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An ${\mathbb{R}}^d$-valued process $(X_t{)}_{t\ge 0}$ is a Brownian motion started at $x$ when:

Put $A_t={\int }_0^t{H}_s^{2}ds$. It is continuous and tends to infinity almost surely, so $A_{T_{\sigma }}={\sigma }^2$. Define the right-continuous inverse $T(u)=\inf \{t:A_t>u\}$ andThe time-change theorem for local martingales shows that $W$ is a continuous local martingale in the time-changed filtration, andFor completeness, this proves the required case of the Dambis-Dubins-Schwarz theorem: for every $\theta \in \mathbb{R}$, Itô formula makes $\exp (i\theta W_u+{\theta }^{2}u/2)$ a local martingale; stopping and conditioning show that $W_v-W_u$ has conditional characteristic function $e^{-{\theta }^2(v-u)/2}$. Hence the increments are independent centered normal variables with the Brownian variances, and continuity makes $W$ a Brownian motion. ConsequentlyThus $X_{\sigma }$ is centered Gaussian with variance ${\sigma }^2$.

A real-valued process $(B_t{)}_{t\ge 0}$ is a standard Brownian motion if:

In a Black-Scholes model, the risk-free asset and risky asset satisfywhere $B$ is a Brownian motion, $\sigma >0$, and the coefficients are constant. With $S_0^0=1$,so the risky asset is a geometric Brownian motion.

A process $(X_t{)}_{t\ge 0}$ is Brownian motion in ${\mathbb{R}}^d$ when $X_0=0$, its paths are almost surely continuous, and for $0\le t_0<\cdots <t_k$ the increments $X_{t_j}-X_{t_{j-1}}$ are independent centered Gaussian vectors with covariance $(t_j-t_{j-1})I_d$.

On $\mathbb{H}\setminus (A\cup C)$, the nonnegative harmonic function $h_A+h_C$ dominates the boundary data $\mathrm{Im}z$ on $A\cup C$: at a point of $A$, for example, $h_A(z)=\mathrm{Im}z$ and $h_C(z)\ge 0$. The maximum principle for harmonic functions, or equivalently Brownian motion stopped on the union, givesComparing the coefficients at infinity proves

Set $h_A(z)=\mathrm{Im}(z-g_A(z))$. This is a nonnegative harmonic function on $\mathbb{H}\setminus A$, has boundary value $\mathrm{Im}z$ on the hull boundary, and tends to zero on the real boundary. If $\tau$ is the first exit time of planar Brownian motion from $\mathbb{H}\setminus A$, the optional stopping theorem givesThe hydrodynamic expansion yields $h_A(iy)=\mathrm{hcap}(A)/y+O(y^{-2})$, and therefore the Brownian representation of half-plane capacity

With the notation supplied in the question, ${\tilde{U}}_t={\psi }_t(U_t)$ and $d{U}_t=\sqrt{\kappa }d{B}_t$. The Itô formula and ${\partial }_t{\psi }_t(U_t)=-3{\psi }_t^{\prime \prime }(U_t)$ giveFor $\kappa =6$, the drift vanishes. The resulting continuous local martingale has quadratic variationIf $s=\tilde{a}(t)/2$ is the usual half-plane-capacity time, then $⟨\tilde{U}⟩=6s$. The Dambis-Dubins-Schwarz theorem therefore givesfor a standard Brownian motion $\tilde{B}$. The mapped hulls are consequently an ${\mathrm{SLE}}_6$, which proves locality.

A Brownian motion starts at zero, has continuous paths, and has independent increments with $W_t-W_s\sim N(0,t-s)$. The stated covariance makes every finite collection jointly Gaussian; disjoint increments have zero covariance and hence are independent. Their variances are interval lengths, proving all defining properties.

Setand define $Q$ on ${\mathcal{F}}_T$ byThe Cameron-Martin theorem for a linear drift says thatis a Brownian motion under $Q$. ThereforeThe discounted stock is an exponential Brownian martingale, so $e^{-rt}{S}_t$ is a $Q$-martingale. Thus $Q$ is a Risk-neutral measure for the Black-Scholes model.

For $f_{ϵ}(x)=({ϵ}^2+x^2{)}^{1/2}$, direct differentiation givesSince the quadratic variation of standard Brownian motion is $[B{]}_t=t$, the Itô formula gives

For every finite collection of positive times, $(B_{t_1},\dots ,B_{t_m})$ is a linear transformation of the corresponding values of $W$, so $B$ is a centered Gaussian process. If $0<s\le t$, thenThus $B$ has the covariance of Brownian motion.

Its paths are continuous for $t>0$. As $t\downarrow 0$, put $u=1/t$. The given almost-sure limit givesThe Gaussian-process characterization of Brownian motion now proves the time inversion of Brownian motion:

By continuity of Brownian motion, the event that the last strict crossing of $b$ occurs by time $t$ isApply part (b) with $a$ and $b$ interchanged and its time parameter replaced by $1/t$. This turns the right-hand side intoPart (c), now with drift $b$, barrier $a$, and horizon $1/t$, givesThis is the distribution function of the last passage time above a level for Brownian motion with negative drift. Taking $t\downarrow 0$ givesthe probability that the negatively drifted Brownian path never exceeds $b$, consistent with the infinite-horizon crossing probability for Brownian motion with negative drift.

The one-dimensional Donsker invariance principle says that if $X_1,X_2,\dots$ are IID random variables with mean zero and variance one, then the linearly interpolated processconverges weakly in $C([0,1])$ with the uniform norm to standard Brownian motion.

Planar Brownian motion and the initial law ${\nu }_{0,R}$ are invariant under every rotation about the origin. The hitting time ${\tau }_{0,r}$ is also rotation invariant, so the law of $B_{{\tau }_{0,r}}$ is invariant under every rotation of the circle of radius $r$. Planar Brownian motion hits that circle almost surely by recurrence of planar Brownian motion. The unique rotation-invariant probability measure on the circle is its uniform measure, and therefore

The integrand $\mathrm{sign}({\beta }_s)$ is a bounded previsible process, so $B$ is a continuous local martingale. The quadratic variation of a stochastic integral isbecause the Brownian zero set has zero Lebesgue measure. Since $B_0=0$, the Lévy characterization of Brownian motion shows that $B$ is a standard Brownian motion.

False. The Brownian motion produced by the Dambis-Dubins-Schwarz theorem need not be independent of its clock. Let $W$ be a standard Brownian motion and setIf the Brownian motion $B$ in $M_t=B_{[M{]}_t}$ were independent of the whole quadratic variation process, then conditioning on $[M]$ would give $\mathbb{E}[M_t[M{]}_t]=\mathbb{E}[B_{[M{]}_t}[M{]}_t]=0$. Instead, the fourth-moment formula for a bivariate normal distribution gives $\mathrm{Cov}(W_t^2,W_s^2)=2s^2$ for $s\le t$, and henceThus $B$ and $[M]$ are dependent.

The Dambis-Dubins-Schwarz theorem states that if $M$ is a continuous local martingale with $M_0=0$ and $[M{]}_{\infty }=\infty$, then, forthe process $W_s=M_{{\tau }_s}$ is a standard Brownian motion and $M_t=W_{[M{]}_t}$. If $[M{]}_{\infty }<\infty$, one obtains the same representation after enlarging the probability space and continuing $W$ independently beyond $[M{]}_{\infty }$.

Because independent Brownian motions have zero quadratic covariation, the Itô product rule givesAfter integration, the random variable in the question is $B_t^1{B}_t^2$. Writing $B_t^j=\sqrt{t}{Z}_j$ for independent standard Gaussian random variables $Z_1,Z_2$, its distribution isthe scaled product of two independent standard normal random variables.

Apply the realized absolute covariation theorem to each dyadic partition of an interval. More explicitly, use the continuous increasing clock $C=[M]+[N]$ and the Radon-Nikodym theorem to writeFor each time, let $(U,V)$ have the centered bivariate normal distribution with covariance matrix $\left(\begin{array}{cc}a& c\\ c& b\end{array}\right)$, and defineThis process is continuous and increasing. To prove convergence, localize $M,N$, represent the pair as stochastic integrals against a two-dimensional Brownian motion, and approximate the integrands in $L^2(dC)$ by bounded step previsible processes. For step integrands, the result is the weak law of large numbers applied on each block to independent Gaussian random variables. The Burkholder-Davis-Gundy inequality and the Cauchy-Schwarz inequality make the error uniform on each compact interval in probability. Consequentlyin the sense of uniform convergence on compacts in probability.