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Work with the filtration relative to which the given Brownian motion or Markov jump process has its stated properties. All martingale assertions below concern every finite time interval; no infinite-horizon uniform integrability is claimed.

First put $C={\sup }_x|\sigma (x)|$. The solution has the representation $X_t={\int }_0^t\sigma (X_s)d{B}_s$, and the Itô isometry givesThus $X$ is a square-integrable martingale, not just a local martingale. Applying the Itô formula to its square cancels the finite-variation term and givesThe expected squared integrand integral is bounded byTherefore both $X$ and $Z$ are true square-integrable martingales. This is the bounded diffusion coefficient gives square martingales criterion. Only bounded measurability of the coefficient and existence of the stipulated solution have been used.

For the jump process, the Lévy kernel of a Markov jump process uses displacements: a mark $y$ sends the state $x$ to $x+y$. Let $\Lambda ={\sup }_x\lambda (x)$ and $V={\sup }_{x}v(x)$. The bounded total rate prevents explosion; the number of jumps on a finite interval is dominated by a rate-$\Lambda$ Poisson process. Let $N(ds,dy)$ record the jump displacements. Its predictable compensator of a jump measure isThe Cauchy-Schwarz inequality gives $\int |y|K(x,dy)\le \sqrt{\lambda (x)v(x)}$, so the zero first moment is absolutely meaningful. Since that moment vanishes,The compensated integral is a square-integrable martingale: its isometry follows first for bounded marks from conditional compensation and then by truncation in the squared-integrand norm.

The jump version of the square Itô formula is $X_t^2=2{\int }_0^t{X}_{s-}d{X}_s+{\sum }_{s\le t}(\Delta X_s{)}^2$. HenceThe first term is square integrable becauseThe second is an integrable martingale: the expected absolute uncompensated jump-square sum is at most $Vt$, and the compensator identity, multiplied by any bounded event known at an earlier time, makes its increments conditionally centered. Values $X_s$ and $X_{s-}$ agree for Lebesgue-almost every time, so the displayed compensation is exactly the one defining $Z$. Thus $X$ and $Z$ are true martingales in the jump case as well. A fourth moment of the jumps is unnecessary; in particular we have not asserted that this $Z$ must be square integrable. This is centered finite-rate jump kernel gives square martingales.

For $g(x)=\log |x|$ on ${\mathbb{R}}^2\setminus \{0\}$,The Itô formula, stopped in a compact annulus, therefore givesThis proves the local martingale property up to a possible hitting of the origin. For the annulus exit time $T$, $M_{t\wedge T}$ is bounded between $\log r$ and $\log R$, so it is a true uniformly integrable martingale.

The exit is finite almost surely. Indeed $|B_t{|}^2-2t$ is a martingale, so stopping at $T\wedge t$ gives $2\mathbb{E}(T\wedge t)=\mathbb{E}|B_{T\wedge t}{|}^2-1\le R^2-1$. Letting $t\to \infty$ proves $\mathbb{E}T<\infty$. Bounded convergence in the logarithmic martingale now yieldsIf $p$ is the probability of reaching radius $r$ before radius $R$, continuity givesHitting zero before reaching radius $R$ would require reaching every positive radius $r$ first. Its probability is bounded by $p$, which tends to zero as $r\downarrow 0$. Any finite-time path hitting zero is bounded before that time and hence hits zero before exiting some integer-radius disk. A countable union over those disks proves $B_t\ne 0$ for every $t\ge 0$, almost surely. This is planar Brownian motion avoids a fixed point; it also makes $M$ a globally defined continuous local martingale.

To decide whether it is a true martingale, compute its expectation directly. Condition on $B_0$ if necessary and use rotation to take $B_0=(1,0)$ and write $B_t=B_0+W_t$. Conditional on $\rho =|W_t|$, the angle of $W_t$ is uniform. The angular average of a logarithmic potential isFor $\rho <1$ this follows by taking the real part of the convergent power series for $\log (1+\rho e^{i\theta })$: each nonconstant Fourier term integrates to zero. For $\rho >1$ factor out $\rho$ and apply the same argument to $1/\rho$. The equality at one follows by the integrable limiting logarithmic singularity. All interchanges are valid: the shifted Gaussian density is bounded near the origin, where ${\int }_0^{1}r|\log r|dr<\infty$, and its Gaussian tail controls logarithmic growth.

The radius $\rho$ has density $(\rho /t)e^{-{\rho }^2/(2t)}$. Consequently, for $t>0$,Since $M_0=0$, $M$ is not a martingale, despite integrability at each deterministic time. This is the logarithmic radius of planar Brownian motion example.

The optional hint is consistent with this conclusion. The upper-radius exit $T(2,0)$ is finite and never occurs at zero. After that exit the stopped square is $(\log 2{)}^2$; before it, a radius in $[1/2,2]$ also gives a logarithmic square no larger than $(\log 2{)}^2$, while a radius below $1/2$ contributes precisely the extra indicated term. This proves the hinted inequality without assuming that the unstopped process is a true martingale.

Write $f=u+iv$. The Cauchy-Riemann equations give $v_x=-u_y$, $v_y=u_x$ and $\Delta u=\Delta v=0$. Since the two Brownian coordinates are independent, $[X{]}_t=[Y{]}_t=t$ and $[X,Y{]}_t=0$. The Itô formula therefore becomesStop in balls on which the derivatives are bounded. The stopped integrals are martingales, proving that $U$ and $V$ are local martingales. Their coefficient rows have equal squared length and are orthogonal. ThusThese are quadratic covariations of an analytic Brownian image; orthogonality does not in general imply independence of the two transformed coordinates.

For the exit law, put $c=2a^2$. Squaring sends the positive quadrant to the upper half-plane, sends $a+ia$ to $ic$, and the Möbius transformation $w\mapsto (w-ic)/(w+ic)$ sends the upper half-plane to the unit disk with the starting point sent to zero. Its composition is the displayed conformal map. The exit time is finite: it is the minimum of the two almost surely finite one-dimensional hitting times of zero. There is no hit at the quadrant's corner, since a planar Brownian motion started away from zero avoids that point.

We can justify the uniform image exit law without assuming a conformal-time-change theorem. For each integer $n\ge 1$, the real and imaginary parts of $f(Z_{t\wedge T}{)}^n$ are bounded martingales obtained from harmonic functions. At the start their values are zero. The bounded convergence theorem at the finite exit time gives $\mathbb{E}f(Z_T{)}^n=0$. The boundary image has modulus one, and conjugation gives the vanishing negative Fourier moments as well. Trigonometric polynomials are dense among continuous functions on the unit circle, so these moments characterize the uniform distribution there.

Consequently the real variable $W=Z_T^2$ is the inverse image of a uniform point $e^{i\Theta }$ under that Möbius transformation:The Jacobian of this change of variable gives the centered Cauchy distribution of scale $c$:Positive $w$ corresponds to exit at $\sqrt{w}$ on the real axis; negative $w$ corresponds to exit at $i\sqrt{-w}$ on the imaginary axis. Pulling back by $w=\pm r^2$ therefore gives the complete boundary lawHere $dr$ on each axis denotes length along that axis, not two-dimensional area measure. Each density integrates to $1/2$, with no remaining atoms. Equivalently the chosen axis is a fair Bernoulli choice independent of the radius, andThis is Brownian exit law from a quadrant, obtained as a harmonic measure and not as an absolutely continuous planar density.

Here is a finite-horizon Girsanov theorem statement with the sign fixed explicitly. Let $B$ be a Brownian motion under $P$, and let $\theta$ be predictable with ${\int }_0^T{\theta }_s^{2}ds<\infty$ almost surely. Ifis a true martingale with $\mathbb{E}{L}_T=1$, then the measure $Q$ defined by $dQ=L_{T}dP$ makes $B_t-{\int }_0^t{\theta }_{s}ds$ a Brownian motion up to $T$. The sufficient Novikov condition is $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{\theta }_s^{2}ds)<\infty$. Merely being a local martingale is not a sufficient density condition.

For a deterministic absolutely continuous function $h$ with $h(0)=0$ and derivative $g\in L^2[0,T]$, take $\theta =g$. Novikov's condition holds because its exponential is deterministic and finite. Under $Q$, the canonical process has the law of a Brownian motion plus $h$. Thus for every bounded measurable path functional $F$,This derives the Cameron-Martin theorem density formula from the Girsanov theorem, rather than assuming the two measures have the same drift convention. Replacing $h$ by $-h$ changes the sign of the stochastic integral.

The question's path space is the whole half-line, not just one finite horizon. When $g\in L^2(0,\infty )$, the density martingale satisfiesIt has a uniformly integrable limit $L_{\infty }=\exp ({\int }_0^{\infty }{g}_{s}d{B}_s-\Vert g{\Vert }_2^2/2)$ of mean one, strictly positive almost surely. Integrals here converge in $L^2$ and almost surely. The finite-time identities then identify $L_{\infty }$ as the density on the full path sigma-algebra, since finite-time cylinder events generate it. This is Cameron-Martin shifts on infinite Wiener path space. Finite-horizon absolute continuity of measures alone does not imply infinite-horizon absolute continuity of measures.

Let $B$ be a standard planar Brownian motion started at $z$. Neighbourhood recurrence means that, with probability one, every nonempty open subset of ${\mathbb{R}}^2$ is visited at arbitrarily large times. Not hitting points means that for each fixed $w\ne z$, ${\mathbb{P}}_z(\exists t\ge 0:B_t=w)=0$. If $w=z$, time zero must be excluded, and there is still no return at any positive time. This assertion concerns each prescribed point; it does not say that the path contains no points.

We prove both claims from the planar Brownian annulus hitting probability. Fix a center $w$ and radii $0<r<\rho =|z-w|<R$, and define the stopping times ${\tau }_r=\inf \{t:|B_t-w|=r\}$ and ${\tau }_R=\inf \{t:|B_t-w|=R\}$. Their minimum $\sigma$ is finite almost surely: a coordinate of Brownian motion cannot stay forever in a bounded interval, by the Brownian reflection principle, so exit from the outer disk is finite. On the annulus,The Itô formula therefore makes $\log |B_{t\wedge \sigma }-w|$ a local martingale. It is bounded between $\log r$ and $\log R$, hence is a genuine martingale. The optional stopping theorem at $t\wedge \sigma$, followed by the bounded convergence theorem, givesThusThe exit can only be at one of the two distinct boundaries by path continuity.

For a fixed $r$, let $R\to \infty$. These events increase to $\{{\tau }_r<\infty \}$: any path up to a finite hitting time is bounded and so avoids some sufficiently large outer circle. The ratio tends to one, proving that any closed disk of positive radius is hit almost surely from any starting point outside it. An initial point inside it already counts as a hit.

To prove the stronger recurrence statement, take a closed disk contained in any prescribed nonempty open set. At each integer time $n$, the Markov property and the preceding disk-hitting result imply that some time $t\ge n$ visits this disk with probability one. Intersecting these events over all $n$ proves visits at arbitrarily large times. A countable basis of disks with rational centers and rational radii makes this simultaneous for all nonempty open sets. This is recurrence of planar Brownian motion, or equivalently neighbourhood recurrence.

For point avoidance, keep $R>\rho$ fixed and let $r\downarrow 0$. Hitting $w$ before ${\tau }_R$ requires hitting every inner circle first. The displayed ratio tends to zero, so this point-hitting event has probability zero. Any finite hit of $w$ would occur before exit from some outer circle with integer radius, again because the preceding path is bounded. A countable union now gives $\boxed{{\mathbb{P}}_z(T_w<\infty )=0(z\ne w)}$, the polar point for planar Brownian motion assertion.

Finally, even if the process starts at $w$, for each deterministic $s>0$ its normal distribution has a density, so $\mathbb{P}(B_s=w)=0$. The Markov property at $s$ and the result from other starting points exclude any visit to $w$ at or after $s$. Taking $s=1/n$ excludes all positive-time returns. Thus planar Brownian motion repeatedly approaches every point, while almost surely never hitting any prescribed point at positive time.

Use the physical dynamics of the Black-Scholes model, $d{S}_t/S_t=\mu dt+\sigma d{W}_t$, with bank rate $\rho$ and $\sigma >0$. Write $\lambda =(\mu -\rho )/\sigma$ for the market price of risk. If $w_t$ is portfolio wealth, ${\pi }_t$ the dollar amount in stock, and $c_t\ge 0$ the consumption rate, the budget dynamics areAdmissible controls are progressively measurable, locally integrable in the required drift and diffusion norms, and keep $w_t\ge 0$. Such a solvency restriction excludes doubling-type strategies. In a complete Brownian market, the investment-consumption problem is to maximizewhere the utility functions are increasing and concave. A terminal-only problem omits the integral. Strict concavity gives uniqueness of optimal consumption and terminal wealth when an optimum exists.

The state-price density isThe Itô formula givesThus $H_t{w}_t+{\int }_0^t{H}_s{c}_{s}ds$ is a nonnegative local martingale, hence a supermartingale, yielding the state-price budget constraintConversely, a nonnegative consumption plan and terminal payoff $\xi$ satisfying budget equality and the necessary integrability can be financed in the Brownian complete market. DefineThe Brownian martingale representation theorem for the total future budget supplies a stochastic-integral coefficient; comparing it with $H_t(\sigma {\pi }_t-\lambda w_t)$ defines ${\pi }_t$. This reconstructs the wealth equation with $w_t\ge 0$. Localization gives the representation for integrable positive budgets when an $L^2$ assumption is not available.

For smooth strictly concave utilities with the usual marginal-utility range, let $I_u=(u^{\prime }{)}^{-1}$ and $I_U=(U^{\prime }{)}^{-1}$ be the inverse marginal utility functions. The Lagrange-multiplier solution iswhere $y>0$ makes the budget an equality. This is a global argument, not merely formal differentiation: the concave supporting-tangent inequality giveswith the analogous terminal inequality. Taking expectations, integrating, and applying the budget bound makes the difference in objectives nonpositive. Replication of the candidate budget proves attainability and optimality whenever these expectations are well-defined and finite. If the marginal-utility ranges or integrability conditions fail, existence must be established separately rather than inferred from the first-order equations.

A detailed example is constant relative risk aversion utility $u(c)=c^{1-\gamma }/(1-\gamma )$ and $U(w)=\eta w^{1-\gamma }/(1-\gamma )$, with $\gamma >0$, $\gamma \ne 1$, $\eta >0$, and $\delta \ge 0$. PutThe optimal plans from the marginal-utility equations are $c_t^{\ast }=(y{e}^{\delta t}{H}_t{)}^{-1/\gamma }$ and ${\xi }^{\ast }={\eta }^{1/\gamma }(y{e}^{\delta T}{H}_T{)}^{-1/\gamma }$. The normal distribution of $W_t$ gives, for $a=1-1/\gamma$,The budget therefore becomes $y^{-1/\gamma }b(T)=w_0$, determining the multiplier explicitly. Independent Brownian increments in the conditional budget formula give $w_t^{\ast }=c_t^{\ast }b(T-t)$. The diffusion coefficient of $c_t^{\ast }$ is $(\lambda /\gamma )c_t^{\ast }$, so matching the wealth diffusion determines the dollar stock holding. Consequently the finite-horizon power-utility investment and consumption optimum isAll candidate moments are finite on a finite horizon, since powers of a lognormal state-price density have finite moments. The budget and tangent argument above therefore verify optimality in this example, including admissibility and positive wealth.

For another verification of the formulas, the Hamilton-Jacobi-Bellman equation for a value function with discount measured from the current time isSubstitute $V(w,t)=b(T-t{)}^{\gamma }{w}^{1-\gamma }/(1-\gamma )$. The two concave maximizations give the displayed controls, and the remaining equation is $b^{\prime }(\tau )=1-kb(\tau )$ with $b(0)={\eta }^{1/\gamma }$, exactly solved by the formula above. ThusFor $k=0$, interpret the integral as $\tau$, so $b(\tau )=\tau +{\eta }^{1/\gamma }$. Unlike the infinite-horizon Merton consumption-investment problem, finite horizons do not require $k>0$. Consumption vanishes in the terminal-only formulation; then the integral term in $b$ is absent and the optimal stock fraction remains $(\mu -\rho )/(\gamma {\sigma }^2)$. Logarithmic utility is handled directly with inverse marginal utility $1/z$ and the same wealth-to-consumption formula at $\gamma =1$, with $k=\delta$.

A Gaussian forward-rate field models the whole instantaneous forward rate curve. Let $H$ be a real separable Hilbert space, let $W$ be a cylindrical Brownian motion, and let deterministic $\sigma (t,T)\in H$ satisfy the maturity-integrability and square-integrability conditions needed below. For a deterministic initial curve, specify under $Q$Using an orthonormal basis, the noise is ${\sum }_j{\sigma }_j(t,T)d{W}_t^j$; an infinite-dimensional cylindrical noise need not exist as an $H$-valued random vector. Its covariation kernel across maturities is the positive semidefinite kernel $C_t(T,U)=⟨\sigma (t,T),\sigma (t,U)⟩$. Thus arbitrary suitable maturity-correlation kernels can be represented by Hilbert-space factors, while finitely many factors give finite-rank kernels.

No arbitrage requires a drift restriction. Define the zero-coupon bond price and integrated volatility byDifferentiation of the moving lower endpoint and the Itô formula giveThe discounted bond must have zero drift. Differentiating ${\int }_t^T\alpha (t,u)du=\frac{1}{2}\Vert \Sigma (t,T){\Vert }^2$ with respect to maturity yields the Heath-Jarrow-Morton model restrictionThis is why forward-rate drift and covariance cannot be selected independently in an arbitrage-free random-field model. With deterministic volatilities square-integrable on the finite horizon, the discounted bond stochastic exponential is a true martingale, not only a local martingale.

The solution is explicitly Gaussian:with mean given by the first two terms and covarianceThe entire Gaussian random field can therefore be calibrated through its initial curve and maturity covariance structure. Bond prices are exponential functionals of these Gaussian rates; conditional bond ratios have lognormal laws under appropriate forward measures. Negative rates are possible. A large or infinite volatility space provides rich maturity correlations, but Gaussianity alone supplies neither a low-dimensional Markov state nor positivity of rates.

A continuous adapted process $M$ is a local martingale if there are stopping times ${\tau }_n\uparrow \infty$ almost surely such that each stopped process $M_t^{{\tau }_n}=M_{t\wedge {\tau }_n}$ is a martingale. In this definition $M_0$ is integrable. A continuous adapted process $A$ has finite variation if, almost surely, its paths have finite total variation of a function on each compact time interval:The permitted fact that this total-variation process is continuous and adapted makes its level-hitting times stopping times.

Suppose $M=A$, and put $N=M-M_0$. DefineThese times increase to infinity almost surely. The optional stopping theorem applied to the localized martingale shows that $N^{{\sigma }_n}$ is a martingale; it is also bounded in absolute value by $n$, with total variation at most $n$. Fix $n,t$ and write $L=N^{{\sigma }_n}$. For a deterministic partition of an interval $0=t_0<\cdots <t_r=t$, orthogonality of square-integrable martingale increments givesPathwise the sum on the right is bounded by $n{\max }_j|L_{t_j}-L_{t_{j-1}}|$. For partitions with mesh tending to zero, continuity makes this converge to zero. It is bounded by $2n^2$, so dominated convergence gives $\mathbb{E}{L}_t^2=0$. First use rational $t$, then continuity, to obtain $L\equiv 0$ almost surely. Letting $n\to \infty$ provesThis proves continuous finite-variation local martingale is constant without assuming integrability of the unstopped total variation. Continuity matters: compensated jump martingales can have finite variation without being constant.

The one-dimensional Lévy characterization of Brownian motion states: a continuous adapted process $L$, with $L_0=0$, is standard Brownian motion relative to its filtration if and only if it is a local martingale and $[L{]}_t=t$ for all $t$ almost surely.

For necessity, the centered independent Gaussian increments of Brownian motion give both the martingale property and the martingale $L_t^2-t$. The uniqueness characterization of quadratic variation then gives $[L{]}_t=t$.

For sufficiency fix $\theta \in \mathbb{R}$. The Itô formula givesThe second-order term cancels the time derivative because $d[L{]}_t=dt$. Thus $E$ is a complex local martingale. Its modulus is the deterministic quantity $e^{{\theta }^{2}t/2}$, bounded on every finite horizon, so its real and imaginary parts are true martingales by the bounded local martingale criterion. For $s<t$,Uniqueness of characteristic functions says that the increment has normal distribution $N(0,t-s)$ independently of ${\mathcal{F}}_s$. Indeed multiplying the identity by any bounded ${\mathcal{F}}_s$-measurable variable gives the product characteristic-function identity expressing independence. Iterating over deterministic times gives independent increments. Together with continuity and $L_0=0$, these are precisely the defining properties of standard Brownian motion. If the starting value is a fixed $x$, the same argument identifies $L-x$ as standard Brownian motion.

For each fixed horizon $t$, the stipulated property is the Class D process condition on $[0,t]$, and imposing it for every finite horizon gives a Class DL process.

First suppose $M$ is a true martingale. For every stopping time $T\le t$, the bounded-time optional sampling theorem givesThe conditional expectations of one integrable variable form a uniformly integrable family. To see the uniformity explicitly, put $V=\mathbb{E}(U\mid \mathcal{H})$. For any $r,K>0$, conditional Jensen and Markov inequalities giveChoose $r$ large and then $K$ large. The bound does not depend on the conditioning sigma-algebra. This proves the required uniform integrability. Bounded-time optional sampling here can also be justified directly: approximate $T$ from above by finite-grid stopping times, use discrete optional sampling with terminal value $M_t$, and use this same conditional-expectation uniform integrability and right continuity to pass to the limit.

Conversely, take a localizing sequence ${\tau }_n$ for the local martingale. For fixed $s\le t$, both families $M_{t\wedge {\tau }_n}$ and $M_{s\wedge {\tau }_n}$ are uniformly integrable, because their times are bounded by $t$. They converge almost surely to $M_t$ and $M_s$, and therefore converge in $L^1$. The stopped martingale identitypasses to the limit by $L^1$ contractivity of conditional expectation. Thus $\mathbb{E}(M_t\mid {\mathcal{F}}_s)=M_s$, and all time values are integrable. Hence a local martingale is a true martingale exactly when it is of Class DL. This is a finite-horizon criterion, not a claim of uniform integrability over all times simultaneously.

Denote the density process by $L_t=\exp (M_t-[M{]}_t/2)$. The given identities for all $A\in {\mathcal{F}}_s$ implyThus $L$ is a strictly positive true martingale, not merely a local martingale. The Itô formula gives $dL=LdM$, and henceFirst work with the zero-starting corrected increments $U=X-X_0-[X,M]$. Integration by parts givesThe finite-variation bracket has zero quadratic covariation with $L$, so no other term appears. Thus $LU$ is a P-local martingale.

To justify the change of measure, stop $U$ at ${\tau }_n=\inf \{t:|U_t|\ge n\}$, with optional further localization if needed. The product $L_t{U}_{t\wedge {\tau }_n}$ remains a P-local martingale: before the stopping time the cancellation above applies, and afterwards only its bounded constant factor multiplies $dL$. Its absolute value is bounded by $n{L}_t$. On a finite horizon the stopped values of $L$ are uniformly integrable, so Q1(c) upgrades this product to a true martingale. The Bayes formula for conditional expectation now givesFinite-horizon equivalence of the measures transfers the localizing sequence to Q. Therefore $U$ is a continuous Q-local martingale. If $X_0\in L^1(\mathbb{Q})$, adding its constant-in-time process provesThis proves the Girsanov theorem assertion under the usual zero or deterministic initial-value convention. More generally the normalization $M_0=0$ makes $L_0=1$, so P and Q agree on ${\mathcal{F}}_0$ and the required initial integrability is automatic.

There is a literal initial-integrability qualification if neither convention is imposed. On positive integers let $P(n)=2^{-n}$, $Q(n)=c/n^2$, with $c=({\sum }_{n\ge 1}{n}^{-2}{)}^{-1}$, and let every filtration equal the full sigma-algebra. Constant processes $M_t=\log (c{2}^n/n^2)$ and $X_t=n$ are P-martingales, and $e^{M_t}$ is precisely the prescribed density. But ${\mathbb{E}}_Q|X_0|=\infty$ and $[X,M]=0$, so the asserted Q-local martingale cannot have an integrable initial value. This is why initial integrability under a density change is part of the precise formulation.

If $X$ is a P-Brownian motion, its starting value is deterministic, so Q5(a) applies. The correction $[X,M]$ has finite variation and does not alter quadratic variation; Q1(b) gives $[\tilde{X}{]}_t=t$ under Q. The Lévy characterization of Brownian motion therefore identifies $\tilde{X}$ as a Q-Brownian motion with the same starting value.

For the likelihood, use Wiener measure ${\mathbb{Q}}^x$ as the reference measure and write its coordinate as ${\omega }_t=x+W_t$. Setand let $L=\mathcal{E}(M)$ be its stochastic exponential. We must prove it is a true martingale; unverified large-horizon Novikov estimates would not suffice.

Fix $T$ and stop at ${\tau }_n=\inf \{s:|{\omega }_s|\ge n\}$, with $n>|x|$. The bounded integrand satisfies the Novikov condition, so $L_{T\wedge {\tau }_n}$ defines a probability $P_n$ on ${\mathcal{F}}_T$. By the result just established,is Brownian under $P_n$ on this horizon. Therefore $d{\omega }_s=d{W}_s^{(n)}-\mu 1_{\{s\le {\tau }_n\}}{\omega }_{s}ds$. The Gronwall inequality givesThe Doob L2 maximal inequality then bounds its squared expectation by $C_T=2e^{2|\mu |T}(x^2+4T)$, uniformly in $n$. In particular $P_n({\tau }_n\le T)\le C_T/n^2$. On $\{{\tau }_n>T\}$ the stopped and unstopped densities agree, soContinuous Brownian paths are bounded on $[0,T]$, and the indicators increase to one. The monotone convergence theorem proves $\mathbb{E}{L}_T=1$. Since a nonnegative local martingale is a supermartingale, constant expectation at every time makes $L$ a true martingale.

Under its tilted measure, ${\hat{W}}_t=W_t+\mu {\int }_0^t{\omega }_{s}ds$ is Brownian and $d{\omega }_t=d{\hat{W}}_t-\mu {\omega }_{t}dt$. The explicit linear solution gives uniqueness of this law, identifying it with ${\mathbb{P}}^x$. Consequently the Ornstein-Uhlenbeck likelihood relative to Wiener measure isThis is a finite-horizon density statement; it does not assert absolute continuity of the full infinite-horizon path laws.

Let $P_{s,t}>0$ denote the time-$s$ price of a zero-coupon bond paying one at maturity $t$, with $P_{t,t}=1$. Information available at $s$ is the sigma-algebra ${\mathcal{F}}_s$ in the market filtration. For a curve differentiable in maturity, the instantaneous forward rate and its diagonal short rate areThe last identity follows by integrating the maturity derivative and using $P_{s,s}=1$. The continuous-time bank account reinvesting at the short rate is $B_s=\exp ({\int }_0^s{R}_{u}du)$, and its reciprocal is the discount factor. Modelling forward rates means specifying a random maturity-indexed curve with an evolution consistent with the same pricing measure for every bond. The expectation below is under that money-market risk-neutral measure, not an unspecified physical measure.

For fixed maturity $t$, if $P_{s,t}/B_s$ is a true martingale on $0\le s\le t$, its terminal value is $1/B_t$. ConsequentlyConversely, this representation gives $P_{s,t}/B_s={\mathbb{E}}_Q[B_t^{-1}\mid {\mathcal{F}}_s]$, a conditional-expectation martingale by the tower property. For example, for $r<s$, conditioning the right side on ${\mathcal{F}}_r$ returns ${\mathbb{E}}_Q[B_t^{-1}\mid {\mathcal{F}}_r]$. These claims assume ${\mathbb{E}}_Q{B}_t^{-1}<\infty$. A positive discounted local martingale alone need not equal the conditional expectation of its terminal value, so the word “martingale” cannot be weakened without an extra integrability argument.

In the Vasicek model, the Brownian motion $W$ and coefficients in $d{R}_s=\alpha (\beta -R_s)ds+\sigma d{W}_s$ must be understood under this pricing measure. Multiplying by $e^{\alpha s}$ and integrating gives, conditionally on ${\mathcal{F}}_s$,Assume first $\alpha \ne 0$, and put $\tau =t-s$ and $B(\tau )=(1-e^{-\alpha \tau })/\alpha$. Integrating the explicit solution over $s\le u\le t$ yieldsThe interchange of the stochastic integrals is justified by their deterministic square-integrable kernels on the finite triangular domain; equivalently approximate the kernels by step functions, interchange finite sums, and pass to the limit using the Itô isometry. The last integral conditionally has a centered normal distribution independent of ${\mathcal{F}}_s$, with varianceA deterministic Brownian integral is Gaussian because its step approximations are sums of independent normal increments; their variances converge by the Itô isometry. Completing the square in the normal density proves $\mathbb{E}{e}^{-Y}=e^{-m+v/2}$ for a normal variable of mean $m$ and variance $v$. Hence the zero-coupon bond price is exponential-affine:An equivalent familiar form isAt $t=s$, both coefficients are zero and the bond value is one. These formulas are valid for negative as well as positive nonzero $\alpha$ on finite horizons, though a genuinely mean-reverting Vasicek model has $\alpha >0$. For $\alpha =0$, direct integration gives $B(\tau )=\tau$, $J(\tau )={\tau }^3/3$ andAll the exponential moments used above are finite, so the prices defined by the conditional expectation do give true discounted martingales.