Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 202 (Stochastic Calculus and Applications)

Work on a filtered probability space with the usual conditions for a filtration. A continuous martingale belongs to the class of L2-bounded continuous martingales whenThis means a uniform bound over the entire time interval, not merely being square-integrable at each individual time. The L2 martingale convergence theorem gives a terminal variable $M_{\infty }$ and almost sure convergence and convergence in the Lebesgue space $L^2$; moreover $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]$ by conditional expectation.

The predictable sigma-algebra $\mathcal{P}$ on $\Omega \times [0,\infty )$ is the smallest sigma-algebra making every left-continuous adapted process measurable. Equivalently, it is generated bywhere the sample coordinate comes first. A previsible process is precisely a process measurable for this predictable sigma-algebra. Values at time zero matter for its definition, although they make no contribution to integration against a continuous quadratic variation starting at zero.

Write $N_2=\Vert M_{\infty }{\Vert }_2$ and $N_{\ast }=\Vert {\sup }_{t\ge 0}|M_t|{\Vert }_2$. The L2 martingale convergence theorem ensures the existence of $M_{\infty }$ and yields $|M_{\infty }|\le {\sup }_t|M_t|$ almost surely. Hence $N_2\le N_{\ast }$.

For a finite horizon $T$, the Doob L2 maximal inequality for the submartingale $|M|$ givesThe inequality does not require $M_0=0$. Send $T$ to infinity: the left side increases to $N_{\ast }^2$ by the monotone convergence theorem, and the right side tends to $4N_2^2$ by convergence in the Lebesgue space $L^2$. Thus the equivalent terminal and maximal norms for L2-bounded continuous martingales satisfyBoth are genuine norms on L2-bounded continuous martingales modulo indistinguishability of stochastic processes: a zero terminal norm gives $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]=0$, first at rational times and then at all times by continuity. This proves that the two norms are equivalent norms.

A simple predictable process has the form $H_s={\sum }_{j=0}^{m-1}{h}_j{1}_{(t_j,t_{j+1}]}(s)$, where $0=t_0<\cdots <t_m<\infty$ are deterministic and $h_j$ is bounded and ${\mathcal{F}}_{t_j}$-measurable. A possible ${\mathcal{F}}_0$-measurable value at zero is irrelevant to this Itô integral. DefineEach coefficient is known before the corresponding increment. The conditional expectation of each subsequent increment is zero, so the result is a continuous martingale starting at zero. It is an L2-bounded continuous martingale because it is a finite sum of bounded coefficients times stopped increments of an L2-bounded continuous martingale. This definition is independent of the chosen subdivision: splitting an interval simply splits its increment into a telescoping sum. General admissible predictable processes are then integrated by completion using the Itô isometry; unbounded step coefficients are allowed when the weighted $L^2$ condition in the quadratic-variation measure in the next part holds.

Fix a L2-bounded continuous martingale $M$. Its quadratic-variation measure on the predictable sigma-algebra isIt is finite, since ${\nu }_M(\Omega \times [0,\infty ))=\mathbb{E}[M{]}_{\infty }=\mathbb{E}{M}_{\infty }^2-\mathbb{E}{M}_0^2$. The source Hilbert space is $L^2(M)=L^2(\Omega \times [0,\infty ),\mathcal{P},{\nu }_M)$: predictable processes with finite $\mathbb{E}\int H^{2}d[M]$, identified when equal ${\nu }_M$-almost everywhere. Its norm is $\Vert H{\Vert }_{L^2(M)}^2=\mathbb{E}\int H^{2}d[M]$.

The target ${\mathcal{M}}_0^2$ consists of L2-bounded continuous martingales starting at zero, identified up to indistinguishability of stochastic processes, with norm $\Vert K{\Vert }_{{\mathcal{M}}_0^2}=\Vert K_{\infty }{\Vert }_2$. It is a Hilbert space: terminal values belong to the closed linear subspace of $L^2({\mathcal{F}}_{\infty })$ whose conditional expectation given ${\mathcal{F}}_0$ is zero and whose associated martingales have continuous versions. Closure of that continuous-version subspace follows from the Doob L2 maximal inequality and an almost surely uniformly convergent subsequence.

The Itô isometry says that integration extends uniquely from simple predictable processes to a linear isometry $L^2(M)⟶{\mathcal{M}}_0^2$, withFor every finite $t$, the same identity holds with the integral restricted to $[0,t]$ and the left side $\mathbb{E}|(H\cdot M{)}_t{|}^2$. No claim of surjectivity onto all of ${\mathcal{M}}_0^2$ is needed: that would require an additional Martingale representation theorem.

Put $S=H\cdot M$ and $A_t={\int }_0^t{H}_s^{2}d[M{]}_s$. The candidate $A$ is continuous, adapted, nondecreasing, starts at zero and has integrable terminal value. We prove $S^2-A$ is a martingale, rather than assuming the desired quadratic variation of a stochastic integral.

The assumed boundedness of $M$ also makes $M^2-[M]$ a true martingale on every finite horizon. Indeed, apply its local martingale identity at a localizing sequence; bounded stopped squares and the Fatou lemma show that $[M{]}_t$ is integrable. The stopped expressions are then dominated by ${\sup }_s|M_s{|}^2+[M{]}_t$, so the dominated convergence theorem gives the unrestricted conditional identity on that horizon.

First let $H$ be a bounded simple predictable process. On each interval with coefficient $h$, an increment of $S$ is $h(M_t-M_a)$. The defining quadratic variation identity for $M$, together with the martingale property of $M$, makesa martingale for $t\ge a$. Expanding $S_t^2-S_a^2$ gives a term $2S_{a}h(M_t-M_a)$ and the square term above; their conditional expectations yield the desired identity on that interval. Integrability follows because the variables are square-integrable; concatenating the finitely many intervals gives $S^2-A$ a martingale.

For general bounded predictable $H$, choose bounded simple predictable processes $H^n\to H$ in $L^2(M)$, which is possible since they generate the predictable sigma-algebra. Set $S^n=H^n\cdot M$ and $A_t^n={\int }_0^t(H_s^n{)}^{2}d[M{]}_s$. The Itô isometry and Doob L2 maximal inequality imply $S_t^n\to S_t$ in $L^2$; the Cauchy-Schwarz inequality gives $(S_t^n{)}^2\to S_t^2$ in $L^1$ andPass the conditional expectation identity for $(S^n{)}^2-A^n$ to the $L^1$ limit. Thus $S^2-A$ is a martingale. Uniqueness in the defining quadratic variation property now gives

Use polarization identity of the quadratic variations:This is continuous and adapted, starts at zero, and has finite variation on every compact interval because it is a difference of two nondecreasing processes. The defining quadratic variation property gives that $(M+N{)}^2-[M+N]$ and $(M-N{)}^2-[M-N]$ are continuous local martingales; their difference divided by four is $MN-[M,N]$. This proves existence without assuming a product formula involving an as yet undefined quadratic covariation.

If $C$ and $D$ both satisfy the requirements, $C-D$ is a continuous finite-variation process and a local martingale. By continuous finite-variation local martingale is constant, it is constant; its initial value is zero, so $C=D$ up to indistinguishability of stochastic processes.

There is a minor initial-value convention in this characterization. The displayed uncentred local martingale identity is directly valid when $M_0=N_0=0$, or when the initial terms have the requisite integrability. With arbitrary finite initial values, centre the quadratic variation construction and use $MN-M_0{N}_0-[M,N]$, which starts at zero. Adding back $M_0{N}_0$ requires it to be integrable under the usual definition of a local martingale; the quadratic covariation itself is unaffected by centring.

For probability measures on $(\Omega ,\mathcal{F})$, absolute continuity of measures means preservation of null sets in one direction:Equivalently, by the Radon-Nikodym theorem, there is an $\mathcal{F}$-measurable Radon-Nikodym derivative $Z\ge 0$ with ${\mathbb{E}}_{\mathbb{P}}Z=1$ and $\tilde{\mathbb{P}}(A)={\mathbb{E}}_{\mathbb{P}}[Z{1}_A]$. The derivative may vanish, so absolute continuity of measures does not require equivalence of the two probability measures. In particular, the reverse null-set implication is not part of the definition.

The relevant fact is uniform convergence on compacts in probability under an absolutely continuous measure change. Write $Z=d\tilde{\mathbb{P}}/d\mathbb{P}$. For any event $E$ and $K>0$,The last term tends to zero as $K\to \infty$ because $Z$ is integrable. Consequently, $\mathbb{P}(E_n)\to 0$ implies $\tilde{\mathbb{P}}(E_n)\to 0$. Apply this to $E_n=\{{\sup }_{t\le T}|C_t^n-[M,N{]}_t|>\epsilon \}$ for each finite $T$ and $\epsilon >0$. The given dyadic sums therefore have the same uniform convergence on compacts in probability limit under the new probability measure.

This proves quadratic covariation under an absolutely continuous measure change:Taking $N=M$ proves the corresponding quadratic variation assertion directly; it has not been assumed. The ceiling in the printed sum includes one grid interval extending beyond $t$. This does not affect the limit: on $[0,T]$ the resulting extra product is bounded by the product of the two path oscillations on mesh $2^{-n}$, which tends to zero by uniform continuity on $[0,T+1]$.

The bracket under the new probability measure must be understood as the quadratic covariation of continuous semimartingales. Semimartingale stability under an absolutely continuous measure change ensures this class is preserved; the local martingale property need not be. For example, on a finite horizon, weighting Brownian motion by $\exp (\mu B_T-{\mu }^{2}T/2)$ makes $B_t-\mu t$ a Brownian motion by the Girsanov theorem, so $B$ has nonzero drift under the new measure even though its quadratic variation is still $t$. The PDF additionally assumes that $M,N$ remain local martingales under the new probability measure, so its bracket is also defined directly by the local martingale characterization.

For continuous semimartingales, the Stratonovich integral adds half the quadratic covariation to the Itô integral:One can see the factor directly from symmetric endpoint sums. On a partition of an interval, replacing $X$ at the left endpoint by the average of its two endpoint values adds $\frac{1}{2}\sum \Delta X\Delta Y$; its limit is $\frac{1}{2}[X,Y]$. Thus the correction depends only on the continuous local martingale parts of the semimartingales. If either integrator or integrand has finite variation, the quadratic covariation vanishes and these two integrals coincide. Here the printed notation $\partial Y$ denotes Stratonovich integral integration, equivalently $\circ dY$.

By the Itô formula,To compute the Stratonovich integral correction, apply the Itô formula also to $f^{\prime }$, which is twice continuously differentiable because $f$ is $C^3$. The continuous local martingale part of $f^{\prime }(X)$ is $\int f^{\prime \prime }(X_s)d{L}_s$, where $L$ is the continuous local martingale part of $X$. A finite-variation process has zero quadratic covariation with $X$, and the quadratic variation of a stochastic integral together with polarization identity givesAll these statements can be localized to compact ranges of $X$, so unbounded derivatives create no global integrability requirement. Substitute this identity into the Stratonovich integral to obtain the Stratonovich chain rule:This is the integrated meaning of the requested differential identity; there is no extra second-order term after the Stratonovich integral correction has been included.

Specify an initial value or initial probability distribution. A strong solution of a stochastic differential equation uses a prescribed Brownian motion on a prescribed filtered probability space. It is continuous, adapted to the completed filtration generated by that Brownian motion and the initial value, and satisfiesalmost surely for every $t$, with ${\int }_0^t(|b(X_s)|+\sigma (X_s{)}^2)ds<\infty$ almost surely. The given initial variable is independent of future Brownian motion increments. Some courses allow a larger prescribed filtration in their definition of a strong solution of a stochastic differential equation; the Brownian-generated convention states explicitly what absence of extra randomness means.

A weak solution of a stochastic differential equation consists of a filtered probability space, a Brownian motion relative to its filtration, and a continuous adapted process $X$ satisfying the same integrability and integral equation. Here the space and noise are part of the unknown; the filtration may contain randomness beyond the initial variable and the driving path.

Pathwise uniqueness means that on any common filtered probability space, any two solutions with the same driving Brownian motion and the same initial value are indistinguishable. Uniqueness in law means that all weak solutions with the same initial probability distribution have the same law as path-valued random variables, even on different spaces. Pathwise uniqueness compares shared-noise paths; uniqueness in law compares path distributions. Neither uniqueness notion by itself asserts existence.

The Tanaka equation is an example:It has weak solutions and uniqueness in law, but lacks pathwise uniqueness. The value $s(0)=1$ is deliberate: using the sign function with value zero at zero would admit the identically zero solution and would change the example.

For a quick verification, take a Brownian motion $W$ and put $B=\int s(W)dW$. Its quadratic variation is $t$, so it is a Brownian motion by the Lévy characterization of Brownian motion, and associativity of stochastic integration gives $W=\int s(W)dB$. Every weak solution likewise has quadratic variation $t$ and therefore the law of Brownian motion. On this same enlarged filtration, both $W$ and $-W$ solve the equation driven by $B$: $s(-W)=-s(W)$ away from zero, and the Brownian zero set has zero time measure, so the discrepancy at zero contributes nothing to the Itô integral. These solutions differ with positive probability at any positive time. This proves the claimed distinction without relying on a choice of zero convention left unstated.

The Exponential martingale for Brownian motion is a true martingale: independent increments and the normal distribution identity $\mathbb{E}{e}^{aN}=e^{a^{2}v/2}$ for $N\sim N(0,v)$ give $\mathbb{E}[Z_t\mid {\mathcal{F}}_s]=Z_s$. Its second moment isThe uniform integrability from bounded second moments criterion now applies: for $K>0$,Thus $\{Z_t:0\le t\le T\}$ has uniform integrability. In fact, all values stopped at stopping times bounded by $T$ have uniform integrability, since $Z_{\tau }=\mathbb{E}[Z_T\mid {\mathcal{F}}_{\tau }]$ and uniform integrability of conditional expectations applies. This is a finite-horizon assertion, not an assertion of uniform integrability over all $t\ge 0$.

Define a probability measure on the same full measurable space byThe Radon-Nikodym derivative is positive and has expectation one, so this is a probability measure and even equivalent to $\mathbb{P}$, in particular ${\mathbb{P}}^T\ll \mathbb{P}$. For $t\le T$ its density on ${\mathcal{F}}_t$ is $\mathbb{E}[Z_T\mid {\mathcal{F}}_t]=Z_t$.

The finite-horizon Girsanov theorem says that if $W$ is a Brownian motion and $\mathcal{E}(\int hdW)$ is a true martingale through $T$ defining the density, then $W_t-{\int }_0^t{h}_{s}ds$ is a Brownian motion under that density for $t\le T$. Here $h_s=\mu$, and the stochastic exponential is exactly $Z$. The previous part verifies the necessary martingale property; alternatively the Novikov condition is $e^{{\mu }^{2}T/2}<\infty$. Hence $B_t-\mu t$ is a standard Brownian motion on $[0,T]$ relative to the same filtration under ${\mathbb{P}}^T$.

No absolutely continuous probability measure can remove this nonzero drift on the whole half-line. By the strong law for Brownian motion, the eventhas $\mathbb{P}$-probability one. If ${\mathbb{P}}^{\infty }\ll \mathbb{P}$, absolute continuity of measures gives ${\mathbb{P}}^{\infty }(A)=1$. But if $W_t=B_t-\mu t$ is a Brownian motion under ${\mathbb{P}}^{\infty }$, the same strong law for Brownian motion gives $B_t/t=W_t/t+\mu \to \mu$ with ${\mathbb{P}}^{\infty }$-probability one. Since $\mu >0$, these two limiting events are disjoint, a contradiction.

This is infinite-horizon singularity of Brownian motion with constant drift. The finite-horizon densities are consistent on ${\mathcal{F}}_t$ but cannot define a globally absolutely continuous measure with the requested property. Indeed,because $t^{-1}\log Z_t\to -{\mu }^2/2$. Hence $Z$ lacks uniform integrability on $[0,\infty )$. A drifted law does exist on canonical path space, but it is mutually singular with Wiener measure on the infinite-horizon path sigma-algebra.

The printed assumptions omit the orthogonality needed for the claimed local martingales. Merely placing two Brownian motions on the same probability space does not make them independent. Interpret both as Brownian motions relative to a common filtration, and put $C_t=[B,\vartheta {]}_t$. For $X=e^B\cos \vartheta$ and $Y=e^B\sin \vartheta$, the two-variable Itô formula givesThe pure second derivatives cancel because $[B{]}_t=[\vartheta {]}_t=t$, while the mixed partial derivatives are $X_{b\vartheta }=-Y$ and $Y_{b\vartheta }=X$. Thus the mixed quadratic covariation term cannot be omitted.

For a counterexample allowed by the PDF, take $\vartheta =B$. Then $C_t=t$ and $X$ has drift $-Ydt$, while $Y$ has drift $Xdt$. In particular $X_0=1$ makes the latter drift nonzero near zero, so $Y$ is not a local martingale. Nor is $X$: $Y$ cannot be identically zero, so its accumulated continuous drift is not identically zero. Uniqueness of the continuous semimartingale decomposition excludes cancellation by another local martingale term.

The corrected hypothesis is $[B,\vartheta ]\equiv 0$, for example independent Brownian motions in their joint filtration. Conversely, if both were local martingales, uniqueness of the continuous semimartingale decomposition would force $YdC=XdC=0$; since $X^2+Y^2=e^{2B}>0$, this forces $dC=0$. Thus the correction is exactly the condition needed for both local martingales. Under this hypothesis the drift terms vanish and the intended stochastic differential equations are $dX=XdB-Yd\vartheta$ and $dY=YdB+Xd\vartheta$. The continuous integrands are locally bounded, so localization makes these Itô integrals continuous local martingales. Subsequent intended conclusions are proved under this explicitly stated correction.

With the printed assumptions alone, the general quadratic covariation calculation from the two noise coefficients givesA continuous finite-variation process contributes no quadratic covariation, so the drift terms in the previous part do not alter these formulas. Taking $\vartheta =B$ gives $d[X,Y]=e^{2B}\cos (2B)dt$, which is nonzero near zero since $B_0=0$. Also $d([X]-[Y])=-4XYdt$ is not identically zero. Thus the requested assertions are false as printed.

Under the corrected assumption $C=0$, $X^2+Y^2=e^{2B}$ yieldsThe pair is therefore a pair of orthogonal continuous local martingales. This equality also identifies the clock for the intended time change of a continuous process.

Under the corrected orthogonality hypothesis, the preceding quadratic variation calculation gives $[X{]}_t=A_t={\int }_0^t{e}^{2B_s}ds$. We justify the needed infinite occupation time of one-dimensional Brownian motion, rather than inferring an infinite integral from recurrent visits alone.

Let $T_0=0$, and recursively let $S_n$ be the first exit of $B$ from $(-1,1)$ after $T_n$, and $T_{n+1}$ its first return to zero after $S_n$. Recurrence of one-dimensional Brownian motion makes these stopping times finite almost surely. The Strong Markov property shows that $D_n=S_n-T_n$ are independent and identically distributed random variables, each strictly positive. Choose $\epsilon >0$ with $\mathbb{P}(D_n\ge \epsilon )>0$. The Borel-Cantelli lemma gives infinitely many such durations and hence ${\sum }_n{D}_n=\infty$ almost surely. These intervals are disjoint, and $|B_s|\le 1$ on each, soThis proves the intended conclusion. Without the orthogonality correction, the printed setup gives instead the more general quadratic variation in the previous part; it does not justify identifying that quantity with $A$. The proof here is explicitly for the corrected setup.

Under the corrected hypothesis, $N_t=X_t-1$ is a continuous local martingale with $N_0=0$, $[N{]}_t=A_t={\int }_0^t{e}^{2B_s}ds$, and $A_{\infty }=\infty$ almost surely. The Dambis-Dubins-Schwarz theorem states that, for such a process,defines a standard Brownian motion relative to the filtration ${\mathcal{F}}_{{\tau }_u}$, and $N_t=W_{[N{]}_t}$. The divergence of the quadratic variation ensures every ${\tau }_u$ is finite, so no extension of the probability space is needed. Here $A$ is continuous and strictly increasing because its derivative is $e^{2B_t}>0$, so the inverse is particularly straightforward. Consequently,The constructed Brownian motion need not be independent of its random clock. Under the literal PDF assumptions $X$ need not be a local martingale, as the counterexample in part (a) shows, so this Dambis-Dubins-Schwarz theorem conclusion cannot be asserted without the correction.

A locally defined stochastic process is a pair $(X,T)$, where $T$ is a lifetime stopping time and $X_t(\omega )$ is defined for $0\le t<T(\omega )$, that is, on a stochastic interval. For continuous local problems one uses stopping times $T_n\uparrow T$ with $T_n<T$ on $\{T>0\}$, so the process stopped at each $T_n$ has an ordinary continuous adapted process version. This is an announcing sequence for a stopping time; no value at $T$ or after $T$ is implicit in the pair.

A local solution of a stochastic differential equation on an open domain $U$ takes its values in $U$ before $T$ and, on every such stopped interval, satisfieswith the local integrability needed for the ordinary and Itô integrals. A maximal local solution of a stochastic differential equation has no extension to a strictly later lifetime that agrees with it before $T$. For locally Lipschitz continuous coefficients on $U$, take a nested compact exhaustion $K_n$ of $U$ and the successive exit times: their increasing limit is the maximal lifetime, and on $\{T<\infty \}$ the solution eventually leaves every compact subset of $U$. To obtain the announcing sequence with finite stops even when the path never exits a compact set, cap these exit times by $n$; the capped times still increase to the maximal lifetime. Thus a finite maximal lifetime means exit from the domain or explosion; it need not mean divergence to infinity. For $U=(0,\infty )$, reaching the boundary zero terminates the local solution even if an absorbing extension could be defined for a different domain.

Before the lifetime, $X>0$, so apply the Itô formula to $g(x)=\sqrt{x}$ after stopping inside compact subintervals of $(0,\infty )$. The derivatives are $g^{\prime }(x)=1/(2\sqrt{x})$ and $g^{\prime \prime }(x)=-1/(4x^{3/2})$. Since the quadratic variation of $X$ satisfies $d[X{]}_t=X_{t}dt$, we getThe negative drift is the second-order correction in the Itô formula. This is a Lamperti transform up to a constant scale: the square root transformation makes the noise coefficient constant. Its drift is singular at zero, which is why this stochastic differential equation is stated only before the boundary lifetime.

Integrating the stopped stochastic differential equation for $Y$ and then taking a common full-probability event for a localizing sequence gives, simultaneously for every $t<T$,The integral is finite on each compact time interval before $T$: the continuous positive path has a positive minimum there. Its integrand is nonnegative, soThis is a pathwise comparison using the same Brownian motion, not a comparison of independent distributions. Continuity lets the identities established initially at rational stopped times hold on the entire stochastic interval.

Let $\tau =\inf \{t\ge 0:B_t=-2\sqrt{x_0}\}$, the first time $W$ reaches zero. Recurrence of one-dimensional Brownian motion implies $\tau <\infty$ almost surely. On the common event where the comparison from part (c) holds, $\tau <T$ would imply $0<Y_{\tau }\le W_{\tau }=0$, a contradiction. ThereforeThe argument only uses the comparison at a time strictly before $T$; it does not assume that the singular drift integral is finite at $T$. The given lifetime characterization identifies this finite endpoint with approach to zero. Any extension that absorbs at zero would be a different, globally defined stochastic process, not the positive-domain maximal local solution of a stochastic differential equation specified here.

For $0<\alpha <1$, use the power transformation $Y=X^{1-\alpha }$ on the positive stochastic interval. The Itô formula, with $d[X]=X^{2\alpha }dt$, givesThus $Y_t\le x_0^{1-\alpha }+(1-\alpha )B_t$ simultaneously for $t<T$. The right side reaches zero almost surely by recurrence of one-dimensional Brownian motion, and the same positive-path contradiction proves $T$ cannot exceed that Brownian first-passage time. This establishes the finite lifetime threshold for a power diffusion throughout $0<\alpha <1$.

At $\alpha =1$, the solution is the geometric Brownian motionIt is finite and positive at every finite time. On every compact time interval it has a positive minimum, so the hitting times of $1/n$ tend to infinity. Although the strong law for Brownian motion implies $X_t\to 0$ as $t\to \infty$, this is not a finite lifetime. ThereforeThe power transformation is a rescaled Lamperti transform; at $\alpha =1$ the corresponding transformation is the logarithm. No claim about pathwise uniqueness after adjoining the boundary zero is needed: the coefficients are locally Lipschitz continuous inside the positive domain, which is the domain of the given maximal local solution of a stochastic differential equation.

Set $V(x)=(1+x^2{)}^{-1}$ and $A_t={\int }_0^{t}V(B_s)ds$. On a fixed horizon $T$, the Itô formula and Itô product rule giveThe drift vanishes by the partial differential equation. This makes $M$ a continuous local martingale. Since $0\le A_t\le T$,The right side is a finite deterministic bound under the stated boundedness assumption, so the dominated convergence theorem removes a localizing sequence and $M$ is a true martingale. If the derivatives are also bounded, the Itô isometry alternatively proves the stochastic integral is square-integrable because $\mathbb{E}{\int }_0^T{e}^{2A_s}|u_x(T-s,B_s){|}^{2}ds\le T{e}^{2T}\Vert u_x{\Vert }_{\infty }^2$.

For the bounded parabolic differentiability class, distinguish global bounds from bounds on each finite time slab. Only the latter are needed here; a single global bound on the classical solution $u$ would conflict with its requested long-time growth in part (c).

For the true martingale of part (a), $M_0=u(T,x)$ and $M_T=f(B_T)e^{A_T}$. Taking expectations gives the Feynman-Kac formula with a bounded potential:The right side is finite since its absolute value is at most $e^t\Vert f{\Vert }_{\infty }$. This also proves uniqueness among classical solutions bounded on every finite time slab: applying the formula to their difference with zero initial data gives zero. The positive sign of the potential in the partial differential equation explains the positive exponent; changing it to a negative exponent would solve a different equation.

The infinite occupation time of one-dimensional Brownian motion impliesFor a start $x\ne 0$, first wait until $B$ hits zero, which happens in finite time by recurrence of one-dimensional Brownian motion, and apply the Strong Markov property. The repeated exit-and-return argument in Question 4(c) then proves the infinite occupation integral. Since $f\ge C>0$, the Feynman-Kac formula gives $u(t,x)\ge C{\mathbb{E}}_x{e}^{A_t}$. The random variables $e^{A_t}$ increase to infinity almost surely; the monotone convergence theorem therefore yields the intended positive-potential growth from Brownian recurrence:Under global boundedness, no such solution exists when the initial function is bounded below by a positive constant. The PDF prints $C_b^{1,2}({\mathbb{R}}_{+}\times \mathbb{R})$; if the subscript means bounded over the whole product, its hypotheses with $f\ge C>0$ are incompatible, rather than an example of growth within that class. The precise correction is to require $u$ and its indicated derivatives to be bounded on $[0,T]\times \mathbb{R}$ for every finite $T$. The derivation above then gives the requested limit. Already $f\equiv 1$ satisfies the printed initial-data assumption but cannot have a globally bounded classical solution, by this formula. Also the PDF's $u(x,t)$ in this clause reverses the argument order from $u(t,x)$ used in the equation; the limit above keeps the equation's time-first convention.