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The boundedness of $a^{\prime }$ makes $a$ globally Lipschitz, so the stock stochastic differential equation has a nonexplosive strong solution. Since $a$ is bounded, $S_t-S_0={\int }_0^{t}a(S_u)d{W}_u$ is a square-integrable martingale on every finite horizon. Apply Itô formula to the option price:Thus both the stock and the option are local martingales under the original probability measure, while cash is constant. The relevant no-arbitrage theorem is the sufficient direction of the fundamental theorem of asset pricing: a common equivalent measure under which all numéraire-denominated traded prices are local martingales excludes arbitrage by admissible trading strategies. In this case the original measure itself is an equivalent local martingale measure. Hence the augmented market has no admissible arbitrage. The nonnegative option-value assumption also makes the replicating wealth below admissible.

SetDifferentiating the pricing equation in the stock-price variable givesThe limiting terminal derivative is $1_{\{S>K\}}$ away from the strike. Its value at the strike may be assigned as the stated $1_{\{S\ge K\}}$: the smooth positive diffusion coefficient gives a continuous terminal distribution, so that point has probability zero. This is the call delta equation for a driftless local volatility diffusion.

Hold ${\pi }_t$ shares and ${\eta }_t=V(t,S_t)-{\pi }_t{S}_t$ units of cash. The portfolio value is $V(t,S_t)$, and the preceding Itô calculation gives $dV={\pi }_{t}d{S}_t+{\eta }_{t}d{B}_t$, with $d{B}_t=0$. It is therefore a self-financing portfolio, with initial wealth $V(0,S_0)$ and terminal wealth $(S_T-K{)}^{+}$. This proves replication by the stated delta hedge.

A further application of Itô formula, now to $U$, yieldsThe density process is the strictly positive stochastic exponentialIts integrand is bounded, so the Novikov condition proves that it is a true martingale. The product formula for $M=Z\pi$ givesThe drift in $Zd\pi$ cancels this quadratic covariation term, leavingThus $M$ is a local martingale; the expression also identifies its stochastic integral integrand explicitly.

Under the stated assumption that $M$ is a true martingale on $[0,T]$, its terminal value givesSince $0\le Z_T{1}_{\{S_T\ge K\}}\le Z_T$ and $\mathbb{E}[Z_T\mid {\mathcal{F}}_t]=Z_t>0$, division givesThis is the derivative-weighted call delta martingale argument. It uses true martingality through maturity, not just local martingality before the terminal date.

For $w\ne 0$ in ${\mathbb{R}}^3$, the radial function $r(w)=\Vert w\Vert$ hasThe multidimensional Itô formula, initially stopped away from zero, givesThe martingale term has quadratic variation $dt$, since ${\sum }_i(W_t^i/R_t{)}^2=1$. The Lévy characterization of Brownian motion says that a continuous local martingale starting at zero with quadratic variation $t$ is a standard Brownian motion. Consequently the radial equation isOn $[\epsilon ,M]$ the drift $1/r$ is Lipschitz continuous. Extend it to a bounded globally Lipschitz function on the real line. The standard existence and pathwise uniqueness theorem for stochastic differential equations with globally Lipschitz coefficients applies to this extension, and gives uniqueness in law up to the first exit from $[\epsilon ,M]$. Both this radial process and the process in part (a) solve the same equation from $x$. Thuswhere each exit time is computed from its own coordinate path.

For completeness, the radial equation is valid globally: the three-dimensional Bessel process does not hit zero. Apply the Itô formula to $1/R$ on the annulus. Its drift is zero, and the stopped process is a bounded martingale. At the first exit,soThe exit time is finite because one coordinate of the underlying Brownian motion reaches $M$ in finite time. Letting $\epsilon \downarrow 0$ and then considering arbitrarily large $M$ proves that zero is never reached. The radial martingale therefore has quadratic variation $t$ on the whole time axis, completing the Lévy characterization of Brownian motion argument.

Before $T(x)$, the centered image satisfies the stochastic differential equationChanging the sign of the Brownian motion shows that $X/\sqrt{\kappa }$ is a Bessel process of dimension $1+4/\kappa$. Here is a direct proof of the relevant hitting property, including finite lifetime.

Put $p=1-4/\kappa >0$. By the Itô formula, the drift of $X^p$ isThus $X^p$ is a local martingale and a scale function of a one-dimensional diffusion. Fix $R>x$. The nonnegative functionvanishes at both endpoints and satisfies $\mathcal{L}{u}_R=-1$, for $\mathcal{L}=(\kappa /2){\partial }_{vv}+(2/v){\partial }_v$. Stop first on $[ϵ,R]$, and also at a deterministic time. The Itô formula gives $\mathbb{E}(t\wedge {\tau }_{ϵ,R})\le u_R(x)$. Let $t\to \infty$ and then $ϵ\downarrow 0$. It follows that the first exit ${\tau }_{0,R}$ is finite almost surely, not merely that the path approaches a boundary at infinite time.

Optional stopping applied to $X^p$ on this bounded interval givesOn $\{{\tau }_0=\infty \}$ the process must therefore hit every $R>x$ before zero. Letting $R\to \infty$ makes that probability zero. ConsequentlyAt $\kappa =4$ the dimension is two; zero is not hit from a positive starting point. This endpoint must not be included in the finite-swallowing assertion.

The time-changed diffusion has generator $\mathcal{A}=(\kappa /2){\partial }_{zz}+2(1/z+1/(z-1)){\partial }_z$. For $1<z<R$, its boundary hitting probability from a diffusion scale function isTo see this directly, stop the scale local martingale on $[1+ϵ,R]$ and pass to $ϵ\downarrow 0$. Exit from $(1,R)$ occurs in finite clock time: the scale density behaves as $(z-1{)}^{-4/\kappa }$ and the speed density as $(z-1{)}^{4/\kappa }$, so the usual scale-times-speed exit integral is finite at $1$. More explicitly, with $m(v)=2/(\kappa s_{\kappa }^{\prime }(v))$ the finite-interval diffusion exit Green kernel givesThe equation $\mathcal{A}u=-1$ follows by differentiation and the derivative jump at $v=z$, so stopping the corresponding Itô identity gives this exit bound. The integrand near $1$ is $O(v-1)$; elsewhere it is bounded. At large $z$ the drift is bounded, excluding explosion at infinity in finite clock time.

Here is why these exits describe the original swallowing event. In clock time,If $\hat{Z}$ hits $1$ at finite $u$, its integrated stochastic differential equation shows ${\int }_0^{{\tau }_1}({\hat{Z}}_v-1{)}^{-1}dv<\infty$: the Brownian term has a finite limit and both positive drift integrals must then have finite limits. Consequently $D({\tau }_1)>0$, $X$ hits zero and $Y$ stays positive. This gives $T(x)<T(y)$.

Conversely $T(x)$ is finite by part (b). If its clock ends at a finite value, the diffusion must hit $1$; an interior endpoint with positive $D$ would leave $X$ positive. If the clock runs forever without hitting $1$, the standard one-dimensional scale classification gives escape to infinity whenever $S_{\kappa }<\infty$. This cannot represent a strict swallowing event, which has $D\to Y_{T(x)}>0$ and $Z\to 1$. Hence the surviving event represents $T(x)=T(y)$. This reasoning also explains why a simultaneous collision must be treated separately rather than assigned the boundary value at $1$.

For $\kappa \ge 8$, the scale is unbounded. The maximal inequality for a nonnegative supermartingale gives $\mathbb{P}(\sup s_{\kappa }(\hat{Z})\ge s_{\kappa }(R))\le s_{\kappa }(z)/s_{\kappa }(R)$; letting $R\to \infty$ rules out escape. Finite-interval exit then forces a hit at $1$. For $4<\kappa <8$, taking $R\to \infty$ givesThe strict-event probability is also positive, since $0<s_{\kappa }(z)<S_{\kappa }$ for finite $z$.

For $\kappa \ge 8$, first intersect the probability-one strict-order events over rational $0<a<b$. The real boundary flow gives nondecreasing swallowing times on $(0,\infty )$. For any $0<x<y$, choose rational $x<a<b<y$; then $T(x)\le T(a)<T(b)\le T(y)$. This proves the simultaneous assertionThe range is positive points, as in the definition of $F$; reflection reverses the ordering on the negative axis. At the printed endpoint $\kappa =4$, each fixed positive point has $T=\infty$. Thus $\mathbb{P}(T(x)=T(y))=1$ if equality of extended times is allowed, but there is no finite simultaneous-swallowing assertion there. The preceding formulas assume $\kappa >4$.

For $0\le t\le T$, set $Z_t=\cosh x{Y}_t$. Part (b) gives $Z_0=1$, $\mathbb{E}{Z}_T=1$, and $Z_t>0$. Its stochastic differential isThus $Z$ is the stochastic exponential of $\int \eta dW$. The Novikov condition also holds, since $\exp (\frac{1}{2}{\int }_0^T{\eta }_t^{2}dt)\le e^{T/2}$.

The Girsanov theorem says that under the measure with Radon-Nikodym derivative $Z_T$, the process $W_t-{\int }_0^t{\eta }_{s}ds$ is a Brownian motion up to $T$. Substituting the sign of $\eta$ and the original stochastic differential equation givesThe density is strictly positive, so $\mathbb{Q}$ and $\mathbb{P}$ are equivalent probability measures.

Assume the usual positive initial asset values. The coefficients determine a unique normalized local martingale deflator. The state-price density and local deflator distinction matters here: calling it a state-price density uses the local convention, while true expectation pricing needs an additional qualification, addressed below.

Let $ZB$ and $ZS$ be local martingales, with $Z_0=1$. The Brownian martingale representation theorem says that every local martingale in the usual natural Brownian filtration is continuous and is a stochastic integral against $W$. Applying it to $ZB$, and dividing by the positive finite-variation account, forcesfor a locally square-integrable predictable $\eta$. The Itô product rule for $ZS$ has driftIt must vanish. Since $S,Z,\sigma$ are positive, this yieldsConversely this drift choice makes both $ZB$ and $ZS$ local martingales. Continuity and strict positivity of $\sigma$ make $\lambda$ bounded along each path on every finite time interval, so its pathwise square integral is finite. The unique linear SDE solution isUniqueness follows from the forced drift and diffusion coefficients and uniqueness of this linear stochastic differential equation.

Continuity alone does not make the density a true martingale. For a true equivalent martingale measure, the stochastic exponential $D_t=Z_t{B}_t/B_0$ must have expectation one, for example under the Novikov condition on each horizon. True martingale pricing of all desired deflated payoffs also requires the relevant integrability. These stronger conclusions do not follow just from pathwise continuity.

An explicit counterexample to the stronger reading uses a three-dimensional Bessel process $R$ with $R_0=1$ and $dR=dW+R^{-1}dt$, which is a positive strong solution in the Brownian filtration. Take $B=1$ and $S=R$. Then $r=0$, $\mu =R^{-2}$ and $\sigma =R^{-1}$ are continuous and $\sigma >0$. The unique local candidate is $Z=R^{-1}$, with $ZS=1$. The reciprocal three-dimensional Bessel strict local martingale is not a true martingale. To verify the loss of expectation, use the standard Bessel transition densityIntegrating $p_t(1,y)/y$ gives $\mathbb{E}[R_t^{-1}]=2\Phi (1/\sqrt{t})-1<1$ for $t>0$, where $\Phi$ is the standard normal cumulative distribution function. A true pricing density for the constant bank account would have expectation one. Hence the printed hypotheses establish the local deflator statement, while a true-density reading needs an additional condition and is false as stated.

Multiply the linear stochastic differential equation by $e^{-at}$. The Itô product rule givesThe integrand is deterministic, so the Itô integral has a centered normal distribution. Its variance, by the Itô isometry, isHenceAt $a=0$ the correct continuous-limit formula is $N(z,b^{2}t)$. A zero variance, for example when $b=0$, denotes the deterministic distribution. For $a<0$ the variance is still positive because both numerator and denominator in its quotient are negative. This is the explicit Ornstein-Uhlenbeck solution, allowing either sign of the linear drift coefficient.

The reference Brownian motion must start at $x$, and both laws must use the same stopped-path rule. Under the reference probability measure let $R_s=x+W_s$, let $\tau$ be its exit time from $(a,b)$, and put $T=t\wedge \tau$ and $c=(\delta -1)/2$. The Girsanov density for a stopped Bessel process, as a functional of the stopped reference path, isUse the predictable process ${\theta }_s=c{R}_s^{-1}{1}_{\{s\le \tau \}}$, with any harmless choice at the endpoint. On $[0,t]$, $\int {\theta }_s^{2}ds\le c^{2}t/a^2$. The Novikov condition holds; under the density $D$, the Girsanov theorem makes ${\tilde{W}}_s=W_s-{\int }_0^s{\theta }_{u}du$ a Brownian motion. Before exit, $d{R}_s=c{R}_s^{-1}ds+d{\tilde{W}}_s$, which is the required stochastic differential equation. The drift is a locally Lipschitz function on $(0,\infty )$, so pathwise uniqueness and uniqueness in law hold up to this exit time.

For a version involving just the endpoint and an ordinary integral, the Itô formula for $\log R$ givesThusThe stopped density is measurable from the stopped path, so it is also the Radon-Nikodym derivative of the stopped-path laws. The stopping times in this formula are evaluated on the reference path, rather than taken from a different realization of $X$. A zero-start reference Brownian motion would have a different deterministic initial value and could not dominate this law.

The Itô formula for $f(t,x)=e^{x+at}$ gives the stochastic differential equationWrite $N_t={\int }_0^t{Z}_{u}d{B}_u$ for its continuous local martingale part. Its clock is the quadratic variationBecause $Z_t>0$, $A$ is strictly increasing. For $s<A_{\infty }$, its inverse ${\tau }_s$ satisfies $A_{{\tau }_s}=s$ and $d{\tau }_s/ds=Z_{{\tau }_s}^{-2}$. The Dambis-Dubins-Schwarz theorem, or the Lévy characterization of Brownian motion after the optional time-change theorem, makes ${\beta }_s=N_{{\tau }_s}$ a Brownian motion for the time-changed filtration. With $Y_s=Z_{{\tau }_s}$,This is the Bessel process equation with dimension $\boxed{2+2a}$, on the positive stochastic interval. This establishes the Exponential Brownian-to-Bessel time change.

The lifetime convention is important: the inverse is defined only for $s<A_{\infty }$. If $a<0$, the strong law for Brownian motion gives $B_t/t\to 0$, so $Z_t\to 0$ exponentially and $A_{\infty }<\infty$. Then $Y_s\to 0$ as $s\uparrow A_{\infty }$, and this is its first hit of zero after continuous extension. If $a>0$ the clock diverges by the same strong law. If $a=0$ it diverges by infinite occupation time of one-dimensional Brownian motion on a bounded interval, since $e^{2B_t}$ is bounded below there. Thus the lifetime is infinite when $a\ge 0$. Negative dimensions are interpreted by the same positive-state equation killed at zero; no post-zero Bessel process behavior is asserted.

The printed assertion is false as a supremum bound uniform over shrinking bandwidths. The supplied hint proves a pointwise stochastic order with constants uniform in $x$. Taking the supremum inside the probability changes the problem: a growing number of spatial windows produces a logarithmic cost. We first establish the valid consequence of the hint, then give a counterexample satisfying the printed assumptions.

For $I=I_{x,h}=[x-h,x+h]$, let $p=\mu (I)$, $A_T={\int }_0^T{1}_I(X_t)dt$ and $N_T={\int }_0^T{1}_I(X_t)\sigma (X_t)d{W}_t$. The Invariant distribution of an Itô diffusion has a probability density function bounded above and bounded away from zero on the required compact interval. Choosing $h_0<M/4$ therefore gives constants $c_0,C_0>0$ such that $c_{0}h\le p\le C_{0}h$, uniformly for $|x|\le M/2$ and $0<h<h_0$.

For the centered function $f=1_I-p$, $\int |f|d\mu =2p(1-p)\le 2p$. Outside $[-M,M]$, $f=-p$. The given second-moment estimate and the Chebyshev inequality yieldOn the event $A_T\ge Tp/2$, the drift coefficient contributes an average whose error is at most $R{h}^{\alpha }$, by Hölder continuity. The Itô isometry and stationarity give $\mathbb{E}{N}_T^2\le \Vert \sigma {\Vert }_{\infty }^{2}Tp$. Another use of the Chebyshev inequality provesThis also accounts for the zero-denominator convention through the first exceptional event. The bound proves pointwise $R{h}^{\alpha }+O_{\mathbb{P}}((Th{)}^{-1/2})$, uniformly in the location's tail probabilities. It supplies no bound for the probability of a supremum over all locations.

For an explicit counterexample, take $\sigma \equiv 1$ and the bounded globally Lipschitz functionIt has Lipschitz constant $R=1$, exponent $\alpha =1$, and satisfies the inward-drift conditions with $M=2$, $\gamma =2$. The stochastic differential equation has a unique strong solution of a stochastic differential equation. Its Invariant distribution of an Itô diffusion has densityThe exponent is zero on $[-1,1]$, equals $-(|x|-1{)}^2$ when $1<|x|<2$, and equals $3-2|x|$ outside $[-2,2]$. This verifies integrability, positivity and a constant density ${\rho }_0=Z^{-1}$ on the central interval. Start the Itô diffusion in this Invariant distribution of an Itô diffusion to obtain the required stationary process.

Set $h=T^{-1/3}$ and choose $N=\lfloor (4h{)}^{-1}\rfloor$ windows with centers $x_j=-1/2+(4j-2)h$, $1\le j\le N$. Their intervals $I_j=[x_j-h,x_j+h]$ are disjoint and lie in $[-1,1]$, where the drift is zero. WriteThen ${\hat{b}}_T(x_j,h)=M_j(T)/A_j(T)$ whenever $A_j(T)>0$. The same occupation bound as above gives $\mathbb{E}(A_j(T)-q{)}^2\le C_{1}T{h}^2$ for $T\ge 1$. With $\delta =T^{-1/6}$, a union bound and the Chebyshev inequality implyThe quadratic variations are $⟨M_j⟩=A_j$, and the cross quadratic covariations vanish since the windows are disjoint. Each clock tends to infinity almost surely: for any fixed window, the occupation estimate at times $2^k$, followed by the Borel-Cantelli lemmas, gives $A_j(2^k)/(2^k\mu (I_j))\to 1$. The Knight theorem for orthogonal martingales therefore represents $M_j(t)=B_j(A_j(t))$ using independent standard Brownian motions $B_1,\dots ,B_N$. This theorem is applied to the finite collection for each $T$; no growing-dimensional central limit theorem is being assumed.

On the preceding clock event, compare each time-changed Brownian motion with its value at deterministic time $q$. The Brownian reflection principle and a union bound show that, for every fixed $\epsilon >0$,This bound does not require the Brownian motions to be independent of their clocks. In particular, ${\max }_j|M_j(T)-B_j(q)|/\sqrt{q}\to 0$ in probability. The variables $B_j(q)/\sqrt{q}$ are independent $N(0,1)$ random variables. Their Gaussian maximum exceeds $\sqrt{\log N}$ with probability tending to one: for $r\ge 1$, integration of the normal density over $[r,r+1/r]$ gives $\mathbb{P}(|Z|>r)\ge c{r}^{-1}{e}^{-r^2/2}$, and henceSince $A_j(T)\le (1+\delta )q$ on the clock event, it follows that, for some $c_3>0$,But $\sqrt{Th}Rh=1$ for this bandwidth. Thus the error after subtraction of $Rh$ and multiplication by $\sqrt{Th}$ is not bounded in probability, contradicting the printed stochastic order. The valid pointwise bound above is provable; the uniform shrinking-bandwidth assertion is not. This counterexample shows why a uniform rate needs at least a $\sqrt{\log (1/h)}$ enlargement of the noise scale. If $h$ is instead held fixed and constants may depend on $h$, this counterexample does not contradict that different interpretation.

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.

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.

On the right-hand side, ${\mathbb{E}}_x$ means that the reference Brownian motion starts at $B_0=x$; on the left it means the stochastic differential equation starts at $X_0=x$. Since $h$ is smooth and compactly supported, $h^{\prime }$ is bounded and globally Lipschitz continuous. Define the stochastic exponentialThe Novikov condition holds on every finite horizon $T$, because $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T{h}^{\prime }(B_s{)}^{2}ds)\le \exp (T\Vert h^{\prime }{\Vert }_{\infty }^2/2)$. Thus $D$ is a true martingale. Under $d\mathbb{Q}=D_{T}d\mathbb{P}$, the Girsanov theorem makes ${\hat{B}}_t=B_t-x-{\int }_0^t{h}^{\prime }(B_s)ds$ a standard Brownian motion for $t\le T$. Consequently $B$ under $\mathbb{Q}$ solves the prescribed stochastic differential equation; the existence and pathwise uniqueness theorem for a stochastic differential equation identifies its law with that of $X$.

The Itô formula gives ${\int }_0^t{h}^{\prime }(B_s)d{B}_s=h(B_t)-h(x)-\frac{1}{2}{\int }_0^t{h}^{\prime \prime }(B_s)ds$. Hence, for any bounded measurable $f$,The finite-horizon argument applies separately to every $t\ge 0$.

Let $X,Y$ be two continuous solutions on the same filtered probability space, driven by the same Brownian motion and with $X_0=Y_0$ almost surely. Write $D=X-Y$ and choose a common Lipschitz constant $L$ for $b,\sigma$. Introduce the stopping time ${\tau }_n=\inf \{t\ge 0:|X_t|+|Y_t|\ge n\}$. On $\{{\tau }_n=0\}$ the stopped difference is zero; before ${\tau }_n$, both coefficients and the difference are bounded.

The Itô formula and the zero expectation of the stopped square-integrable Itô integral yieldThe Gronwall inequality implies $\mathbb{E}{D}_{t\wedge {\tau }_n}^2=0$. Continuous solution paths are bounded on each compact time interval, so ${\tau }_n\uparrow \infty$ almost surely. Letting $n\to \infty$ proves $X_t=Y_t$ almost surely at each fixed time, and a countable dense set of times together with continuity gives indistinguishability of stochastic processes. Therefore the stochastic differential equation has pathwise uniqueness.

Start the Bessel process at $x>0$ and let ${\tau }_0$ be its first hit of zero. Away from zero its stochastic differential equation isApply the Itô formula to $f(x)=x^{2-\delta }$, localizing first to a compact interval in $(0,\infty )$. Its drift cancels:Thus the Bessel power local martingale satisfiesFor $\delta =2$, this power is simply the constant one. For $\delta >2$, it is a continuous local martingale for all time, as zero is inaccessible.

Here is a proof of that inaccessibility and of the stronger minimum assertion. For $0<r<x<R$, stop the power at ${\tau }_r\wedge {\tau }_R$. The stopped process is bounded, so the optional stopping theorem givesExit from the bounded interval occurs almost surely. Taking $r\downarrow 0$ with $R$ fixed shows that zero cannot be hit before $R$, hence cannot be hit in finite time. Taking $R\to \infty$ instead givesIf the all-time infimum were zero, every level $r>0$ below $x$ would be hit, by continuity. Letting $r\downarrow 0$ proves the all-time minimum of a transient Bessel process assertion:The first claim in the printed question needs a qualification when $\delta <2$: it holds before the first hit of zero, or for the process stopped there, rather than for the usual reflecting continuation. For example, at $\delta =1$ the proposed power is $X$ itself, a Reflected Brownian motion; the Tanaka formula contains a nonzero local time of a semimartingale term, so it is not a local martingale after reflection. A positive starting point is also needed for the negative power when $\delta >2$; an entrance process started at zero would give an infinite value at time zero.

Use positive initial asset prices and the usual augmentation of the natural Brownian filtration. Define the market price of riskContinuity and strict positivity of $\sigma$ imply ${\int }_0^T{\lambda }_t^{2}dt<\infty$ on every finite horizon almost surely, because each path has a positive minimum of $\sigma$ there. Hence the strictly positive stochastic exponentialis defined and has $Y_0=1$ and $d{Y}_t=-Y_t(r_{t}dt+{\lambda }_{t}d{W}_t)$. The Itô product rule givesso it is a local martingale deflator.