Past exam of the stochastic differential equations and Brownian motion in complex analysis course of the University of Oxford / 2009 (C10.1, DMAC 2483)

Let $V={\sum }_{n\ge 1}\Vert X_n{\Vert }_{\infty }$. It is a nonnegative extended random variable, and $\mathbb{E}V<\infty$ implies $\mathbb{P}(V<\infty )=1$. Otherwise its expectation would be infinite. Intersect this event with the countably many probability-one events on which the individual sample paths are continuous.

For every path in that intersection, ${\sum }_n\Vert X_n(\cdot ,\omega ){\Vert }_{\infty }<\infty$. The Weierstrass M-test gives absolute and uniform convergence of ${\sum }_n{X}_n(t,\omega )$ on the entire stated time domain. A uniform limit of continuous functions is continuous. Therefore $Y$ has continuous paths almost surely. The finite expectation is used to obtain pathwise summability before applying this deterministic theorem.

For $\lambda >1$, use $x/\lambda \ge 1$ on the tail:This proves the requested Gaussian tail bound.

The subsequent printed maximum inequality is false as stated. For $n=1$ its integrand never reaches one, and its right-hand side isBut for two independent absolute standard normal variables, using their common distribution function $F(x)=2\Phi (x)-1$ and the standard normal density $\phi$,The middle step is integration by parts with ${\phi }^{\prime }=-x\phi$. Thus no proof of the displayed source inequality can hold for all its stated indices.

The correct tail integral formula for expectation isA valid union bound givesIn particular, for $x>1$ this is at most $\min (1,C_n{e}^{-x^2/2})$, where $C_n=2^{n+1}/\sqrt{2\pi }$. This corrected estimate proves the intended summability in part (c), without silently replacing the source formula.

Use the valid Gaussian maximum estimate from part (b), withSplit the tail integral formula for expectation at $a_n$. Bounding the first part by $a_n$ and applying the Gaussian tail estimate to the second givesThus $\mathbb{E}{M}_n=O(\sqrt{n})$. The finitely many small indices also have finite expectation; the bound above already covers them.

By Tonelli's theorem, nonnegativity permits interchanging sum and expectation:The final series converges because its exponential decay dominates the factor $\sqrt{n}$. This establishes the requested conclusion independently of the false bound printed in (1).

In the Lévy construction of Brownian motion, each level is a continuous piecewise linear sum of tents with disjoint interiors. Its supremum norm is therefore the largest absolute tent height, not the sum of the heights. With the standard construction's height standard deviation $2^{-(n+2)/2}$,Part (c) makes $\mathbb{E}{\sum }_n\Vert X_n{\Vert }_{\infty }$ finite, and part (a) proves almost sure uniform convergence to a continuous path on $[0,1]$. The separately chosen linear endpoint term completes the construction; its Gaussian finite-dimensional distributions and covariance must also be checked to identify the limit as Brownian motion. Independent copies on successive unit intervals extend it to the half-line.

The source calls $2^{-(n+2)/2}$ a variance. Literally using that variance would give height standard deviation $2^{-(n+2)/4}$ instead. The same $O(\sqrt{n})$ maximum estimate still proves uniform convergence, since ${\sum }_n{2}^{-n/4}\sqrt{n}<\infty$, but the resulting covariance is not that of the standard Lévy construction. The standard midpoint Brownian bridge variance on an interval of length $2^{-n}$ is $2^{-n}/4=2^{-(n+2)}$. Thus continuity follows under either summable scaling, while Brownian covariance requires the correct variance.

For a continuous semimartingale $X$ and a $C^{1,2}$ function $f$, Itô's formula isFor Brownian motion $B$, $d[B{]}_t=dt$. With $f(t,x)=e^{-t/2}\cosh x$, the drift cancels:ConsequentlyThe stochastic-integral representation makes this a local martingale; localization handles unbounded $B$. In fact it is the average of the two positive exponential Brownian martingales $e^{B_t-t/2}$ and $e^{-B_t-t/2}$, so it is a true martingale on every finite time interval.

The two-dimensional Itô formula gives the local martingale part of $u(X_t,Y_t)$ as ${\int }_0^t{u}_{x}dX+{\int }_0^t{u}_{y}dY$, with all derivatives evaluated along $(X_s,Y_s)$. Independent Brownian coordinates have brackets $[X{]}_t=[Y{]}_t=t$ and $[X,Y{]}_t=0$. Hence the quadratic variation isSimilarly, the quadratic covariation with $v(X,Y)$ isSufficient pointwise conditions are equal gradient lengths and perpendicular gradients:The Cauchy-Riemann equations $u_x=v_y$, $u_y=-v_x$ imply both. These are the quadratic covariations of an analytic Brownian image; harmonicity is needed to remove the drift, but the two bracket identities themselves follow just from the displayed derivative relations.

The solution is the continuous Doléans-Dade exponential:Here $[X]$ is the continuous quadratic variation starting from zero, denoted by the angle-bracket notation in the source. If $X=M+A$ with $A$ continuous of finite variation, then $[X]=[M]$.

Set $U=X-X_0-\frac{1}{2}[X]$. Its quadratic variation is still $[X]$. Applying Itô's formula to $e^U$ givesThe initial value is $S_0=1$. This verification applies to the full semimartingale, including its finite-variation part.

A uniformly bounded local martingale is a true martingale: stop at a localizing sequence and use dominated convergence in its conditional-expectation identities. Its deterministic bound also gives uniform integrability. The uniformly integrable martingale convergence theorem therefore yields an integrable limit $Z_{\infty }$ with convergence in $L^1$ andSince $Z_{\infty }\ge 0$, the Radon-Nikodym derivative defines a positive measure of total mass one:The limit can vanish on a set of positive $\mathbb{P}$ measure, so global equivalence is not required. At each finite time, however, the density is the strictly positive $Z_t$.

Write the integral with an explicit dummy time:On every compact time interval, strict positivity and continuity of $Z$ make its reciprocal locally bounded; the covariation has locally finite variation. Thus $X$ is a well-defined continuous semimartingale.

Apply the Itô product rule. Subtracting a finite-variation process does not change covariation, so $[Z,X]=[Z,M]$. The finite-variation terms cancel:Thus $ZX$ is a local martingale under $\mathbb{P}$.

For the transfer step, choose stopping times increasing to the horizon that bound $X$, $M$, reciprocal $Z$, and the brackets needed for the stochastic integrals. For each such time $\sigma$, the same product calculation with $X^{\sigma }$ shows that $Z_t{X}_{t\wedge \sigma }$ is a true $\mathbb{P}$ martingale. The part after $\sigma$ is simply the bounded held value $X_{\sigma }$ times the density martingale. By the Bayes formula for conditional expectation, for $s\le t$,The stopping times exhaust the horizon also under $\mathbb{Q}\ll \mathbb{P}$. Hence $X$ is a continuous $\mathbb{Q}$ local martingale, proving this form of the Girsanov theorem without assuming a positive terminal density everywhere.

Conformal invariance of planar Brownian motion means invariance after changing the clock, not preservation of the original speed. If $f:D\to D^{\prime }$ is a conformal bijection and $Z$ is planar Brownian motion stopped on exiting $D$, putThen $f(Z_t)$ has the law of planar Brownian motion started at $f(Z_0)$ and run at time $A_t$, up to the corresponding lifetime. Equivalently, composing with the inverse clock produces ordinary Brownian motion in $D^{\prime }$.

Indeed, the real and imaginary parts of a holomorphic function are harmonic, so Itô's formula gives zero drift. The Cauchy-Riemann equations give equal coordinate brackets $A_t$ and zero cross bracket. The Lévy characterization of multidimensional Brownian motion after time change identifies the image. In particular the harmonic measure of boundary arcs is carried to their conformal images.

Let $a=1/r\in (0,1)$ and $\theta ={\tan }^{-1}((1-a^2)/(2a))\in (0,\pi /2)$. The Möbius transformation sends the semicircle endpoints toAlso $f(-1)=-1$. Therefore the left semicircle maps to the unit-circle arc centered at $-1$ with angular width $2\theta$.

By conformal invariance of harmonic measure, its exit probability from the starting point $a$ equals the probability of this image arc for Brownian motion started at zero. Rotational symmetry makes the latter exit angle uniform. ThusFor scaling, if $Z$ starts at $1/r$, then ${\hat{Z}}_t=r{Z}_{t/r^2}$ is planar Brownian motion started at one. Its unit-disk exit becomes radius-$r$ exit, and multiplication by the positive number $r$ preserves the sign of the real part. This proves exactly the same formula for ${\mathbb{P}}_1(\mathrm{Re}{\hat{Z}}_{T_{D_r}}<0)$.

Continuity and the regularity of zero for a one-dimensional Brownian coordinate make $S$ the first contact with the imaginary axis. The Strong Markov property and symmetry of the subsequent real increments imply that reversing those increments after $S$ preserves the Brownian law. The imaginary coordinate is unchanged, so $\tilde{Z}$ is again planar Brownian motion started at one.

Reflection preserves distance to the origin, hence ${\tilde{T}}_{D_r}=T_{D_r}$ pathwise. On $T_{D_r}<S$, the path exits through the right semicircle and reflection has not yet acted. On $S<T_{D_r}$, reflection reverses the sign of the exit real part. ThusThe two circular points on the imaginary axis have zero exit probability, so they cause no extra term. Subtracting proves the reflection identity for Brownian exit from a half-disc requested in the question.

Both processes have the same exit law. With $p={\pi }^{-1}{\tan }^{-1}((r^2-1)/(2r))$ from part (b),The final trigonometric identity follows from ${\tan }^{-1}((r^2-1)/(2r))=\pi /2-2{\tan }^{-1}(1/r)$ for $r>1$.

A continuous nondecreasing function $\Phi$ with $\Phi (0)=0$ has moderate growth if for some fixed $a>1$ and finite $K$,Iteration and monotonicity then give such an inequality, with a different constant, for every fixed dilation factor. Powers $\Phi (x)=x^p$, $p>0$, are standard examples.

The Burkholder-Davis-Gundy inequalities say that for a real continuous local martingale $M$ with $M_0=0$ and any stopping time $\tau$, there are positive finite constants $c_{\Phi },C_{\Phi }$, depending only on $\Phi$, such thatHere $M_{\tau }^{\ast }={\sup }_{s\le \tau }|M_s|$. Expectations may be infinite, with the inequalities understood by localization. For $\Phi (x)=x^p$, these compare the $p$th moment of the running maximum with the $p/2$th moment of quadratic variation.

Set $a=\delta \lambda$, $t_0={\lambda }^2$, and $h=({\beta }^2-1){\lambda }^2$. On the event $\{{\tau }^{1/2}>\beta \lambda ,X_{\tau }^{\ast }\le a\}$, one has $\tau >t_0$ and the Brownian path remains in $[-a,a]$ throughout the next $h$ time units after $t_0$.

For Brownian motion started at $x\in (-a,a)$, let $H$ be its first exit time from $(-a,a)$. The martingale $B_t^2-t$, stopped at $H\wedge n$, gives ${\mathbb{E}}_x(H\wedge n)\le a^2-x^2$. Monotone convergence first proves finite mean exit time, and dominated convergence for the bounded stopped position then givesThus Markov's inequality yields ${\mathbb{P}}_x(H>h)\le a^2/h$. Remaining in the closed interval for a fixed positive duration has the same probability as avoiding boundary exit: Brownian motion crosses immediately after first hitting an endpoint.

Condition on ${\mathcal{F}}_{t_0}$. The event $\{\tau >t_0\}$ is measurable there because $\tau$ is a stopping time, and future Brownian increments are independent of this sigma-algebra. The preceding bound is uniform in the current point inside the interval. ThereforeThis is a good-lambda bound for Brownian stopping time and maximum.

Put $Y=\sqrt{\tau }$ and $V=X_{\tau }^{\ast }$. Splitting according to whether $V>\delta \lambda$, the good-lambda bound for Brownian stopping time and maximum givesIntegrate against the positive measure $d\Phi (\lambda )$. The tail integral formula for expectation in Stieltjes form yieldsBy moderate growth, choose $K_{\beta }$ with $\Phi (\beta x)\le K_{\beta }\Phi (x)$. ThenFix $\beta >1$ and choose $\delta$ small enough that $K_{\beta }\eta <1/2$. Absorb the last term and apply moderate growth once more to the fixed dilation $1/\delta$. This provesTo avoid subtracting infinities, perform the argument first for $\tau \wedge n$, where Brownian maxima have all finite polynomial moments and moderate growth is at most polynomial, and then use monotone convergence. Applying the Dambis-Dubins-Schwarz theorem transfers this Brownian estimate to continuous local martingales, giving the lower-maximum side of the Burkholder-Davis-Gundy inequalities.

Let $d=|z_0|$, the initial distance from the annulus center, and $T=T_{A_{r,R}}$. The function $z\mapsto \log |z-z_0|$ is harmonic on the annulus, so its value along stopped planar Brownian motion is a local martingale by Itô's formula. It is bounded between $\log r$ and $\log R$ before exit, hence is a true bounded martingale.

By the assumed almost sure finiteness of $T$, continuity, and dominated convergence,where $p$ is the inner-boundary exit probability. Solving gives the harmonic measure of an annulus:The two probabilities sum to one, as the exit must lie on one of the two circles.

Suppose first that $d=|z_0|>ϵ$. Apply part (a) with inner radius $ϵ$ and outer radius $R>d$:Each event implies finite hitting time of the $ϵ$ circle, so that hitting probability is one. If $d=ϵ$, the hitting time is zero. If $d<ϵ$, it is the exit time from a bounded disk. The martingale $|Z_t-z_0{|}^2-2t$, stopped before that exit, gives a finite mean exit-time bound $({ϵ}^2-d^2)/2$. Thus the asserted hitting time is almost surely finite in this case too.

For alternating visits, choose two closed smaller disks $K_i$ strictly inside the given open disks $D_i$, so their distance is positive. Define successive first hitting times of $K_1,K_2,K_1,K_2,\dots$. Translation invariance and the Strong Markov property apply the just-proved hitting result from every successive starting point. A countable intersection makes all these times finite almost surely, with the required disk memberships.

These increasing times cannot accumulate at a finite time. On a compact time interval a continuous Brownian path is uniformly continuous, whereas every consecutive visit must cross the fixed positive distance between $K_1$ and $K_2$. ThereforeThis is the recurrence of planar Brownian motion in neighborhoods; it does not assert that every individual point is hit.

Write $Z_t=X_t+i{Y}_t$ and $u=\mathrm{Re}g$. Since $g$ is entire, $u$ is a harmonic function, so $u_{xx}+u_{yy}=0$. The two-dimensional Itô formula givesThe drift vanishes. Stop on successively larger disks to bound the derivatives and define the stochastic integrals without global growth assumptions. ThusThis is the harmonic functions of Brownian motion principle.

Assume that a nonconstant polynomial $f$ has no zero. Then $g=1/f$ is entire. The leading polynomial term makes $|f(z)|\to \infty$ as $|z|\to \infty$, so $g(z)\to 0$ there. On any remaining compact disk, continuity and absence of zeros bound $1/|f|$. Consequently $g$ is bounded on the entire plane.

Part (c)(i) makes $u(Z_t)=\mathrm{Re}g(Z_t)$ a bounded local martingale, hence a true bounded martingale. The martingale convergence theorem says it converges almost surely as $t\to \infty$.

The harmonic function $u$ cannot be constant. Otherwise the Cauchy-Riemann equations would make both derivatives of $\mathrm{Im}g$ zero, so $g$ and then $f$ would be constant. Choose points $z_1,z_2$ with $u(z_1)<u(z_2)$. Continuity supplies two disjoint disks on which the ranges of $u$ are separated by a positive gap. By part (b), the Brownian path visits these disks alternately at times tending to infinity. The corresponding subsequences of $u(Z_t)$ then remain separated, contradicting its almost sure convergence.

The zero-free assumption is impossible. Hence
Every nonconstant complex polynomial has a complex root.
This proves the fundamental theorem of algebra using recurrence and martingale convergence, without invoking Liouville's theorem as a prior algebraic shortcut.