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A share paying a continuous dividend rate ${\delta }_t$ produces dollar gains $d{S}_t+{\delta }_{t}dt$. If its price follows a geometric Brownian motion with drift ${\mu }_{\mathrm{price}}$ and its dividend yield ${\delta }_t/S_t$ is constant, its drift in an investment-consumption problem is their sum. The self-financing portfolio equation uses this total return, even if the agent consumes the dividends rather than reinvesting them. Omitting the dividend yield changes both the optimal stock fraction and the equilibrium interest rate.

In the basic structural model, the firm's asset value $V$ follows geometric Brownian motion with constant volatility and a single debt obligation of face value $F$ matures at $T$. Default at maturity occurs when $V_T<F$. Equity is a call on firm assets; debt is the risk-free face claim minus a put on those assets, since $\min (V_T,F)=F-(F-V_T{)}^{+}$. This links debt valuation to asset value, volatility and capital structure. The basic model does not permit default before maturity; richer structural models add a default barrier or more detailed obligations.

On a filtered probability space carrying a standard Brownian motion $W$ with independent increments relative to the filtration, a strictly positive security price follows geometric Brownian motion with constant drift $\mu$ and volatility $\sigma$ if it solves the stochastic differential equationHere $W_t-W_s$ is independent of ${\mathcal{F}}_s$ and has the normal distribution $N(0,t-s)$ for $s<t$. The solution isIndeed applying the Itô formula to this exponential gives the specified drift and volatility. Consequently the price is positive and continuous, andThus disjoint log-return increments are independent and stationary. The parameter $\mu$ is the instantaneous expected relative return under the physical probability measure, not the mean log-return. Under the pricing risk-neutral measure it becomes the risk-free rate for a non-dividend-paying stock.

One alternative is a stochastic volatility model, such as the Heston model:The parameters satisfy $\kappa ,\theta ,\xi >0$ and $|\rho |\le 1$, and $v_0\ge 0$. The square-root Itô diffusion has a nonnegative solution; $2\kappa \theta \ge {\xi }^2$ is a standard sufficient condition for the positive initial variance not to hit zero. Mean reversion of $v_t$ allows persistent high- and low-volatility episodes, and negative $\rho$ can produce the empirical association between falling prices and rising volatility. Mixtures of conditional return distributions produce richer tails and option implied volatility shapes than constant-volatility geometric Brownian motion.

For data showing volatility clustering and option smiles, this is often a more useful fit. It is not universally superior: more parameters create estimation and calibration uncertainty, the model is more expensive to use, and its continuous price paths still exclude price jumps. With two independent noise directions and only a stock and a bank account, the market is generally incomplete, so a volatility risk premium or an extra traded instrument is needed for unique contingent claim pricing. The alternative improves flexibility at the cost of calibration and hedging complexity.

Assume a frictionless market with no taxes or transaction costs, unrestricted short selling and borrowing/lending at the same constant rate $r$, continuous trading, and absence of arbitrage. The non-dividend-paying stock has geometric Brownian motion dynamics $dS=\mu Sdt+\sigma SdW$ with constant $\sigma >0$, and the bank account satisfies $dB=rBdt$. A contingent claim with no interim cash flows has sufficiently regular value $f(t,S)$, meaning $f\in C^{1,2}$ before maturity; terminal financial payoff kinks can be handled by the smooth value for earlier times. Trading strategies are adapted, self-financing portfolios, and admissible trading strategies so that doubling strategies are excluded.

The Itô formula givesChoose $\Delta =f_S$ shares and put the remaining value $f-S{f}_S$ in the bank account. Its self-financing gain isMatching the contingent claim's gain to this replicating gain cancels the random increment and equates the drifts:This is the Black-Scholes equation. The argument uses the self-financing gain $\Delta dS$, not a product differential that incorrectly treats a changing $\Delta$ as free cash. Rebalancing purchases are financed from the bank account. Conversely, a smooth solution with suitable growth and boundary conditions defines a delta hedge whose value replicates the contingent claim; law of one price identifies this value with the contingent claim price. Here $S$ is the current stock price, $t$ is time, $\sigma$ is proportional volatility, $r$ is the risk-free rate, and $f$ is the contingent claim price. The physical expected return $\mu$ cancels because exposure to price risk has been hedged.

For strike $K>0$ and maturity $T$, a European call option and a European put option have boundary conditionsThe Black-Scholes equation is solved backward from these conditions. At zero stock price, the zero state is absorbing in geometric Brownian motion, soAt the other boundary, with $\tau =T-t>0$,These conditions and a suitable growth bound select the financial solution and provide boundary data for numerical truncations. In the constant-coefficient model, under the risk-neutral measure,For $d_1=[\log (S/K)+(r+{\sigma }^2/2)\tau ]/(\sigma \sqrt{\tau })$ and $d_2=d_1-\sigma \sqrt{\tau }$, the exercise event is $Z>-d_2$. Completing the square in $e^{\sigma \sqrt{\tau }Z}$ times the normal density gives the truncated first moment and hencewhere $\Phi$ is the standard normal cumulative distribution function. These are the Black-Scholes formula values and satisfy the specified terminal and boundary conditions.

Put-call parity is the relationA call minus a put has terminal financial payoff $S_T-K$. A portfolio of one share minus a zero-coupon bond paying $K$ has the same financial payoff, so absence of arbitrage and law of one price equate their current values. The formula here assumes the same strike and maturity and no dividends; known dividends modify the stock-side value.

Separate $L$ into a diffusion generator and the constant killing term $-1/2$. The appropriate stochastic differential equation and its explicit geometric Brownian motion solution areThe second-order term in the Itô formula cancels the drift of $\log |X|$ when $x\ne 0$; if $x=0$, the solution stays zero.

Fix a terminal time $t$. Apply the Itô formula to $e^{-s/2}u(t-s,X_s)$ for $0\le s\le t$. Its drift isAfter localization this is a local martingale, and it is bounded because $u$ is bounded. It is therefore a true martingale. Taking expectations at its endpoints gives the Feynman-Kac formula

For $t>0$ and $x\ne 0$, the change of variable $y=x{e}^z$ in the normal density produces the killed geometric Brownian heat kernel with respect to Lebesgue measure:For $xy>0$, an equivalent form isIn particular,The apparent missing $1/y$ in this last expression is absorbed into the completed square, together with the killing factor; it is not a transcription error. The total mass is $e^{-t/2}$, as expected for killing at rate $1/2$. At $x=0$ the fundamental kernel is instead the measure $e^{-t/2}{\delta }_0(dy)$; it has no Radon-Nikodym derivative with respect to Lebesgue measure. Thus $u(t,0)=e^{-t/2}g(0)$. At $t=0$ the kernel is ${\delta }_x$, and continuity of bounded $g$ gives the initial condition. This also specifies the fundamental solution at the degenerate point omitted by a density-only formula.

The power-wealth investment with running utility objective is homogeneous of degree $\gamma$ in wealth: scaling both wealth and positions by $a>0$ scales the reward by $a^{\gamma }$. Accordingly set $J(w,t)=A(t)w^{\gamma }$. ThenFor completeness the remaining equation is $A^{\prime }+\kappa A+e^{-\rho t}=0$, $A(T)=1$, whereThis confirms $J_{ww}<0$. The constant optimal risky fraction also makes wealth a positive geometric Brownian motion when started positive.

Financially the relative risk aversion coefficient is $1-\gamma$. Increasing $\gamma$ reduces risk aversion, raises the risky fraction, and decreases the bond fraction: its derivative is $-(\mu -r)/[{\sigma }^2(1-\gamma {)}^2]<0$. If the risky fraction is below one, some wealth is lent; at one, none is invested in bonds; above one, $B^{\ast }<0$ represents borrowing to leverage the risky position. As $\gamma \uparrow 1$ the risky fraction becomes unbounded, consistent with the absence of a finite unconstrained optimum for linear utility and positive excess drift. As $\gamma \downarrow 0$ the limiting fraction is $(\mu -r)/{\sigma }^2$, the logarithmic-utility allocation obtained by the appropriately rescaled power-utility limit; the literal objective at $\gamma =0$ would be constant and would not select a unique strategy.

For the Cox-Ross-Rubinstein diffusion approximation, fix calendar maturity $T$, take $h=T/n$, and choose $u=e^{\sigma \sqrt{h}}$, $d=e^{-\sigma \sqrt{h}}$, and the one-step bank factor $b=e^{\rho h}$. For sufficiently fine mesh, $d<b<u$. The risk-neutral probability expands asEach stock log-increment is $\pm \sigma \sqrt{h}$. Its mean under $Q$ is $(\rho -{\sigma }^2/2)h+O(h^2)$ and its variance is ${\sigma }^{2}h+O(h^2)$. Independent increments, their vanishing maximum size, and the Donsker invariance principle for this triangular array give the path limitExponentiation gives the geometric Brownian motion of the Black-Scholes model, and the bank account tends to $e^{\rho t}$.

The physical model is matched by taking $p_h=(e^{\mu h}-d)/(u-d)$. Its corresponding limit has log drift $\mu -{\sigma }^2/2$, so $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t^P$. The likelihood ratios from part (ii) have the matching continuous-time limit. With $\lambda =(\mu -\rho )/\sigma$, a Taylor expansion of their two possible log-multipliers giveswhere ${\epsilon }_k$ is $+1$ or $-1$. Thus the limiting Radon-Nikodym derivative is the Girsanov theorem density $\exp (-\lambda W_T^P-{\lambda }^{2}T/2)$, turning the physical stock drift into $\rho$.

The binomial risk-neutral pricing formula converges to $e^{-\rho T}{\mathbb{E}}_Q[f(S_T)]$ for bounded continuous terminal payoffs by convergence in distribution. For ordinary call and put payoffs, uniform stock second-moment bounds give the required uniform integrability and hence convergence despite unboundedness. Bounded payoffs with discontinuities only at events of zero limiting probability are handled by the same weak-convergence criterion. Corresponding continuous path functionals can be treated at the process level. Arbitrary discontinuous path claims do not follow from weak convergence without further conditions. In the smooth limiting model, the discrete hedging difference quotient becomes the delta hedge $V_x$, and its value satisfies the Black-Scholes equation.

Use the usual Black-Scholes model with $\sigma >0$ and the filtration generated by $W$. SetFor $u>t$, independence and the Gaussian distribution exponential moment formula give ${\mathbb{E}}_P[Z_u/Z_t\mid {\mathcal{F}}_t]=1$. Thus $Z$ is a strictly positive martingale of mean one, defining an equivalent measure $Q$ on every finite horizon by $dQ/dP{|}_{{\mathcal{F}}_t}=Z_t$.

To verify the measure change directly, put ${\hat{W}}_t=W_t+\lambda t$. For any real $a$ and $h=u-t$,The conditional characteristic function is that of $N(0,h)$ and does not depend on past information. Therefore $\hat{W}$ has independent increments with Gaussian distribution and continuous paths, and is a Brownian motion under $Q$. Substituting $dW=d\hat{W}-\lambda dt$ gives $d{S}_t=S_t(rdt+\sigma d{\hat{W}}_t)$. Hence $S_t/B_t$ is a geometric Brownian motion that is a martingale and $Q$ is an equivalent martingale measure.

The form is also forced. By the Brownian martingale representation theorem, any density process that is a martingale has $d{Z}_t=a_{t}d{W}_t$. The change of measure identity makes $Z_t{S}_t/B_t$ a local martingale under $P$. Its drift, calculated by Itô formula, is $(S_t/B_t)[Z_t(\mu -r)+\sigma a_t]dt$, which must vanish. Thus $a_t=-\lambda Z_t$. Solving $d{Z}_t=-\lambda Z_{t}d{W}_t$ with $Z_0=1$ gives exactly the displayed stochastic exponential. Equivalence here is on each finite horizon, as usual for Black-Scholes pricing.

For constant relative risk aversion utility, $I(z)=z^{-1/R}$. Define the Merton consumption constantThe Gaussian exponential moment givesWhen ${\gamma }_M>0$, the budget is $w_0=y^{-1/R}/{\gamma }_M$. Conditional pricing then gives $c_t^{\ast }={\gamma }_M{w}_t^{\ast }$, and the Merton consumption-investment problem hasThe resulting wealth is a geometric Brownian motion, so positive initial wealth remains positive. Substitution in the Hamilton-Jacobi-Bellman equation verifies the first-order conditions and the value; the budget argument in part (a) supplies global optimality.

The problem has a finite value precisely when ${\gamma }_M>0$. For $0<R<1$ and ${\gamma }_M\le 0$, the supremum is $+\infty$. Indeed, use the displayed risky fraction and consumption $c_t=k{w}_t$, $k>0$. Its expected utility isFor ${\gamma }_M<0$, sufficiently small $k$ already gives infinite utility. For ${\gamma }_M=0$, the finite values tend to infinity as $k\downarrow 0$.

For $R>1$ and ${\gamma }_M\le 0$, every admissible stream has utility $-\infty$. To see this, put $q=(R-1)/R$ and apply Holder inequality on $[0,T]\times \Omega$:The left side diverges as $T\to \infty$, forcing the positive utility-cost integral to diverge. Thus there is no finite-value optimization problem in either ill-posed case.

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.

Interpret the exponential distribution change time as an exogenous time independent of the stock's Brownian motion, or, equivalently for this calculation, as having constant conditional intensity $\lambda$ while the rate remains high. Its marginal distribution alone would not imply that conditional intensity if the change time carried information about stock returns. Write $r_1=r+\epsilon$, $p=1-R$, $B_1=B(r_1)$, and let $V_1$ be the prechange value function.

Over a short interval $dt$, a rate change occurs with probability $\lambda dt+o(dt)$ and changes continuation value from $V_1(w)$ to $V_0(w)$. The dynamic programming equation is consequentlyThis jump term distinguishes the exponential interest-rate switch from a permanent higher bank rate.

By the homogeneity of CRRA utility, put $V_1(w)=K_1{w}^p/p$, with $K_1>0$. Differentiating the strictly concave expressions in $c$ and $\pi$ givesThe optimized consumption terms contribute $R{K}_1^{1-1/R}{w}^p/p$, and the optimized investment terms contribute $K_1{B}_1{w}^p$. With $K_0={\gamma }_0^{-R}$, the Hamilton-Jacobi-Bellman equation reduces toEquivalently,The left side is continuous and strictly increasing for ${\gamma }_1>0$, with limits zero and infinity. Thus $D_1>0$ gives one and only one positive root. This scalar equation is an explicit characterization for arbitrary real $R$; elementary closed forms are not generally available. For example, when $R=2$,Use $({\pi }_1^{\ast },{\gamma }_1)$ until the rate changes, and $({\pi }_0^{\ast },{\gamma }_0)$ afterwards. If the two rates coincide, the root is ${\gamma }_1={\gamma }_0$; as $\lambda$ tends to infinity, it also tends to ${\gamma }_0$.

For a direct attainability check, use the proposed prechange constant fractions. The geometric Brownian motion moment is $E[w_t^p]=w^p\exp [p(B_1-{\gamma }_1)t]$ on the prechange evolution. Conditioning on the independent exponential distribution time therefore givesThe denominator is positive because the scalar equation gives $D_1+p{\gamma }_1={\gamma }_1+\lambda K_0/K_1>0$. Hence the candidate policy actually achieves the stated value.

One can also verify the upper bound for both signs of $p$ without relying on a sign-sensitive terminal-value argument. Let $j_t$ be the current regime, $w^{\ast }$ the candidate portfolio wealth, andIf $N_t=1_{\{\tau \le t\}}$ and ${\tilde{N}}_t=N_t-\lambda (t\wedge \tau )$, the Itô formula and the scalar equation givewhere the jump term stops after the switch and continuous coefficients are understood predictably. Thus $M$ is a positive martingale deflator: for any admissible portfolio wealth $w$, the process $M_t{w}_t+{\int }_0^t{M}_s{c}_{s}ds$ is a nonnegative local martingale. It follows thatFor the candidate policy, $E[M_t{w}_t^{\ast }]\to 0$. Indeed its surviving prechange term decays at rate $D_1+p{\gamma }_1>0$, and its postchange terms decay at rate ${\gamma }_0>0$; integrating over the switch time preserves convergence to zero. The finite-time moment computations also make its budget equality exact. Consequently $E{\int }_0^{\infty }{M}_t{c}_t^{\ast }dt=M_0{w}_0$. Apply the concave supporting-tangent inequalityand integrate in time and expectation. The budget inequality proves that no admissible policy exceeds the candidate. This verifies the value function, including when $R>1$.

Both ${\gamma }_0>0$ and $D_1>0$ are needed for a finite prechange answer. If the postchange value is already infinite, that problem persists after the almost surely finite switch. If ${\gamma }_0>0$ but $D_1\le 0$, then for $p>0$ a small constant prechange consumption fraction makes the expected utility function before the switch diverge, or makes its supremum diverge when $D_1=0$. For $p<0$, the optimal no-consumption terminal moment before the switch is $w^p{e}^{p{B}_{1}t}$; withdrawals only increase this negative-power moment. The expected magnitude of the negative postchange continuation utility function therefore contains the divergent integral $\lambda K_0{w}^p{\int }_0^{\infty }{e}^{-D_{1}t}dt/|p|$. Thus these cases yield respectively $+\infty$ or $-\infty$, rather than a positive-root solution.

Solving the geometric Brownian motion equation over $[t,T]$ givesThe Brownian increment is independent of ${\mathcal{F}}_t$ and has a normal distribution with variance $T-t$. Its exponential moment-generating function therefore givesThe required deterministic price function isThis includes $s=0$, $v=0$ and $t=T$. For $s>0$ and $t<T$, it is continuous and strictly decreasing in $v\ge 0$, with range $(0,\sqrt{s}]$.

Set $\theta =(\mu -r)/\sigma$. The positive densityhas mean one, and the Girsanov theorem makes $W_t^Q=W_t+\theta t$ a Brownian motion under the equivalent measure $Q$. HenceThe Black-Scholes equation makes $e^{-rt}V(t,S_t)$ a local martingale. Bounded $V$ on the finite horizon makes it a true martingale, so its terminal condition gives the risk-neutral valuationThe geometric Brownian motion solution under $Q$ isThe Brownian increment is independent of ${\mathcal{F}}_t$ and equals $\sqrt{T-t}Z$ in distribution, where $Z$ has the standard normal distribution. ThereforeThe expectation on the right is only an integral against the standard normal law; the physical drift $\mu$ has disappeared.

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

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

The Itô formula verifies the geometric Brownian motion solutionThis is valid for negative $x$ as well as positive $x$; $x=0$ gives the zero solution. Pathwise uniqueness follows from the Lipschitz continuity of the linear coefficients. The moment-generating function of a normal distribution gives $\mathbb{E}{e}^{k\sigma B_t}=e^{k^2{\sigma }^{2}t/2}$, and thereforeThese moments are finite on every finite horizon. If the convention $0\in \mathbb{N}$ is used, the zeroth moment is $1$.

The price process is a geometric Brownian motion withPutand define the equivalent measure $Q$ on ${\mathcal{F}}_T$ by the exponential Brownian martingaleThe Cameron-Martin-Girsanov theorem says that $W_t^Q=B_t+\vartheta t$ is Brownian motion under $Q$. Thereforeso $e^{-rt}{S}_t$ is a $Q$-martingale. This is the unique equivalent martingale measure, since cancellation of the single Brownian risk forces the displayed drift shift.