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A uniformly bounded local martingale is a true uniformly integrable martingale: a localizing sequence gives uniformly dominated stopped values, so their conditional identities pass to the limit. A deterministic bound on each compact time interval gives a true martingale on each such interval, even when the bound grows with time.

A continuous local martingale $M$ with $M_0=0$ and deterministically bounded quadratic variation on every finite horizon is a square-integrable martingale, with $\mathbb{E}{M}_t^2=\mathbb{E}[M{]}_t$. Localization, the Doob L2 maximal inequality, and the resulting integrable bound on the squared supremum justify removing the localizing sequence.

An integrable discrete-time local martingale is a true martingale. Its stopped increments have the form $1_{\{\tau \ge t\}}(X_t-X_{t-1})$, and the dominated convergence theorem removes a localizing sequence from the conditional increment identity.

Subtract $M_0$, which does not affect either the integral or its quadratic variation. Let $N=M-M_0$. Choose a localizing sequence ${\tau }_n$ for $N$. By local boundedness of the previsible process $H$, there are increasing stopping times ${\rho }_n\uparrow \infty$ and finite deterministic constants $K_n$ with $|H_s|\le K_n$ on $0<s\le {\rho }_n$. PutContinuity and finiteness on compact intervals imply ${\sigma }_n\uparrow \infty$ almost surely. The stopped process $N^{{\sigma }_n}$ is a bounded square-integrable martingale, its bracket is bounded by $n$, and its stopped integrand is bounded by $K_n$. The indicator $1_{\{s\le {\sigma }_n\}}$ is predictable, so the stopped stochastic integral identity is valid.

Fix $n$ and abbreviateThe Itô isometry gives $\mathbb{E}{J}_t^2=\mathbb{E}{Q}_t\le K_n^{2}n$, so $J$ is a square-integrable martingale. For $F\in {\mathcal{F}}_s$, apply the same isometry to the predictable integrand $1_F{1}_{(s,t]}H$, stopped at ${\sigma }_n$. This yieldsAlso $\mathbb{E}[J_t-J_s\mid {\mathcal{F}}_s]=0$. Expanding $J_t^2-J_s^2$ therefore proves that $J^2-Q$ is a true martingale. The process $Q$ is continuous adapted increasing and starts at zero. By the characterization of continuous martingale quadratic variation as its increasing square compensator, and uniqueness of an increasing square compensator, we have $[J]=Q$.

Stochastic integration and quadratic variation commute with stopping; for the latter this follows either from squared-increment sums or from the same square-compensator characterization. Hence the desired identity holds up to every ${\sigma }_n$. Since these times increase to infinity, they exhaust every finite time interval almost surely, and the continuous identities patch to giveThis is the localized isometry proof of stochastic-integral quadratic variation. No unstopped second-moment or bracket-integrability assumption has been made.

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.

For the integrable terminal bracket criterion, assume $M_0=0$, or more generally $M_0\in L^2$. This initial-value assumption is necessary if it is not already part of the notation: an integrable but non-square-integrable random constant process has zero quadratic variation and still fails square-integrability.

For the zero-starting case, choose a localizing sequence ${\tau }_n$ bounding both $|M|$ and its quadratic variation. The stopped processes are true square-integrable martingales, and the Itô isometry givesFor each fixed time these variables are uniformly integrable by their bounded second moments. Almost sure convergence as $n\to \infty$ is therefore convergence in $L^1$. Passing to the limit in the stopped conditional-expectation identity gives $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$: the local martingale is a true martingale. The Fatou lemma also givesso $M$ belongs to the global ${\mathcal{M}}_c^2$ space. For general square-integrable $M_0$, apply this argument to $M-M_0$ and use $|M_t{|}^2\le 2|M_0{|}^2+2|M_t-M_0{|}^2$.

Relative to a filtration satisfying the usual conditions, a local martingale is an adapted càdlàg process $X$ for which there exist increasing stopping times $S_n\uparrow \infty$ almost surely such that every stopped process $X_t^{S_n}=X_{t\wedge S_n}$ is an integrable martingale. Thus, for $s\le t$,The sequence is a localizing sequence. A continuous local martingale additionally has continuous sample paths. The usual definition entails integrability of the initial value; local martingales need not be integrable at later times without stopping.

Choose a localizing sequence of discrete-time stopping times ${\tau }_j\uparrow \infty$ for $X$. For fixed integer $t$, every stopped value is bounded in absolute value by the finite sumThat sum is integrable under the hypothesis. Since $X^{{\tau }_j}$ is a martingale,The dominated convergence theorem, including its conditional version, now removes the stopping. Thus $\mathbb{E}[X_t\mid {\mathcal{F}}_{t-1}]=X_{t-1}$. The process is adapted and integrable by hypothesis, so $X$ is a true discrete-time martingale. This is the integrable discrete-time local martingale is a martingale criterion. The finite sum dominating stopped values is the crucial discrete-time feature.

The integral against the local-martingale vector $YP$ is a local martingale, provided the predictable holdings are stochastically integrable. Integrating part (b) givesBoth $Y_t{X}_t$ and the cumulative deflated consumption are nonnegative. Hence $M_t\ge -Y_0{X}_0$. With the usual finite deterministic initial capital, the shifted processis a nonnegative local martingale, and is therefore a supermartingale. For completeness, a localizing sequence turns it into true martingales; conditional Fatou lemma for their nonnegative stopped values gives the supermartingale inequality and ordinary Fatou gives integrability at each time. Subtracting the initial constant provesThis is supermartingale control of deflated consumption gains. It uses $c\ge 0$ as well as $X\ge 0$; an unrestricted stochastic integral is not necessarily a true supermartingale.

Let $K={\sup }_T\mathbb{E}[e^{M_T/2}]<\infty$, with $T$ ranging over finite stopping times. To locate the half-threshold for the exponential-martingale Hölder bound, put $u=\sqrt{pq}>\sqrt{q}$. Since $u(u-1)$ is increasing for $u>1$,On the other hand, taking $q=1+\epsilon$ and $p=1+{\epsilon }^2$ gives $C(p,q)\to 1/2$ as $\epsilon \downarrow 0$. Hence

Fix $0<a<1$. Choose $p,q>1$ with $aC(p,q)<1/2$, and put $\lambda =2aC(p,q)\in (0,1)$. Part (a) and the Jensen inequality give, for every finite stopping time,Thus the entire stopped family has a uniform $L^p$ bound for some $p>1$. By uniform integrability from an Lp bound, it is uniformly integrable.

To check that this local martingale is a true martingale, stop it by a localizing sequence. At each fixed time the stopped variables are uniformly integrable by the same bound; taking limits in their conditional martingale identities proves the unstopped identity. The uniform bound over all finite stopping times then makes it a uniformly integrable martingale, with terminal expectation one. Therefore

Write ${\delta }_n=2^{-n}$ and let $Q_n(U)$ denote the printed squared-increment sum for a process $U$, on grid times $k{\delta }_n$. We construct its limit first for bounded martingales, then use localizing sequences and the finite variation part of a semimartingale.

Continuity and monotonicity of the bounded-martingale limit. Let $U$ be a uniformly bounded continuous martingale. The given result supplies a limit $R^U$ in uniform convergence on compacts in probability. Each $Q_n(U)$ is continuous and adapted. A subsequence converges almost surely uniformly on each compact time interval, by choosing summable error probabilities and diagonalizing. Its limit therefore has a continuous version; in the usual completed filtration this version is adapted.

The sums $Q_n(U)$ themselves can decrease between grid points, because the last partial increment is being squared. To establish monotonicity, use instead the quadratic variation from completed grid incrementsThese step processes are nondecreasing, and pathwise uniform continuity giveswhere ${\omega }_U(\delta ;T)=\sup \{|U_t-U_s|:s,t\le T,|t-s|\le \delta \}$. Thus the same almost surely uniform subsequence of $S_n$ converges to $R^U$, proving that $R^U$ is nondecreasing. Also $R_0^U=0$.

Localization of a continuous local martingale. Subtract the initial value, which does not affect increments, so that $M_0=0$. SetContinuity gives ${\tau }_m\uparrow \infty$ almost surely and makes $M^{{\tau }_m}$ bounded. The bounded local martingale criterion makes it a true martingale. Let $R^m$ be its continuous, adapted, nondecreasing limit.

The discrete sums commute exactly with stopping:For $\ell \le m$, uniqueness of the probability limit and stability of uniform convergence on compacts in probability under stopping giveup to indistinguishability. Taking a common null set for the countably many pairs, patch these processes into $R^M$ by setting $R_t^M=R_t^m$ when $t\le {\tau }_m$. Compatibility makes this definition independent of $m$. It is continuous, adapted, and nondecreasing. For each $T,\epsilon >0$,Let $n\to \infty$ and then $m\to \infty$. This proves the localization and patching of quadratic variation.

Adding finite variation. Decompose the continuous semimartingale as $X=X_0+M+A$, where $M$ is a continuous local martingale starting at zero and $A$ is continuous, adapted, and locally of finite variation. For every compact interval, the sum of the absolute increments of $A$ is at most its total variation of a function. Hence, pathwise,The Cauchy-Schwarz inequality bounds the mixed increment sum byThe first factor is bounded in probability, since $Q_n(M)\to R^M$ uniformly on compacts in probability; the second tends to zero almost surely. Thus the mixed term tends to zero in probability. Expanding the square provesThe constructed $R=R^M$ is continuous, nondecreasing, and adapted. This is the quadratic variation of $X$, and expresses the fact that finite-variation terms do not change quadratic variation.

A continuous martingale is an adapted process with almost surely continuous paths, $\mathbb{E}|M_t|<\infty$ for every $t$, andA continuous local martingale is continuous and adapted and has a localizing sequence of stopping times $T_n\uparrow \infty$ almost surely such that every $M_t^{T_n}=M_{t\wedge T_n}$ is a true martingale. The initial value is normally integrable; the frequently used class ${\mathcal{M}}_{c,\mathrm{loc}}$ is normalized to start at zero.

The usual conditions for a filtration are completeness and right continuity. Completeness means that ${\mathcal{F}}_0$ contains every subset of each $\mathbb{P}$-null set in the ambient $\mathcal{F}$; right continuity means ${\mathcal{F}}_t={\bigcap }_{u>t}{\mathcal{F}}_u$. These conditions concern the filtered probability space, whereas continuity of a martingale concerns its sample paths.

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

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.

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).

Let $U_t=(1,0,0)+(B_t^1,B_t^2,B_t^3)$, with three independent standard Brownian motions, and set $R_t=|U_t|$ and $M_t=R_t^{-1}$. The radial stochastic process $R$ is a three-dimensional Bessel process, which never hits zero by the Hitting-zero classification for a Bessel process. In three dimensions $u\mapsto |u{|}^{-1}$ is a harmonic function away from zero: direct differentiation gives $\Delta (|u{|}^{-1})=0$.

For the stopping times $T_n=\inf \{t:R_t\notin (1/n,n)\}$, $n\ge 2$, the Itô formula givesThe coefficients are bounded by $n^2$, so each stopped Itô integral is a true martingale on every finite horizon. The stopping times form a localizing sequence; as permitted, its divergence need not be proved here. Thus $M$ is a nonnegative continuous local martingale.

To see that $M$ is not a true martingale, integrate the three-dimensional multivariate normal density of $U_t$ in spherical coordinates. The angular integral yieldswhere $\Phi$ is the standard normal distribution function. Therefore the reciprocal of a three-dimensional Bessel process is a strict local martingale.

Use the standard definition of a local martingale, including an integrable initial value, and let $T_n\uparrow \infty$ be a localizing sequence. If $X\ge 0$, the Fatou lemma gives $\mathbb{E}{X}_t\le {\mathrm{liminf}}_n\mathbb{E}{X}_{t\wedge T_n}=\mathbb{E}{X}_0<\infty$. For $s\le t$, the Conditional Fatou lemma givesFor clarity, the Conditional Fatou lemma itself follows by setting $U_m={\inf }_{n\ge m}{U}_n$ for nonnegative $U_n$, applying the conditional version of the monotone convergence theorem to $U_m\uparrow \mathrm{liminf}{U}_n$, and observing $\mathbb{E}(U_m\mid {\mathcal{F}}_s)\le {\inf }_{n\ge m}\mathbb{E}(U_n\mid {\mathcal{F}}_s)$. Hence a nonnegative local martingale is a supermartingale.

If $|X_t|\le C$ simultaneously for all $t$, with a deterministic $C<\infty$, the dominated convergence theorem applied to the stopped martingale identity gives, for each $A\in {\mathcal{F}}_s$,This is exactly the conditional expectation identity, so a bounded local martingale is a true martingale. The same proof works for deterministic bounds on every finite time interval; merely having a finite random pathwise bound supplies no dominating integrable random variable and is a different convention.

Take the stopping timesContinuity gives $|M_{t\wedge S_n}|\le n$, with no overshoot at the first crossing. Stopping preserves the local martingale property, and the bounded case proved in question 4(i) makes each $M^{S_n}$ a continuous true martingale.

Every continuous path is bounded on each compact time interval, so $S_n\uparrow \infty$ almost surely. Hence this is also a localizing sequence consisting of bounded stopped martingales.

Interpret the Stratonovich integral in its standard extension to continuous semimartingale integrands:This extension is needed because $f^{\prime }(X)$ is generally a semimartingale rather than a local martingale. Also the introductory prose in the PDF reverses integrand and integrator relative to its displayed definition; the displayed definition fixes the convention used here.

Apply the Itô formula to $f^{\prime }(X)$, using $f\in C^3$:The continuous finite-variation process in the second term has zero quadratic covariation with $X$, while the quadratic covariation rule for an Itô integral givesA second application of the Itô formula, now to $f(X)$, therefore yieldsAll unbounded coefficients are handled by stopping $X$ on compact spatial intervals, then removing the localizing sequence.