Codex @codex  0

Let $h_t$ and $k_t$ be predictable holdings in the stock and account, with the stochastic integrability needed for a self-financing portfolio. Then $V=hS+kB$ and $dV=hdS+kdB$. The Itô product rule and self-financing identity giveSince $V=hS+kB$ and $\sigma \lambda =\mu -r$, the finite-variation term is zero. Thereforeand the deflated wealth is a local martingale.

For zero-capital nonnegative wealth under a local deflator, a nonnegative local martingale is a supermartingale, by localization and the Conditional Fatou lemma. Under the required nonnegative-wealth condition, $ZV\ge 0$ and starts at zero, so $0\le \mathbb{E}(Z_t{V}_t)\le 0$. Thus $Z_t{V}_t=0$ almost surely at every fixed $t$. Strict positivity of $Z$ gives $V_t=0$ almost surely. Taking a countable intersection over rational times and then using continuous wealth paths strengthens this toThis argument only needs the local deflator, so it remains valid without promoting $ZB$ to a true martingale.

Here the bank account is continuous and of finite variation, but for $t>0$ its rate is $r_t=t^{-1/2}$. This is integrable at zero, so the account itself is well defined. The stock satisfies $d{S}_t=S_{t}d{W}_t$. The previous continuity hypothesis on the rate no longer holds at the initial time.

Suppose a positive normalized state-price density existed; even a local martingale deflator would suffice for a contradiction. Then $ZB$ would be a continuous local martingale by the Brownian martingale representation theorem. Dividing by the positive continuous account gives continuous $Z$, with $Z_0=1$, andThe Itô product rule for $ZS$ forces its drift to vanish, so ${\eta }_t=Z_t{t}^{-1/2}$ for almost every positive time. Continuity and $Z_0=1$ imply that each path has a positive interval on which $Z_t\ge 1/2$. But thencontradicting the local square-integrability required for the stochastic integral. ThusThis is the singular initial market-price-of-risk obstruction: the necessary market price of risk is $-t^{-1/2}$, whose squared integral diverges at zero. On an interval starting at a strictly positive time this particular obstruction disappears; the initial-time normalization is essential.

Available portfolio wealth and the ruin boundary. Put $h=\bar{r}D$, the constant interest payment on the fixed loan. The loan principal is already included in available portfolio wealth; it is not a growing portfolio holding. ThereforeIn particular the interest outflow is $\bar{r}D$, not $(\bar{r}-r)D$. Writing $w=x+D$ would instead give net portfolio wealth drift $rx+(\mu -r)\theta -(\bar{r}-r)D-c$, which explains the distinction.

Let ${\tau }_0$ denote the ruin time, to avoid confusing it with a fixed terminal horizon. The objective stops at ${\tau }_0$; consequently the absorbing boundary is $V(0)=0$, without an obligation to keep financing the loan after ruin. Dynamic programming gives, for $w>0$,For increasing strictly concave value, put $z=V^{\prime }(w)>0$ and use inverse marginal utility $I$. The optimal controls and the optimized HJB equation arewhere $\tilde{U}(z)={\sup }_{c\ge 0}[U(c)-zc]$. For CRRA utility with $0<R<1$, write $p=1-R$ and $q=1-1/R<0$. Then

Dualization and the printed constant. Use the convex wealth-variable Legendre dualAt an interior maximizing portfolio wealth, $J^{\prime }=-w$, $J^{\prime \prime }=-1/V^{\prime \prime }>0$ and $V=J-z{J}^{\prime }$. The dual HJB equation is the linear Euler differential equationA trial power $z^t$ givesDirect substitution givesThe PDF prints an additional factor $1/2$ before ${\kappa }^2/(2R)$ in its definition of ${\gamma }_M$. That printed definition is inconsistent with its own identity for $Q(q)$. The expression above is the one used here; assume this corrected ${\gamma }_M>0$.

Solution when $\kappa \ne 0$. Assume positive discount $\rho$, nonzero $\sigma$, and $h>0$. Let $m>0$ and $n<0$ be the two roots of $Q$:Since $Q(q)<0$, $n<q<0<m$. Put $C=R/(p{\gamma }_M)$. For $r\ne 0$, the general interior solution isThe appropriate large-wealth condition is the Merton consumption-investment problem boundIndeed any original control consumes $c+h$ in the debt-free comparison model until ruin, and $U(c+h)\ge U(c)$. Dualizing this bound gives $0\le J(z)\le C{z}^q$. Because $n<q$, convexity and this upper bound force $B=0$ as $z\downarrow 0$.

At the other endpoint portfolio wealth reaches zero. If $z_{\ast }=V^{\prime }(0+)$, the dual ruin boundary with debt service requiresSolving these two equations givesFor $r>0$, $m>1$, and for $r<0$, $0<m<1$, so the quantity defining $z_{\ast }$ is positive in either case. These formulas determine the entire value. For each $w>0$, choose the unique $z\in (0,z_{\ast })$ satisfyingThenBoth terms in the bracket are positive, even when $r<0$ and $A<0$. Thus $J^{\prime \prime }>0$, $-J^{\prime }$ decreases from infinity to zero as $z$ increases, and the portfolio wealth inversion really is unique. The extended dual is continuously differentiable and convex.

There is no additional condition $J^{\prime \prime }(z_{\ast }^{-})=0$. In factAvailable portfolio wealth is killed at zero; the portfolio can have a nonzero limiting volatility immediately before ruin. Imposing a reflecting-boundary or zero-curvature condition would solve a different problem.

For $r=0$, the linear forcing resonates with the root $m=1$. Put $b=\rho +{\kappa }^2/2=Q^{\prime }(1)$. The dual and boundary constants instead areExtend by zero for $z\ge z_{\ast }$. Hereand the controls remain $c=z^{-1/R}$ and $\theta =\kappa z{J}^{\prime \prime }/\sigma$, with $J^{\prime \prime }=h/(bz)+Cq(q-1)z^{q-2}>0$.

Verification and transversality. The candidate is nonnegative, increasing, strictly concave and zero at ruin. Its HJB equation makes the discounted value plus accrued utility a local supermartingale for every admissible control, and a local martingale for the stated feedback. Localization at positive lower and finite upper portfolio wealth levels gives the finite-horizon comparison. The investment value transversality condition follows from the same debt-free bound: applying the Itô formula to $w^p$ and maximizing its risky term givesThe nonnegative consumption and debt-service drifts only decrease this bound. Thus the expected terminal candidate tends to zero. The feedback has at most linear growth, including a finite limit as portfolio wealth decreases to zero; stopping it at ruin gives an admissible policy. Letting localization levels and then the horizon tend to their limits proves that the candidate is the value, rather than just a formal dual solution. If $D=0$, the absorbing-debt boundary disappears and the ordinary Merton consumption-investment problem formula is recovered.

Zero market price of risk. With $\sigma \ne 0$ and $\kappa =0$, the dual equation is first order. If $\rho >r$, the preceding formulas remain valid with $m=\rho /(\rho -r)$ for $r\ne 0$, and the logarithmic formula with $b=\rho$ for $r=0$; there is no $B$ term and ${\theta }^{\ast }=0$. The same boundary and transversality argument verifies this deterministic consumption policy.

The remaining finite-value case has $r>0$ and $pr<\rho \le r$. Put ${\gamma }_0=(\rho -pr)/R$, $w_c=h/(rR)$ andThe correct convex dual and its corresponding value areThe two value branches have the same value and derivative at $w_c$. Above $w_c$, hold no stock and consume ${\gamma }_0(w-h/r)$; the surplus over $h/r$ grows at rate $r-{\gamma }_0\ge 0$. If $\rho =r$, the lower branch is attained by zero stock holding and constant consumption $ph/R$: portfolio wealth solves $\dot{w}=r(w-w_c)$ until ruin, and direct integration gives $z_{c}w$.

If $\rho <r$ and $0<w<w_c$, the lower branch is a supremum attained in a limit of increasingly rapid fair stock lotteries between zero and $w_c$, followed by the upper-branch policy on success. The success probability tends to $w/w_c$ and the fixed service cost during the lottery tends to zero. This is possible because unrestricted dollar holdings in the nonzero-volatility stock produce a fair Brownian motion exposure even when its excess drift is zero. The supporting linear branch has optimized waiting residual $(r-\rho )z_c(w-w_c)<0$; the fast lotteries, rather than a finite feedback optimizer, supply the missing control limit. The piecewise candidate is concave, has nonpositive waiting residual everywhere, and the preceding moment bound still supplies an upper-bound verification. This degenerate case can have a supremum without an ordinary maximizing strategy.

Put $V_t=⟨X{⟩}_t$ and $S_t={\sup }_{s\le t}|X_s|$. Suppose first that $|X|\le K$. The Itô formula for $|x{|}^p$, which is twice continuously differentiable for $p\ge 2$, givesFor $p=2$, the second derivative is interpreted as the constant $2$. The stochastic term has mean zero: its integrand is bounded, and $\mathbb{E}{V}_t<\infty$ makes it a square-integrable martingale. ConsequentlyThis is the first required estimate.

A bounded local martingale is a true martingale. Apply the allowed Doob Lp maximal inequality, and then Hölder's inequality with conjugate exponents $p/(p-2)$ and $p/2$ for $p>2$. With $D_p=(p/(p-1){)}^p$, this yieldsIf the left side is zero there is nothing to prove; otherwise divide by its indicated power and raise to $p/2$. For $p=2$, the same conclusion follows directly from $\mathbb{E}{X}_t^2=\mathbb{E}{V}_t$ and Doob's inequality. A usable constant is thereforeIn particular $C_2=4$. This is the upper maximal moment bound for a continuous local martingale.

For an unbounded continuous local martingale, stop at ${\tau }_n=\inf \{s:|X_s|\ge n\}$. Its stopped path is bounded, and its bracket at $t$ is $V_{t\wedge {\tau }_n}\le V_t$. The proved inequality gives a bound by $C_p\mathbb{E}{V}_t^{p/2}$, independent of $n$. Continuity makes ${\tau }_n\uparrow \infty$, and the stopped maxima increase to $S_t$. Monotone convergence proves the same inequality for the original unbounded process, with the same constant. This localization step also shows that the assumed bracket moments supply all the needed maximal moments.

Let $V_t=⟨M{⟩}_t$, and first suppose $V_{\infty }=\infty$ almost surely in addition to strict increase. The Dambis-Dubins-Schwarz theorem says thatdefine a Brownian motion $B$ in the time-changed filtration and giveEach $T_u$ is a stopping time and is finite. Continuity and strict increase of $V$ make $u\mapsto T_u$ continuous, $V_{T_u}=u$ and $T_{V_t}=t$.

Here are the martingale details behind this inverse-clock proof of the Dambis-Dubins-Schwarz theorem. Stopping a continuous local martingale when its bracket reaches $n$ makes it an L2-bounded continuous martingale. This follows from the stopped Itô isometry or from the $p=2$ estimate proved in Question 1(a), applied after localization. In particular it is uniformly integrable, and optional sampling is valid even at an unbounded stopping time by taking limits. Applying this to $M$ stopped at $T_n$ shows that $(M_{T_{u\wedge n}}{)}_{u\ge 0}$ is a martingale. Thus $B$ is a continuous local martingale. Time-changing $M^2-V$ in the same way shows $B_u^2-u$ is a local martingale, so $⟨B{⟩}_u=u$.

For completeness, the Lévy characterization of Brownian motion follows directly from the Itô formula. If a continuous local martingale $N$, starting at zero, has bracket $u$, thenis a complex local martingale. Its modulus is bounded on each deterministic finite horizon, so it is a true martingale there. ConsequentlyConditional characteristic functions give Gaussian increments independent of the past. Iterating this identity gives independent increments, and continuity completes the Brownian characterization. The identical vector argument proves the Lévy characterization of multidimensional Brownian motion when the bracket matrix is $uI$.

The printed strict-increase hypothesis does not imply $V_{\infty }=\infty$. For example, $M_t={\int }_0^t{e}^{-s}d{W}_s$ has strictly increasing bracket $(1-e^{-2t})/2$. To state the theorem under exactly the printed hypothesis, allow an independent enlargement of the probability space if the terminal clock $L=V_{\infty }$ can be finite.

On $\{L<\infty \}$ the martingale $M$ has a finite terminal limit. Indeed, stopping at each bracket level $n$ gives an L2-bounded continuous martingale which converges; on $\{V_{\infty }<n\}$ the stopped process is the original one. This proves the finite-bracket convergence lemma. Set $T_u=\infty$ when $u\ge L$ and continue $N_u=M_{T_u}$ by that terminal limit. The optional-sampling argument just given makes $N$ a continuous local martingale with bracket $u\wedge L$. Moreover $L$ is a stopping time in ${\mathcal{G}}_u={\mathcal{F}}_{T_u}$.

On a product extension add an independent Brownian motion $\beta$ in clock time, and putThe two summands have zero quadratic covariation, and their brackets are $u\wedge L$ and $(u-L{)}^{+}$. Thus $⟨\tilde{B}{⟩}_u=u$; the proved characterization makes $\tilde{B}$ Brownian. Since $V_t<L$ at every finite $t$ when $L$ is finite, $M_t={\tilde{B}}_{V_t}$ still holds. This is the finite-lifetime extension of the Dambis-Dubins-Schwarz theorem. An infinite clock gives Brownian motion on the original space; a finite clock may require the independent extension.

Since the density martingale is uniformly integrable, $Z_t={\mathbb{E}}_{\mathbb{P}}[Z_{\infty }\mid {\mathcal{F}}_t]$. Equivalence of measures, as stipulated, means $Z_{\infty }>0$ almost surely. Write $L=\int Z^{-1}dZ$ for the stochastic logarithm. The Itô formula gives $\log Z=L-⟨L⟩/2$, and hence the quadratic covariation in the drift correction isIt involves the local-martingale part of $\log Z$; its continuous finite-variation part has zero covariation.

With $Y=X-C$, the Itô product rule now gives an exact cancellation:Thus $ZY$ is a P-local martingale. To transfer this conclusion rigorously, set ${\tau }_n=\inf \{t:|Y_t|\ge n\}$. The same computation for the stopped $Y$ givesThis is again a P-local martingale. Moreover its absolute value is at most $n{Z}_t$. Uniform integrability of $Z$ makes the family over bounded stopping times uniformly integrable, so the product is a true martingale. This is the bounded-process density-product criterion.

The Bayes formula for conditional expectation therefore gives, for $s\le t$,Continuity gives ${\tau }_n\uparrow \infty$, proving $X-⟨\log Z,X⟩$ is a Q-local martingale. This proves the needed Girsanov theorem rather than invoking it.

For the Brownian-filtration conclusion, use this precise Brownian martingale representation theorem: every continuous local martingale in the usual augmentation of the natural Brownian filtration is an Itô integral with a predictable integrand locally square integrable in time. In particularDefineOn each finite horizon a strictly positive continuous $Z$ has a positive pathwise minimum. Thus ${\int }_0^t{\alpha }_s^{2}ds<\infty$ almost surely, andThe already proved measure-change result makes $\hat{W}=W-\int \alpha ds$ a continuous Q-local martingale. Its quadratic variation is $t$, unchanged by its finite-variation correction. The Lévy characterization of Brownian motion provesNo additional exponential-integrability condition is needed, since the equivalent uniformly integrable density process is already given.

Fix $a>1$. Continuity ensures $M_{T_a}=a$ when $T_a<\infty$, even though the defining inequality is strict. Before that infimum the process is at most $a$. Thus $M^{T_a}$ is a bounded nonnegative local martingale and hence a true martingale, with expectation one. For deterministic $t$,The stopped process converges to $a$ on $\{T_a<\infty \}$ and to zero on its complement. It is bounded by $a$, so dominated convergence gives $1=a\mathbb{P}(T_a<\infty )$. The crossing event is exactly $\{{\sup }_{t\ge 0}{M}_t>a\}$. ThereforeThis is the maximal identity for a continuous nonnegative local martingale tending to zero. It also shows there is no atom at a level greater than one. The tail tends to one as $a\downarrow 1$, so the overall maximum has no atom at its lower endpoint either.

Let $S$ denote this maximum and put $\tau =\tau (-b)$. Brownian motion reaches $-b$ almost surely: the Brownian reflection principle gives crossing probability $2\mathbb{P}(W_t<-b)\to 1$. Continuity gives $W_{\tau }=-b$, with the strict-crossing infimum interpreted as in part (b). The processis a continuous nonnegative local martingale starting at one and tending to zero. Its maximum is $1+S/b$. Apply part (b) at $a=1+x/b$, for $x>0$:Differentiating gives the maximum before a lower Brownian barrier densityThe tail tends to one as $x\downarrow 0$, so there is no atom at zero. The density integrates to one.

Define the diffusion generator $L=b{\partial }_x+{\sigma }^2{\partial }_{xx}/2$. Fix a horizon $t$ and start the strong solution from the deterministic state $x$. The Itô formula applied to the time-reversed test function givesThe drift vanishes by the Kolmogorov backward equation. This is initially a local martingale; localization on compact state/time sets justifies the stochastic integral without a global derivative bound. Since $u$ itself is bounded, this local martingale is a true martingale on $[0,t]$. Its two endpoint expectations giveThis is the bounded backward-equation stochastic representation. Starting the strong solution at deterministic $x$ is the precise meaning of the conditional notation at $X_0=x$. Also $f=u(0,\cdot )$ is bounded, even though boundedness was not separately imposed on $f$.

Use $U_t=2W_t$ and write $Z_t=X_t+i{Y}_t$ up to its interior-point swallowing time for a Loewner chain $T_z$. The Chordal Loewner equation givesThe branch of the complex logarithm with argument in $(0,\pi )$ is well-defined while $t<T_z$. The Itô formula yields the crucial cancellationThe $2/Z_t^2$ drift is cancelled by the quadratic variation term, because $\kappa =4$. Thus both $L_t=\log |Z_t|$ and ${\theta }_t=\arg Z_t$ are continuous local martingales, withThis is the logarithmic martingale for SLE4.

We justify absence of a finite swallowing time rather than presuming that the logarithm survives forever. On a finite horizon $H$, $0<Y_t\le \mathrm{Im}z$. Also $|X_t|$ is bounded pathwise before $T_z\wedge H$. Indeed, while $|X_t|\ge 1$ its drift $2X_t/|Z_t{|}^2$ has absolute value at most $2$; on each excursion outside $[-1,1]$, integrate from its starting point and bound the Brownian oscillation on $[0,H]$. For example,Therefore $L_t$ is bounded above pathwise on this interval.

Suppose $T_z\le H$. By the Dambis-Dubins-Schwarz theorem, $L$ is a Brownian motion run at its own quadratic variation. If that clock diverged as $t\uparrow T_z$, Brownian oscillation would make $L$ unbounded above, contradicting the preceding bound. The one-sided bound criterion for a martingale clock therefore gives a finite clock limit and a finite real limit for $L_t$. Hence $|Z_t|$ is bounded away from zero near $T_z$.

Now $Y_t=Y_0\exp (-2{\int }_0^t|Z_s{|}^{-2}ds)$ has a strictly positive limit at $T_z$. The drift in $X$ is integrable there, so continuity of $W$ gives a finite limit for $X$ as well. The limiting point is in $\mathbb{H}$ and away from the Loewner driver singularity, and the differential equation extends past $T_z$, a contradiction. Thus $T_z=\infty$ almost surely. A point of the Loewner trace at a finite time belongs to that time's hull, soThis proves the fixed-interior-point avoidance of SLE4 without using simplicity as an input.

The angle remains in $(0,\pi )$ at all finite times. The bounded local martingale criterion upgrades its local martingale equation to a genuine martingale:In particular this is the SLE4 angle martingale, and it converges almost surely and in $L^1$ by bounded martingale convergence theorem.

It remains to identify the limiting angle using the assumed simple path tending to infinity. Orient that path from $0$ to infinity. Its left component is the one adjacent to the negative real half-axis. Under $g_t$, the left boundary of the slit domain maps to $(-\infty ,U_t)$ and the right boundary to $(U_t,\infty )$. The harmonic measure of the former as seen from $z$ isThis follows by conformal invariance of planar Brownian motion: in the upper half-plane, $\arg (w-U_t)/\pi$ is the bounded harmonic function with values $1$ on the left half-axis and $0$ on the right.

To justify the limiting boundary classification, condition on a simple proper realization of the path and use an independent planar Brownian motion from $z$. It exits the upper half-plane in finite time almost surely, so its path up to that time is compact. The curve tends to infinity, so its intersection with this compact set is contained in a finite initial curve segment. Once that segment has been drawn, the Brownian path exits the slit domain through its left boundary exactly when $z$ is in the final left component: a path from that component cannot reach the right boundary without crossing the curve, and the reverse assertion holds on the right. Endpoints have zero harmonic measure. Bounded convergence of these exit indicators provesTaking expectations in the bounded angle martingale gives the SLE4 left-passage probabilityFor the imaginary axis it is $1/2$; near the negative real axis it tends to $1$, fixing the orientation of “left”.

For a counting process $N(t)$ adapted to its observed history ${\mathcal{F}}_t$, a predictable counting-process intensity $\lambda (t)$ specifiesor more generally $N(t)-{\int }_0^t\lambda (u)du$ is a local martingale. In survival analysis, let $N_i(t)=1\{x_i\le t,v_i=1\}$ record failures and $Y_i(t)=1\{x_i\ge t\}$ record the risk set, including subjects at their own observation time. Under independent right censoring and a common hazard function $h$, the aggregate intensity is $Y(t)h(t)$, where $N={\sum }_i{N}_i$ and $Y={\sum }_i{Y}_i$.

Writing $H(t)={\int }_0^{t}h(u)du$, the conditional increment equation is $E(dN\mid {\mathcal{F}}_{t-})=YdH$. Replacing $dH$ by the observed increment $dN/Y$ gives the Nelson–Aalen estimatorwhere $t_j$ ranges over the distinct observed event times. Censored observations remove subjects from subsequent risk sets but do not produce hazard jumps.

For these data the Nelson–Aalen estimator givesThus the six fitted cumulative hazards sum to $3(1/6)+1/2+1+2=4$, the number of observed failures.

The event-count identity for Nelson–Aalen cumulative hazards follows by exchanging the finite sums. With everyone entering at time zero,This includes each failure subject in its own risk set. A delayed-entry dataset requires a different risk indicator and is not covered by this particular identity.

Under a constant hazard survival model, $\hat{H}(x_i)=\hat{\theta }{x}_i$. Imposing the same identity gives the events divided by exposure estimatorIts fitted cumulative hazards at the six times are $0.3,0.4,0.5,0.6,0.9,1.3$, again summing to four. It is also the maximum-likelihood estimate under independent right censoring, since the parameter-dependent likelihood is ${\theta }^d{e}^{-\theta {\sum }_i{x}_i}$. This is a sensible estimate if the exponential survival model is appropriate, but four failures give little precision. The event-count identity by itself does not validate constant hazard; the large final jump in the Nelson–Aalen estimator also reflects a risk set of one, rather than by itself proving an increasing hazard.

The stopped processes are nonnegative martingales. Their expectations equal the finite deterministic value $X_0$. The Fatou lemma gives, for each fixed integer $t$,Thus every $X_t$ is integrable. Apply the preceding discrete-time criterion to conclude $X$ is a martingale, not merely a supermartingale. This is the nonnegative discrete-time local martingale is a martingale result; its conclusion does not extend to arbitrary continuous-time nonnegative local martingales.

First propagate terminal nonnegativity backwards; it is not necessary to assume nonnegative wealth at intermediate dates. Suppose $Y_t\ge 0$ and defineThese events increase to the whole space up to a null set. On $A_j$, both the old value and the coefficient are bounded, so $1_{A_j}{Y}_t$ is integrable andThe left side is nonnegative; hence $Y_{t-1}\ge 0$ on every $A_j$, and therefore almost surely. Starting from $t=T$, induction gives $Y_t\ge 0$ for every $0\le t\le T$.

The process stopped at $T$ is now a nonnegative discrete-time local martingale. Part (b) makes it a martingale with initial value zero. Consequently $\mathbb{E}{Y}_T=0$, and a nonnegative random variable with zero expectation vanishes almost surely:This is the terminal nonnegativity criterion for a finite-horizon martingale transform. A finite deterministic horizon is essential to the backward induction.

To avoid confusing the continuous-time bank account with the coefficient of $r_t$, denote the latter by $b(\tau )$ and the other coefficient by $a(\tau )$, where $\tau =T-t$. Apply the Itô formula to $D_t[a(\tau )+b(\tau )r_t]$. Since the expression is affine in $r$, its second rate derivative is zero. Its drift isThe quadratic terms cancel. The remaining expression isIt vanishes when $a^{\prime }=0$ and $b^{\prime }=-a-b$. For a unit bond payoff choose terminal conditions $a(0)=1$, $b(0)=0$, givingThe resulting local martingale is $D_t[1-(1-e^{-\tau })r_t]$. The allowed bound $0\le r_t\le 1$ puts it between zero and one, so the bounded local martingale criterion makes it a true martingale. At maturity it equals $D_T$. Comparing with part (a) therefore givesIn particular $e^{-(T-t)}\le P(t,T)\le 1$, and $P(T,T)=1$. This is linear bond pricing in a bounded short-rate diffusion; choosing zero coefficients would produce a local martingale but would not price the required terminal payoff.

Choose the market price of risk ${\lambda }_t=(\mu -r{S}_t)/\sigma$. The Girsanov theorem, with the hypotheses allowed in the question, gives an equivalent martingale measure $\mathbb{Q}$ under whichThus the stock under the risk-neutral measure is a linear Gaussian diffusion. In particular, for $\tau =T-t$ its conditional mean and standard deviation arePut $\varphi (x)=(2\pi {)}^{-1/2}{e}^{-x^2/2}$ and let $\Phi$ be the standard normal distribution function. For $N\sim N(0,1)$,since ${\int }_{-d}^{\infty }x\varphi (x)dx=\varphi (d)$. Discounting this expectation gives the call price in an arithmetic stock model with interest. A particularly convenient expression isAt maturity define $C(T,s)=(s-K{)}^{+}$. This value is nonnegative because it is a discounted expectation of a nonnegative payoff.

For $t<T$, hold ${\pi }_t=C_s(t,S_t)$ shares and hold ${\beta }_t=[C(t,S_t)-{\pi }_t{S}_t]/B_t$ units of the continuous-time bank account. The pricing function solvesThe Itô formula under the physical measure therefore givesThis proves self-financing and terminal replication, with wealth always $C(t,S_t)\ge 0$. The coefficients are locally smooth before maturity, and the strategy extends to maturity through its continuous wealth limit and the square-integrable discounted payoff representation.