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On the positive half-line, divide by the stochastic exponential $F_t=e^{\alpha B_t-{\alpha }^{2}t/2}$. The Itô product rule reduces the equation to the random Bernoulli differential equation $d(X/F)/dt=F^{\delta -1}(X/F{)}^{\delta }$. For $\delta \ne 1$ this gives $X_t=F_t[x_0^{1-\delta }+(1-\delta ){\int }_0^t{F}_s^{\delta -1}ds{]}^{1/(1-\delta )}$, until the bracket vanishes; for $\delta =1$, $X_t=x_0{F}_t{e}^t$.

For asset dynamics $dS=\mathrm{diag}(S)(\mu dt+\sigma dW)$, let $\lambda ={\sigma }^{-1}\mu$ and $Z=\mathcal{E}(-\int {\lambda }^{T}dW)$. The Itô product rule cancels the drift of each $Z{S}^i$, leaving the displayed stochastic integral. When $\mu$ is bounded and $\sigma {\sigma }^T\ge \epsilon I$, $\lambda$ is bounded and the Novikov condition makes $Z$ a true martingale on every finite horizon. If $\sigma$ is also bounded, each $Z{S}^i/S_0^i$ satisfies the Novikov condition as a stochastic exponential with bounded integrand ${\sigma }_i-\lambda$, so it too is a true martingale.

Suppose $X$ is a continuous local martingale starting at zero, with conditionally symmetric increments, and terminal conditional expectations have continuous martingale versions. ThenTo prove this, set $M_t=\mathbb{E}[e^{i\theta X_T}\mid {\mathcal{F}}_t]$. Symmetry makes $e^{-2i\theta X_t}{M}_t$ a martingale, and the Itô product rule gives $d⟨M,X⟩=i\theta Md⟨X⟩$. Consequently $e^{-i\theta X_t-{\theta }^2⟨X{⟩}_t/2}{M}_t$ is a bounded local martingale, hence a martingale. Evaluating at the endpoints proves the formula. If also $X_T\sim N(0,T)$ for all $T$, the values of the bracket Laplace transform at $1$ and $2$ force $⟨X{⟩}_T=T$ almost surely. Continuity and the Lévy characterization of Brownian motion then identify $X$ as Brownian motion.

A self-financing investor with consumption satisfies $X=H\cdot P$ and $dX=H\cdot dP-cdt$. The Itô product rule and $d[Y,X]=H\cdot d[Y,P]$ give the displayed equation. For a positive local martingale deflator $Y$, the first term is a stochastic integral against the vector of deflated asset-price local martingales, not their current levels.

In the completed natural Brownian filtration, suppose $d{S}_t/S_t={\mu }_{t}dt+{\sigma }_{t}d{W}_t$ with ${\sigma }_t\ne 0$, and let $d{\zeta }_t=-{\zeta }_t(r_{t}dt+{\kappa }_{t}d{W}_t)$, ${\kappa }_t=({\mu }_t-r_t)/{\sigma }_t$. For an affordable nonnegative terminal claim $X$, represent the conditional-expectation martingale $N_t=\mathbb{E}[{\zeta }_{T}X\mid {\mathcal{F}}_t]$ as $d{N}_t=h_{t}d{W}_t$. Then $w_t=N_t/{\zeta }_t$ and the displayed dollar amount in the stock replicate $X$. The Itô product rule gives $d(\zeta w)=\zeta (\sigma \theta -\kappa w)dW$, proving the formula; share holdings are ${\theta }_t/S_t$.

For a bounded domain, a bounded classical solution with boundary data $g$, nonnegative bounded $V$, bounded $f$ and integrable diffusion exit time $\tau$ satisfiesApply the Itô product rule to the discounted stopped solution and add its source integral. The stopped martingale is dominated by $\Vert u{\Vert }_{\infty }+\Vert f{\Vert }_{\infty }\tau$, so dominated convergence permits passage to the exit. For $V=f=0$, this is a boundary-value averaging principle for functions harmonic under $L$.

For $\lambda >0$, set ${\mathcal{G}}_t={\mathcal{F}}_{e^{2\lambda t}}$ and $W_t=(2\lambda {)}^{-1/2}{\int }_1^{e^{2\lambda t}}{u}^{-1/2}d{B}_u$. This is a Brownian motion in the new filtration; its independent Gaussian increments have variance equal to elapsed time. The Itô product rule gives $dY=\sqrt{2\lambda }dW-\lambda Ydt$. Its initial value is $B_1$, independent of future W increments, and its covariance function is $e^{-\lambda |t-s|}$. Consequently it is a stationary Ornstein-Uhlenbeck process. The quadratic variation of the unscaled time change is $e^{2\lambda t}-1$, not $e^{2\lambda t}$.

For bounded $V$ and bounded initial data $f$, a sufficiently regular classical solution bounded on every finite time slab satisfiesTo prove this, the Itô formula and Itô product rule make $u(T-t,B_t)e^{{\int }_0^{t}V(B_s)ds}$ a local martingale. Its deterministic bound on $[0,T]$ makes it a true martingale; taking expectations at the two endpoints gives the formula. This also gives uniqueness in the finite-horizon bounded class. The exponent has the same sign as the potential in the equation.

An Itô process is a continuous semimartingale expressible as the displayed sum, with adapted coefficients locally integrable for the time integral and locally square integrable for the Itô integral. The coefficients may depend on the whole past; an Itô diffusion usually specifies them as functions of the current state and time. A continuously differentiable deterministic process is an Itô process with zero Brownian coefficient. The Itô formula and Itô product rule apply to these processes.

For a left-continuous locally bounded adapted process $H$ and a continuous semimartingale $X$, the elementary predictable left-endpoint approximations tend pointwise to $H$. Decompose $X$ into a continuous local martingale and a continuous finite-variation process. After localization, dominated convergence with respect to the variation measure handles the finite-variation integral. The Itô isometry and Doob L2 maximal inequality handle the martingale integral. The omitted partial last interval is bounded by the local bound on $H$ times the modulus of continuity of $X$. This proves uniform convergence on compacts in probability and is the grid-sum foundation of the Itô product rule.

For continuous local martingales $M$ and $N$, the Itô product rule says thatis a local martingale. If the martingales are square-integrable and converge in $L^2$, then

For a fixed initial value $X_0=x_0$, the Ornstein-Uhlenbeck process has the normal distributionThe two parameters of $\mathcal{N}(m,v)$ are the expected value and variance, not the standard deviation. At $t=0$ the variance is zero and the value is fixed. More generally, its transition kernel over any interval of length $h>0$ has conditional expected value $\mu +(x-\mu )e^{-\theta h}$ and conditional variance ${\sigma }^2(1-e^{-2\theta h})/(2\theta )$ when the current value is $x$. The coefficients depend on elapsed time; the initial probability distribution determines whether the resulting stochastic process is stationary.^[ref]

In the stipulated example $\theta =1$, $\mu =2$, $\sigma =\sqrt{2}$, $x_0=6$, the expected value at time two is $2+4e^{-2}\approx 2.5413$ and the variance is $1-e^{-4}\approx 0.9817$. The initial distance from two has fallen by a factor $e^{-2}$, but the standard deviation has grown to about $0.9908$. Approaching the reference level in expected value is compatible with growing uncertainty from a perfectly known initial value.

The solid probability density functions use times $0.25$, $1$ and $2$; the dashed curve is the limiting normal distribution with expected value two and variance one. The curves are exact model predictions. There is no ordinary density for the fixed value at time zero.

A short derivation uses the integrating factor $e^{\theta t}$. The Itô product rule gives $d[e^{\theta t}(X_t-\mu )]=\sigma e^{\theta t}d{W}_t$, henceThe stochastic integral has zero expected value. The Itô isometry gives its variance as ${\sigma }^2{\int }_0^t{e}^{-2\theta (t-u)}du$. Its deterministic integrand makes it a Gaussian random variable, yielding the displayed normal distribution. Shifting the integration interval gives the same transition kernel from any later time.

For a nonexplosive diffusion process, bounded source $f$, terminal data $g$ and nonnegative bounded potential $V$, a bounded classical solution with $u(T,x)=g(x)$ isThe Itô product rule shows that the discounted solution plus its accumulated discounted source is a local martingale. Its deterministic finite-horizon bound upgrades it to a true martingale, giving the formula and uniqueness without a global derivative bound.

Here is the Brownian change-of-drift form of the Girsanov theorem. Let $W$ be a standard Brownian motion on $[0,T]$ and $\theta$ predictable with ${\int }_0^T{\theta }_s^{2}ds<\infty$ almost surely. Suppose the stochastic exponentialis a true martingale with $\mathbb{E}{Z}_T=1$. Define $dQ=Z_{T}dP$. Then $Q$ is equivalent to $P$ andThe same statement holds in $d$ dimensions with scalar products, squared norms and a vector drift. Novikov's condition $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{\theta }_s^{2}ds)<\infty$ is a standard sufficient condition for the true-martingale hypothesis. Bounded $\theta$ is enough for all applications below.

For the proof, Itô formula gives $d{Z}_t=-Z_t{\theta }_{t}d{W}_t$. The Itô product rule gives the cancellationThe two drift terms cancel because $d[Z,W{]}_t=-Z_t{\theta }_{t}dt$. Stop $W^Q$ when its absolute value reaches $n$, at a time ${\tau }_n$ capped by $T$. The same product calculation makes $Z_t{W}_{t\wedge {\tau }_n}^Q$ a local martingale under $P$. Its absolute value is at most $n{Z}_t$; the density martingale is uniformly integrable over this finite horizon, so the stopped product is a true martingale. Bayes' conditional identity then givesAs $n$ increases, these stopping times exhaust the horizon by continuity. Thus $W^Q$ is a continuous local martingale under $Q$. Adding a finite-variation drift does not change quadratic variation, so $[W^Q{]}_t=t$.

To prove the Brownian conclusion rather than just assert it, for every real $u$ apply Itô formula to $F_t=\exp (iu{W}_t^Q+u^{2}t/2)$. Its differential is $iu{F}_{t}d{W}_t^Q$, and its modulus is bounded on the fixed horizon, so it is a true martingale. ConsequentlyThe conditional characteristic function identifies a centered normal increment with variance $t-s$, independent of the past. Together with continuity and $W_0^Q=0$, this proves the Brownian assertion. For vector $u$, the same argument uses $\Vert u{\Vert }^2$ and gives the multivariate version. If $|\theta |\le K$, stopping $Z$ and applying Itô formula to $Z^2$ gives a uniform second-moment bound $\mathbb{E}{Z}_{t\wedge \tau }^2\le e^{K^{2}t}$ by Gronwall inequality; this proves uniform integrability of the stopped densities and the needed true-martingale property in the bounded case.

In the Black-Scholes model with its usual augmented Brownian filtration, write $d{S}_t=S_t(\alpha dt+\sigma d{W}_t)$, $B_t=e^{\rho t}$, with $\sigma >0$. Take $\theta =(\alpha -\rho )/\sigma$, the market price of risk. Under $Q$, the preceding change of drift givesso $S_t/B_t$ is a martingale. The Martingale representation theorem then makes integrable Brownian-market claims attainable and gives their risk-neutral pricing as discounted conditional expectations. The physical drift $\alpha$ has disappeared from the price dynamics.

For the joint terminal-value and maximum law, first use the Brownian reflection principle at level $a>0$. Reflecting a path after its first hit of $a$ preserves its law as a Brownian motion and sends a terminal value $y<a$ to $2a-y>a$; the symmetry and strong Markov property of the post-hit increments justify this map. Thus, with ${\phi }_t(y)=(2\pi t{)}^{-1/2}{e}^{-y^2/(2t)}$,Take the constant-drift exponential density $e^{\mu W_t-{\mu }^{2}t/2}$. By Girsanov theorem, the coordinate process under this tilted measure has the law of $W_s+\mu s$ under $P$. Consequently, for $t>0$ and $x\le a$,Completing the squares givesIntegrating provesHere $\Phi$ is the standard normal distribution function. The first variable is the terminal Brownian value, distinct from its running maximum.

Set $T=t_0$, $a=\log (b/S_0)/\sigma >0$ and $\mu =(\rho -{\sigma }^2/2)/\sigma$. Under the risk-neutral measure, $\log (S_s/S_0)/\sigma =W_s^Q+\mu s$. Taking $x=a$ in the joint law gives the probability of not hitting the upper level. Complementing it and discounting the unit payment at maturity gives the one-touch option priceThe exponential weight is also $(b/S_0{)}^{2\rho /{\sigma }^2-1}$. The payment occurs at $T$, even if the barrier was hit earlier, so its discount factor is $e^{-\rho T}$, not a hitting-time discount factor.

Under the specified risk-neutral measure, the Itô product rule and $d(B_t^{-1})=-r_t{B}_t^{-1}dt$ giveThus discounted traded zero-coupon bonds have zero drift. Assume the usual integrability and admissibility conditions that make discounted replicating values true martingales, not merely local martingales. If $X$ is an attainable financial payoff with ${\mathbb{E}}_Q|B_T^{-1}X|<\infty$, its self-financing portfolio value $V_t$ has terminal value $V_T=X$ and satisfiesMultiplying by $B_t$ yieldsThis proves the requested conditional pricing formula. In a Brownian filtration, the Martingale representation theorem realizes the discounted conditional expectation as a stochastic integral. If an available bond has nonzero Itô diffusion exposure, its holdings can match that integral, with the remaining value in the bank account, giving replication under the usual square-integrability conditions.

These attainability or market completeness hypotheses matter for a completely arbitrary contingent claim: existence of a risk-neutral measure alone need not imply a unique price for every unspanned financial payoff, and a zero-drift local martingale alone need not satisfy the displayed terminal expectation identity. In the general path-dependent Heath-Jarrow-Morton model, write the value as $V_t$; notation $V(t,r_t)$ is justified only when the current short rate is a sufficient state for the financial payoff and dynamics.

Differentiate the material derivative of the passive scalar component by component. With ${\sigma }_{ij}={\partial }_j{u}_i$,The chain rule along a Lagrangian trajectory $\dot{X}=u(X,t)$ therefore gives the covector evolution of a material scalar gradientThe transpose matters: a gradient is transported as a covector, rather than as a material separation vector. When $u_3=0$, the third row of $\sigma$ vanishes, andLet $k_h=(k_1,k_2{)}^T$ and setFor $c>0$, the matrix exponential gives the exact horizontal solutionThe two projections are onto the eigenspaces with eigenvalues $-c$ and $+c$. If $p_{-}\ne 0$, write $p_{-}=k_0(\cos \varphi ,\sin \varphi {)}^T$ with $k_0=|p_{-}|>0$. ThenThe stated growing asymptotic requires this nonzero projection. An initial gradient entirely in the other eigenspace decays, and when $c=0$ the horizontal gradient is constant.

For constant vertical shear, write $h=({\sigma }_{13},{\sigma }_{23}{)}^T$. Integrating the third component using the exact horizontal solution yieldsConsequently, if $D=h\cdot (\cos \varphi ,\sin \varphi )$ is nonzero, $k_3\sim -k_{0}D{e}^{ct}/c$, andIf $D=0$, exponential growth of $k_3$ is absent, but the ratio still tends to zero on the growing horizontal branch. Thus exponential vertical growth is generic, not unconditional. The scalar-gradient inclination in a steady planar strain is the tangent of the angle between the scalar's normal and the horizontal plane. Equivalently, it is the cotangent of the inclination of an isoconcentration surface to that plane. Large $\alpha$ corresponds to nearly horizontal scalar surfaces; small $\alpha$ corresponds to nearly vertical ones.

For the probability calculation, take $\Gamma ,\Lambda >0$. The independent random variables $a,b$ with normal distributions of variance ${\Gamma }^2$ make $c$ distributed according to the Rayleigh distribution, withThe advertised Gaussian-over-Rayleigh inclination distribution also needs $D$ to be independent of $(a,b)$; its Gaussian marginal alone does not imply this. Under that intended assumption, for $A>0$,The cumulative distribution function is zero for $A\le 0$. Differentiating givesand the probability density function is zero for negative $A$. Its integral is one because the cumulative distribution function tends to one at infinity. Independence is a real hypothesis: $D=(\Lambda /\Gamma )a$ has exactly the stated Gaussian marginal but makes $\alpha \le \Lambda /\Gamma$ almost surely, contradicting the unbounded support of this density. A sufficient physical model is isotropic Gaussian vertical shear, independent of $a,b$, whose projection in any horizontal direction has variance ${\Lambda }^2$.

If the strain varies slowly, the horizontal gradient must align with the current growing direction, and the vertical-to-horizontal ratio must adjust before that strain changes. Both adjustment times are of order $c^{-1}$. Thus the local condition is $c{\tau }_{\sigma }\gg 1$, and the typical condition for the specified statistics isThis is an adiabatic approximation, not a uniform statement about arbitrarily weak instantaneous strains: samples with $c{\tau }_{\sigma }\lesssim 1$ retain memory of earlier strain. The instantaneous eigenvector must likewise turn slowly compared with $c$.

For rapidly fluctuating vertical shear, insert the large-time horizontal solution into its stochastic differential equation:The linear combination $W=\cos \varphi W^{(1)}+\sin \varphi W^{(2)}$ is a Brownian motion: it has independent Gaussian increments and increment variance $({\cos }^2\varphi +{\sin }^2\varphi )dt=dt$. For the signed inclination $\beta =k_3{e}^{-ct}/k_0$, the Itô product rule introduces no extra cross term because the exponential factor is deterministic. It follows thatThis is an Ornstein-Uhlenbeck process, explaining why white-noise vertical shear gives an Ornstein-Uhlenbeck inclination. Its Fokker-Planck equation isIn a stationary state the Fokker-Planck probability current $j=-cBp-(g^2/2)p^{\prime }$ is constant. A nonzero constant current cannot give an integrable, nonnegative density at both ends of the real line, so $j=0$. Integrating $p^{\prime }/p=-2cB/g^2$ and normalizing with the Gaussian integral givesThe stationary variance is $g^2/(2c)$. More explicitly the explicit Ornstein-Uhlenbeck solution has mean $\beta (0)e^{-ct}$ and variance $g^2(1-e^{-2ct})/(2c)$, confirming convergence to this density. Here $c>0$ and $g\ne 0$ are needed for the stated smooth stationary density; at $c=0$ there is no confinement, while $g=0$ gives a point mass at zero.

For the first equation use an integrating factor. The Itô product rule gives $d(e^{\lambda t}{X}_t)=\sigma e^{\lambda t}d{B}_t$, soThe Gaussianity of deterministic Brownian stochastic integrals follows by approximating its deterministic integrand by step functions and using the Itô isometry. HenceThe variance is positive also when $\lambda <0$, since numerator and denominator then both have negative sign. This is the explicit Ornstein-Uhlenbeck solution with initial value one.

For the second equation, take independent standard Brownian motions $U,V$ and define $R_t=\sqrt{(1+U_t{)}^2+V_t^2}$. The associated planar Brownian motion does not hit the origin. To check this rather than silently differentiate at zero, stop its radius upon hitting $\epsilon$ or $L$, with $0<\epsilon <1<L$. The function $\log r$ is harmonic on that annulus, so the optional stopping theorem givesLet $\epsilon \downarrow 0$: hitting zero before $T_L$ has probability zero. Taking integer $L\to \infty$ covers every possible finite hitting time, since continuous paths are bounded on compact intervals. Thus planar Brownian motion avoids a fixed point and $R$ stays positive.

The radial Itô formula is now legitimate and givesThe sum of the two martingale terms has bracket $t$ and starts at zero, so it is a standard Brownian motion $\beta$ by Lévy characterization of Brownian motion. Thus $dR=d\beta +(2R{)}^{-1}dt$ with $R_0=1$. The law-determination assumption permitted in the question identifies the law of $Y$ with that of this dimension-two Bessel process. For independent standard normal random variables $G_1,G_2$,This is the two-dimensional Bessel transition law and answers the requested distribution without needing a special-function density.

For continuous semimartingales $U,V$, the semimartingale integration by parts formula isIndeed each partition increment satisfiesSumming telescopes the left side. The first two sums converge to the stochastic integrals by left-endpoint approximation, and the last sum converges to the quadratic covariation by polarization of the quadratic-increment formula. All limits are in uniform convergence on compacts in probability. Hence the equality holds at every time outside one null set, using continuity. In differential notation it is the Itô product rule $d(UV)=UdV+VdU+d[U,V]$.

For a vector of continuous semimartingales $X=(X^1,\dots ,X^d)$ and $F\in C^{1,2}$, the multidimensional Itô formula isOne route from the product formula is to apply it repeatedly to powers and products of the coordinates, obtaining this expression for polynomials; polarization identifies the mixed second derivatives. Time is another coordinate of finite variation, so it has no quadratic-covariation contribution. Localize to compact ranges, approximate a $C^{1,2}$ function and its indicated derivatives by smooth polynomial approximations, and pass to the limit using stochastic-integral continuity and the variation bounds on the finite-variation terms. Equivalently, second-order Taylor expansion along partitions gives the same limit, with integration by parts accounting for every quadratic term.

Use the Itô product rule with the deterministic factor $e^{-\lambda t}$. Its quadratic covariation with $X$ is zero, givingMultiplication by the integrating factor $e^{\lambda t}$ solves the Ornstein-Uhlenbeck stochastic differential equation:The initial value $B_1$ is standard normal and independent of the future driving Brownian motion. Thus $Y$ is a centered Gaussian process with variance one. For $s\le t$, its independent future noise gives $\mathbb{E}(Y_s{Y}_t)=e^{-\lambda (t-s)}$. Its Gaussian finite-dimensional distributions consequently depend only on time differences, so it is a stationary process. This verifies the exponential Brownian time change to a stationary Ornstein-Uhlenbeck process directly.

Set $F_t=e^{\alpha B_t-{\alpha }^{2}t/2}$, a positive stochastic exponential with $d{F}_t=\alpha F_{t}d{B}_t$. The Itô formula givesFor $Y=X/F$, the Itô product rule and $d[X,F^{-1}{]}_t=-{\alpha }^2{X}_t{F}_t^{-1}dt$ cancel both the Brownian terms and the extra drift. ConsequentlyThis is an ordinary Bernoulli differential equation with continuous random coefficients. For $\delta \ne 1$, separate variables to obtainTherefore the Bernoulli stochastic differential equation has the explicit maximal positive solutionvalid while its bracket remains strictly positive. When $\delta =1$, the reduced equation is $d{Y}_t=Y_{t}dt$, andSubstituting these expressions back through the product rule verifies the original stochastic differential equation, not just the reduced equation.