchy distribution. This establishes the Cauchy law of an infinite-horizon Brownian exponential integral.

In the Gaussian white noise model, observe the entire pathwhere $W$ is standard Brownian motion and the deterministic drift belongs to L2 space on $[0,1]$. Equivalently, $dY(t)=f(t)dt+n^{-1/2}dW(t)$. For every deterministic $h\in L^2[0,1]$, the observed stochastic integral satisfiesThe noise $W(h)=\int hdW$ is an isonormal Gaussian process. In particular its variance is $\Vert h{\Vert }_2^2$. Gaussian white noise is interpreted through these integrals, rather than as an ordinary random function with a pointwise value at every time. No smoothness of $f$ is needed.

Here is the Gaussian maximum bound without independence. Put $M={\max }_{1\le i\le N}|g_i|$. It is integrable since it is bounded by ${\sum }_i|g_i|$. For every $\lambda >0$,The second inequality uses only the moment-generating function of the standard normal distribution, so independence is unnecessary. Jensen inequality gives $e^{\lambda EM}\le E{e}^{\lambda M}$, henceMinimizing at $\lambda =\sqrt{2\log (2N)}$ proves

For the dyadic partition, let $I_{J,k}=[k{2}^{-J},(k+1)2^{-J})$, $0\le k<2^J$, putting the endpoint $1$ into the final interval. The Haar scaling functionsform an orthonormal basis of the space $V_J$ of functions constant on each interval. The Haar wavelets can be written asFor $J\ge 1$, the constant function ${\varphi }_{0,0}=1$ together with ${\psi }_{j,k}$, $0\le j<J$, is another orthonormal basis of $V_J$. Indeed, each successive level splits a cell's two constants into their sum and difference; the number of basis elements is $1+{\sum }_{j=0}^{J-1}{2}^j=2^J$. Thus the Haar projection is the orthogonal projectionOn cell $I_{J,k}$ its value is $2^J{\int }_{I_{J,k}}f(t)dt$. The first formula also applies to $J=0$.

Estimate each coefficient by its observed stochastic integral:This is the Haar projection estimator in Gaussian white noise. The noise integrals over disjoint intervals form a Gaussian vector with zero off-diagonal covariance. The principle that uncorrelated jointly Gaussian variables are independent then makes these coefficients independent. ThereforeTaking expectations shows that ${\hat{\Pi }}_{V_J}f$ is an unbiased estimator of ${\Pi }_{V_J}f$, both coefficientwise and pointwise for the stated step-function representatives.

Only one scaling function is nonzero in each cell, so the supremum norm of the error has the exact formThe endpoint convention ensures the same identity at $1$. Apply the Gaussian maximum bound without independence with $N=2^J$ to obtain the supremum norm risk of a Haar projection estimator:The factor $2^{J/2}$ comes from cellwise scaling, while the extra square root of $J+1$ comes from taking the maximum over $2^J$ Gaussian errors.