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Work first with $M_0=0$. More generally, the assertion needs $M_0\in L^2$ and then applies to $M-M_0$. Without this assumption, an integrable but non-square-integrable initial random variable kept constant in time has zero quadratic variation but is not bounded in L2 space.

Choose increasing stopping timesThey tend to infinity almost surely, because the continuous path and its quadratic variation are finite on every compact time interval. The stopped process is a bounded martingale, and $M_{t\wedge {\tau }_n}^2-[M{]}_{t\wedge {\tau }_n}$ is bounded and a local martingale. Taking expectations givesSuppose $C=\mathbb{E}[M{]}_{\infty }<\infty$. These squared expectations are at most $C$. For each fixed $t$, the stopped values converge almost surely to $M_t$ and are uniformly integrable, since they are uniformly bounded in L2 space. They therefore converge in $L^1$. Passing in the stopped conditional-expectation identity, using contraction of conditional expectation in $L^1$, gives $\mathbb{E}(M_t\mid {\mathcal{F}}_s)=M_s$ for $s\le t$. Thus $M$ is a true martingale. The Fatou lemma gives $\mathbb{E}{M}_t^2\le C$, uniformly in $t$, so it is an L2-bounded continuous martingale.

Conversely, suppose $M$ is a true martingale with ${\sup }_t\mathbb{E}{M}_t^2\le C$. For fixed $t$, bounded-time optional sampling theorem gives $M_{t\wedge {\tau }_n}=\mathbb{E}(M_t\mid {\mathcal{F}}_{t\wedge {\tau }_n})$. Conditional Jensen inequality then yieldsThe brackets increase to $[M{]}_t$. Applying the monotone convergence theorem first in $n$ and then in $t\uparrow \infty$ proves $\mathbb{E}[M{]}_{\infty }\le C<\infty$.

The exact identities and terminal limits can also be justified. On each fixed horizon $t$, the Doob L2 maximal inequality givesThis is an integrable dominating variable for $M_{t\wedge {\tau }_n}^2$, so the dominated convergence theorem and increasing convergence of the brackets remove the stopping:The L2 martingale convergence theorem supplies $M_{\infty }$ with $M_t\to M_{\infty }$ almost surely and in L2 space. Consequently the squares converge in $L^1$, while $[M{]}_t\uparrow [M{]}_{\infty }$. We obtain the integrable terminal bracket criterion with the precise terminal identityFor $M_0\in L^2$ rather than zero, the final identity becomes $\mathbb{E}{M}_{\infty }^2=\mathbb{E}{M}_0^2+\mathbb{E}[M{]}_{\infty }$.

Set $N_t={\int }_0^t{Z}_s^{-1}d{Z}_s$, the stochastic logarithm of $Z$. On each finite horizon, a strictly positive continuous path has its reciprocal bounded; stopping at levels of $Z$, $Z^{-1}$ and $[Z]$ rigorously localizes this stochastic integral. Thus $N$ is a zero-start continuous local martingale, withThe Itô formula for the ordinary logarithm givesAccordingly, with the exponential convention printed in this question,The initial constant in $M$ contributes no quadratic variation. Under the standard normalized Doléans-Dade exponential convention, which subtracts the driver's initial value in the exponent, the same formula is written $Z=Z_0\mathcal{E}(N)$.

For uniqueness, suppose another continuous driver $L$ gives the printed exponential. Evaluating at zero forces $L_0=\log Z_0$. Applying the Itô formula to its exponential gives $d{Z}_t=Z_{t}d{L}_t$. Since $Z_t>0$, associativity of stochastic integration yields $d{L}_t=Z_t^{-1}d{Z}_t$. Thus $L_t=L_0+N_t=M_t$, up to indistinguishability of stochastic processes.

There is an initial-integrability qualification if “local martingale” is used with the usual integrable-initial-value definition. The displayed $M$ is a continuous local martingale when $\log Z_0\in L^1$, in particular when $Z_0$ is a positive deterministic number. An arbitrary positive integrable $Z_0$ does not ensure this. For example, take an ${\mathcal{F}}_0$-measurable positive integer $W$ with $\mathbb{P}(W=n)=6/({\pi }^2{n}^2)$ and put $Z_t=e^{-W}$ for all $t$. This is a bounded strictly positive martingale, but any representing driver must have $M_0=-W$, which is not integrable. Thus the unqualified claim is false under that definition. For general $Z_0$, the always-valid local-martingale statement is the zero-start representation $Z=Z_0\mathcal{E}(N)$; the printed representation additionally requires the stated integrability or the convention that allows arbitrary finite initial constants.

On the finite horizon $[0,T]$, strict positivity gives equivalence of the restricted probability measures: for $A\in {\mathcal{F}}_T$, $\mathbb{Q}(A)={\mathbb{E}}_{\mathbb{P}}(Z_T{1}_A)$ vanishes if and only if $\mathbb{P}(A)$ vanishes. The standard change-of-measure theorem gives semimartingale invariance under equivalent measures: a process is a semimartingale under $\mathbb{P}$ on this horizon if and only if it is one under $\mathbb{Q}$. The sample-path quadratic variation is unchanged, but the local-martingale and finite-variation parts generally change.

More explicitly, set $N=\int Z^{-1}dZ$, the zero-start stochastic logarithm of the density. For every continuous $\mathbb{P}$-local martingale $L$, the Girsanov theorem states thatis a continuous $\mathbb{Q}$-local martingale. Consequently, if the continuous semimartingale decomposition under $\mathbb{P}$ is $X=X_0+L+V$, its decomposition under $\mathbb{Q}$ isThe covariation correction is a continuous finite-variation process, locally well-defined because $Z$ stays strictly positive. In the Brownian case, if $d{Z}_t=Z_t{\theta }_{t}d{B}_t$, thenis a Brownian motion under $\mathbb{Q}$. The inverse finite-horizon density is $1/Z$, giving the reverse change of measure. These assertions apply on each horizon on which the density is strictly positive; they do not assert equivalence on a larger terminal sigma-algebra where the full density $D$ may vanish.

Work on a common stochastic basis on which $B,B^{\prime }$ are independent Brownian motions and the solutions are adapted; independent weak solutions can be put on the product stochastic basis. Independence gives $[B,B^{\prime }]=0$. Put $S=Z+Z^{\prime }\ge 0$ and define the predictable weightsThey are bounded and satisfy $u_t^2+v_t^2=1$ everywhere. DefineThis is a zero-start continuous local martingale, and its quadratic variation is $[\beta {]}_t={\int }_0^t(u_s^2+v_s^2)ds=t$. By the Lévy characterization of Brownian motion, $\beta$ is a Brownian motion in the common filtration.

On $\{S_t>0\}$, $\sqrt{S_t}{u}_t=\sqrt{Z_t}$ and $\sqrt{S_t}{v}_t=\sqrt{Z_t^{\prime }}$. On $\{S_t=0\}$, nonnegativity forces both $Z_t,Z_t^{\prime }$ to be zero, so the same identities hold. Associativity of stochastic integration therefore givesThis proves the addition law for square-root branching diffusions: $Z+Z^{\prime }$ is a weak solution with initial value $z+z^{\prime }$. Specifying the weights at zero is necessary to keep $[\beta {]}_t=t$ after extinction; simply setting both weights to zero would fail to produce a Brownian driver.

The previsible sigma-algebra $\mathcal{P}$ on $\Omega \times [0,\infty )$ is generated by $E\times \{0\}$ with $E\in {\mathcal{F}}_0$ and by $E\times (s,t]$ with $E\in {\mathcal{F}}_s$, $0\le s<t$. Equivalently it is the smallest sigma-algebra making every left-continuous adapted process measurable. A previsible process is a process measurable for $\mathcal{P}$.

A bounded simple predictable process has finitely many time intervals and bounded coefficients known at their left endpoints. One may use deterministic endpoints, or the customary more general elementary predictable process with stopping-time intervals:A bounded ${\mathcal{F}}_0$-measurable value at time zero may be added; it does not affect integration against a continuous martingale. A process with finite support is included by adding a zero coefficient on its terminal interval. Both versions of simple processes satisfy the arguments below. The elementary stochastic integral isWhen a stopping time is infinite, use the terminal value $M_{\infty }$, which exists in $L^2$ and almost surely for an L2-bounded continuous martingale. Refining intervals leaves this definition unchanged.

For a continuous local martingale $M$, its quadratic variation $[M]$ is the unique continuous adapted nondecreasing process starting at zero such that $(M_t-M_0{)}^2-[M{]}_t$ is a local martingale. It is predictable because it is continuous and adapted. To recover it from the path, use refining deterministic grids, for example $t_k^{(n)}=k{2}^{-n}$, and formOnly finitely many increments are nonzero on a compact time interval. ThenIn particular each fixed-time sum converges in probability, and a suitable subsequence converges uniformly on compact intervals almost surely. The partial last grid interval is included in this path approximation.

First justify optional sampling for the square-minus-bracket process. The Doob L2 maximal inequality gives $\mathbb{E}{\sup }_{t\ge 0}|M_t{|}^2\le 4\mathbb{E}|M_{\infty }{|}^2<\infty$. Stop the local martingale $M^2-[M]$ using a localizing sequence and bounded bracket levels. The stopped expectation identity, followed by Fatou, bounds $\mathbb{E}[M{]}_{\infty }$ by the expected squared path supremum. Hence $M^2-[M]$ has an integrable absolute supremum, and localization with conditional dominated convergence makes it a uniformly integrable martingale. The optional stopping theorem stated in part (b) is therefore applicable to both $M$ and $M^2-[M]$.

For ordered stopping times $\sigma \le \tau$, expand the squared increment and condition on ${\mathcal{F}}_{\sigma }$. The mean increment is zero, soThis is the conditional bracket isometry for stopped martingale increments. Apply it with $\sigma =T_j$, $\tau =T_{j+1}$ and multiply by the bounded ${\mathcal{F}}_{T_j}$-measurable variable $h_j^2$. The orthogonality proved in part (b) gives the Itô isometry

To identify the whole bracket process, let $N=H\cdot M$ and $C_t={\int }_0^t{H}_s^{2}d[M{]}_s$. For deterministic $s<t$ and $E\in {\mathcal{F}}_s$, apply the proved isometry to the simple integrand $1_E{1}_{(s,t]}H$. It yieldsThus $\mathbb{E}[(N_t-N_s{)}^2\mid {\mathcal{F}}_s]=\mathbb{E}[C_t-C_s\mid {\mathcal{F}}_s]$. Since $N$ is a martingale, the cross term $2N_s(N_t-N_s)$ has conditional mean zero. Consequently $N^2-C$ is a martingale. The process $C$ is continuous, adapted, nondecreasing and starts at zero, so uniqueness in the definition of quadratic variation provesThis deduction uses only the proved elementary isometry, not a previously assumed formula for the quadratic variation of stochastic integrals.

Define $k(x,t)=h_x(x,t)/h(x,t)={\partial }_x\log h(x,t)$. Positivity and smoothness of $h$ make $k$ continuous and bounded on compact space-time sets. Brownian paths remain in compact spatial sets on each finite horizon, so ${\int }_0^{T}k(X_s,s{)}^{2}ds<\infty$ almost surely for every $T$. Thusis a well-defined continuous local martingale with $M_0=0$. The backward heat equation from part (b) removes the drift in Itô's formula and yieldsBy the allowed fact $Z_0=1$ and the uniqueness in part (a),

The finite-horizon form of the Girsanov theorem used here says: if $M_0=0$ is a continuous local martingale and its stochastic exponential $Z$ is a strictly positive true martingale of mean one on $[0,T]$, then under $d\mathbb{Q}{|}_{{\mathcal{F}}_T}=Z_{T}d\mathbb{P}{|}_{{\mathcal{F}}_T}$ every continuous $\mathbb{P}$-local martingale $N$ becomes the $\mathbb{Q}$-local martingale $N-[N,M]$. Quadratic variation is unchanged by this locally equivalent change of measure.

All the assumptions hold. The integrand defining $M$ is locally square integrable as checked above; $Z$ is strictly positive; and it is already a true martingale because it is the density process. On each finite horizon $Z_t=\mathbb{E}[Z_T\mid {\mathcal{F}}_t]$, so it is uniformly integrable on that horizon. There is no need to establish a separate Novikov condition, nor to assume equivalence on the infinite-time sigma-algebra.

Take $N=X$. Its covariation with $M$ is $[X,M{]}_t={\int }_0^{t}k(X_s,s)ds$. Hence $B_t=X_t-{\int }_0^{t}k(X_s,s)ds$ is a continuous $\mathbb{Q}$-local martingale starting at zero with $[B{]}_t=t$. The Lévy characterization of Brownian motion makes it Brownian under $\mathbb{Q}$. Since this argument holds on every finite horizon for the same process $B$,This is the Brownian change of measure by a positive space-time harmonic function, a Doob h-transform with drift ${\partial }_x\log h$.

Put $p=1-R$, $b=\mu -r1$, $A=\sigma {\sigma }^T$ and $q=R+\epsilon$. First assume the covariance matrix $A$ is invertible and initial portfolio wealth is positive. Introduce the accumulated fee processThe Itô formula and the self-financing portfolio equation giveThis converts the path-dependent payment into a terminal-utility problem with state $y$. The volatility-penalized terminal fee changes its drift as well as its spot volatility.

Let $J(t,y)$ be the manager's value function. Its Hamilton-Jacobi-Bellman equation, with terminal condition $J(T,y)=y^p/p$, isThe homogeneity of CRRA utility suggests $J(t,y)=f(t)y^p/p$. Since $y{J}_y=f(t)y^p$ and $y^2{J}_{yy}=-Rf(t)y^p$, the portfolio-dependent expression isIts coefficient is positive even when $p<0$. Completing the square givesThus the stock fractions are constant. Substitution into the Hamilton-Jacobi-Bellman equation yieldsIn particular, multiplying the fee by $a$ does not affect the optimal investment portfolio.

To prove global optimality rather than merely solve a formal Hamilton-Jacobi-Bellman equation, write $v_t={\sigma }^T{\pi }_t$, $v^{\ast }={\sigma }^T{\pi }^{\ast }$ and let $y^{\ast }$ be the fee under the proposed constant policy. Then $b^T\pi =q{v}^{\ast }\cdot v$. Define a probability $Q$ byThis density is a true martingale because $v^{\ast }$ is constant. By the Girsanov theorem, $W_t^Q=W_t-p{v}^{\ast }t$ is a Brownian motion under $Q$. The geometric Brownian motion formula for the fee ratio givesFor admissible, locally square-integrable exposures with finite accumulated quadratic variation, setThis stochastic exponential is a positive local martingale, so $E_Q{Z}_T\le 1$. Moreover,For $p>0$, the exponent $p/(1+\epsilon )$ lies between zero and one. The Jensen inequality for this concave function gives $E_Q[Z_T^{p/(1+\epsilon )}]\le 1$. For $p<0$, the power is convex and decreasing, so the Jensen inequality gives $E_Q[Z_T^{p/(1+\epsilon )}]\ge (E_Q{Z}_T{)}^{p/(1+\epsilon )}\ge 1$; an infinite moment only makes this inequality stronger. Dividing by $p$, with its corresponding sign, proves $E[U(y_T)]\le E[U(y_T^{\ast })]$ in both cases. Equality is achieved by the constant policy.

For an ordinary terminal-wealth agent with relative risk aversion $q$, the same Hamilton-Jacobi-Bellman equation calculation, with no fee penalty, maximizes $b^T\pi -q{\pi }^{T}A\pi /2$. It therefore gives exactly the same optimal investment portfolio. For $q\ne 1$, the preceding stochastic exponential and Jensen inequality verification applies with penalty zero and exponent $1-q$, proving global optimality for this agent too. The spot volatility penalty adds $\epsilon$ to effective relative risk aversion. If $q=1$, the printed expression for $U_0$ has a zero denominator; its appropriate limit, after subtracting an irrelevant constant, is $U_0(x)=\log x$, which still gives ${\pi }^{\ast }=A^{-1}b$. In this logarithmic case, the terminal log-wealth difference from the candidate equals $\log Z_T$ for a positive stochastic exponential $Z$, so $E[\log Z_T]\le \log E[Z_T]\le 0$ proves its optimality.

If $A$ is singular, the same conclusion holds in the absence of arbitrage, provided $b$ lies in the image of a linear map $x\mapsto Ax$: replace $A^{-1}$ by the Moore-Penrose inverse, and add any vector in the kernel of $A$ to ${\pi }^{\ast }$. These additions change neither returns nor Brownian portfolio exposures. If $b$ has a nonzero component in that kernel, a zero-volatility position has a nonzero excess return. Scaling its profitable sign gives arbitrage, so a finite attained optimum as claimed requires this degeneracy to be excluded.

Treat the two-point prior for $\alpha$ as independent of the driving Brownian motion. On a partition $0=t_0<\cdots <t_n=t$, conditionally on $\alpha =\pm a$, the increments of $X$ are independent normal distributions with means $\pm a\Delta t_j$ and variances $\Delta t_j$. Their likelihood ratio isThus every partition containing $t$ gives the same Bayesian posterior, depending only on the endpoint. Refine partitions through a dense set of observation times. Continuity makes their increasing observation fields generate the full path field ${\mathcal{X}}_t$, and convergence of bounded conditional expectations preserves that Bayesian posterior. Equal prior probabilities and the Bayes' theorem consequently giveIts conditional mean drift isThis is the binary Brownian drift filter.

Define the observed innovation processIt is continuous and adapted to ${\mathcal{X}}_t$. For $s<t$, independence of future Brownian motion increments from the past and the latent drift givesFor every $u\ge s$, the tower property of conditional expectation givesSince $|h|\le a$, conditional integration is justified. Subtracting the integrated identities shows $E[{\hat{W}}_t-{\hat{W}}_s\mid {\mathcal{X}}_s]=0$. Thus $\hat{W}$ is a square-integrable martingale. Subtracting a continuous finite variation process does not change quadratic variation, so $[\hat{W}{]}_t=[X{]}_t=[W{]}_t=t$. The Lévy characterization of Brownian motion now proves that $\hat{W}$ is a Brownian motion in the observation filtration. Therefore

The common Brownian motion cancels from the difference $D_t=Y_t-X_t$. Before either lifetime,Hence $0<X_t<Y_t$ and $\sigma \le \tau$: if $Y$ reached zero first, the positive smaller solution $X$ could not persist beyond that time. WriteUsing the Itô formula for $Y^{-1}$ givesThe difference has finite variation, so there is no cross-variation term in the Itô product rule for $D/Y$. Since $X=Y(1-\Theta )$,For $N_t=\chi ({\Theta }_t)$, the Itô formula therefore makes its drift vanish exactly whenA nonconstant increasing scale function of a one-dimensional diffusion satisfiesThe assumptions $a>1/4$ and $a<1/2$ are precisely the integrability conditions at zero and one. Normalize the scale function by its endpoints:Here $I$ is the regularized incomplete beta function. The normalized function is continuous and strictly increasing on $[0,1]$, with $\varphi (0)=0$ and $\varphi (1)=1$. We have obtained the local martingale

The terminal identification needs care; mere boundedness does not by itself say what happens when both lifetimes coincide. Stop first on compact subsets of $0<X<Y$, and then exhaust the interval $[0,\sigma )$. The bounded local martingale criterion makes $N$ a bounded martingale with a limit $N_{\sigma -}$, and optional sampling theorem gives $\mathbb{E}[N_{\sigma -}]=N_0$. For completeness, applying the Itô formula to each stopped $N^2$ bounds the expected quadratic variation by one. Monotone convergence over the localization sequence impliesSince ${\varphi }^{-1}$ is continuous on $[0,1]$, ${\Theta }_t$ itself has a limit ${\Theta }_{\sigma -}$.

On $\{\sigma <\tau \}$, $Y_{\sigma }>0$ while $X_t\to 0$, so ${\Theta }_{\sigma -}=1$ and $N_{\sigma -}=1$. On $\{\sigma =\tau \}$, the supplied divergent clock isIf ${\Theta }_{\sigma -}>0$, then eventually ${\Theta }_t{\varphi }^{\prime }({\Theta }_t)$ is bounded below by a positive constant. This remains true if its limit is one: ${\varphi }^{\prime }(\theta )$ diverges there, rather than tending to zero. The divergent clock would then force $[N{]}_{\sigma -}=\infty$, contradicting the finite quadratic variation. Thus ${\Theta }_{\sigma -}=0$ and $N_{\sigma -}=0$ on the simultaneous-lifetime event. We have provedTaking expectations yields the complete probability formulaThis is the beta integral for strict ordering of coupled Bessel lifetimes; the complementary probability is the probability of simultaneous swallowing, not of the larger initial point dying first.

To connect the calculation to SLE, let its driver be $W_t=\sqrt{\kappa }{\beta }_t$. For a positive boundary point $r$ before its boundary swallowing time, put $V_t^r=(g_t(r)-W_t)/\sqrt{\kappa }$. The Chordal Loewner equation givesThus two positive boundary points give exactly the coupled Bessel processes above, with $B=-\beta$ and $a=2/\kappa$. The interval $1/4<a<1/2$ corresponds to $4<\kappa <8$. Their boundary swallowing times obeyThe formula quantifies whether the growing hull separates one boundary point from infinity before the other, or swallows them together inside the same boundary interval. It does not identify swallowing with visiting each point. At $\kappa =6$, the two beta parameters are $1/3$ and this becomes the Cardy boundary crossing formula in the corresponding cross-ratio coordinate, explaining its role in the conformal invariance of planar percolation.

Fix $T<\infty$. By uniform continuity of each path of the continuous local martingale on $[0,T+1]$, its modulustends to zero almost surely. For every $s\le T$, comparison of powers of each increment givesThe squared-increment sum converges in probability to $[X{]}_T$, by the dyadic definition of quadratic variation. Including the last full grid interval instead of the last partial interval changes that sum by a quantity bounded by $2{\eta }_n^2$, so the ceiling convention has the same limit. In particular these squared-increment sums are bounded in probability.

Since $p-2>0$, a quantity tending to zero almost surely multiplied by a family bounded in probability tends to zero in probability. HenceThis is uniform convergence on compacts in probability, and proves the superquadratic case of dyadic power variation of a continuous local martingale.

Fix the specified time $t$. With the path modulus ${\eta }_n$ on $[0,t+1]$ as above, comparison of increment powers now givesThe hypothesis makes the second factor eventually bounded almost surely, while the first tends to zero almost surely because $2-p>0$. Thus the squared-increment sums tend to zero almost surely. They also converge in probability to the quadratic variation $[X{]}_t$, so uniqueness of the limit in probability implies $[X{]}_t=0$ almost surely.

The quadratic variation is nondecreasing, hence it vanishes throughout $[0,t]$. To see explicitly that the continuous local martingale must then be constant, localize it to square-integrable stopped martingales. The Itô isometry givesRemove the localization, apply the conclusion at every rational $s\le t$, and use continuity. Since $X_0=0$, this provesThus $X$ is indistinguishable from the zero process on that interval. This is the subquadratic case of dyadic power variation of a continuous local martingale.

No quadratic variation argument is needed. For deterministic $u<v$, the martingale property and square integrability giveConsequentlyApply this identity separately to each consecutive grid increment. The second moments telescope, givingThis uniform bound follows directly from martingale-difference orthogonality, including the final interval specified by the ceiling.

Let $g_t=P_{1-t}(\nabla f)(X_t)$. The triangle inequality for the conditional average givesThe independent coordinate Brownian motions have zero cross-variation, so the quadratic variation of a stochastic integral givesThe Dambis-Dubins-Schwarz theorem says that a continuous local martingale $N$ starting at zero whose quadratic variation diverges admits a Brownian motion $B$, with the inverse-clock filtration, such that $N_t=B_{[N{]}_t}$. If the total clock is finite, one may enlarge the probability space and append independent Brownian motion after the clock terminates. This gives the same identity up to the original lifetime.

Apply this theorem to $N=M-M_0$. The clock bound implies the event inclusionThe latter event is contained in the union of the upper and lower crossing events. Symmetry of Brownian motion givesThis uses only an event inclusion; independence between the Brownian motion and its random clock is neither assumed nor needed.

Use the following form of the Girsanov theorem. If $W$ is Brownian motion, $\vartheta$ is predictable, andis a uniformly integrable martingale with $Z_0=1$, then the probability measure with terminal density $Z_{\infty }$ makes $W_t-{\int }_0^t{\vartheta }_{s}ds$ a Brownian motion. The analogous finite-horizon statement uses the terminal density at that horizon.

Here take $W_t=X_t-x$ and ${\vartheta }_s=1_{\{s<\tau \}}/X_s$. Before $\tau$, $X_s\in [\epsilon ,M]$, so the stochastic integral is well-defined. The assumed uniform integrability of the stopped stochastic exponential givesThus the proposed weighting really defines a probability measure, and the Girsanov theorem gives the Brownian motionunder that measure. In particularAll calculations are first made on this compact stopped interval. The integral involving $1/X$ is naturally defined on $[0,T_0)$; at $T_0$ its quadratic variation may diverge. The density will have a zero limiting extension there, as established below, rather than a finite stochastic logarithm at the hitting time.