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For claim dividends proportional to the underlying stock, put $V(S,t)=S\varphi (t)$. Then $V_S=\varphi$, $V_{SS}=0$ and $V_t=S{\varphi }^{\prime }$, so the partial differential equation reduces to the linear ordinary differential equationThe integrating factor $e^{-Dt}$ gives $(e^{-Dt}\varphi {)}^{\prime }=-e^{-Dt}f(t)$. Integrating backwards from the terminal value,The formula assumes the displayed integral is finite. Under the risk-neutral measure, ${\mathbb{E}}_t[S_u]=S{e}^{(r-D)(u-t)}$. Discounting each cash payment at its own payment time gives the same result:The interest rate disappears because every payment is proportional to the stock price.

For the non-dividend-paying stock, a self-financing portfolio replicating a smooth European option has $\Delta =V_S$ shares and $V-S{V}_S$ in the bank account. Its gain satisfiesThe Itô formula gives $dV=(V_t+\mu S{V}_S+{\sigma }^2{S}^2{V}_{SS}/2)dt+\sigma S{V}_{S}dW$. Matching its diffusion and drift with the portfolio gains cancels $\mu$, yieldingThe financial payoff supplies the terminal value. This is the Black-Scholes equation; the associated risk-neutral measure replaces the physical stock drift by $r$.

Keep the maturity $T$ fixed and write $F(t,T)=\exp L(t,x_t)$, whereAs calendar time increases, ${\partial }_{t}a=\alpha a$, so $L_x=a$, $L_{xx}=0$ and $L_t=\alpha a(x-\tilde{\mu })-{\sigma }^2{a}^2/2$. Applying the Itô formula to $L$ and then to its exponential givesThe Itô formula's exponential correction cancels the displayed drift of $L$. The zero drift also follows from the tower property of conditional expectation: for $t\le u\le T$, ${\mathbb{E}}_t[F(u,T)]={\mathbb{E}}_t[S_T]=F(t,T)$. In particular $F(T,T)=S_T$, while the quote's instantaneous spot volatility approaches $\sigma$ as expiry approaches.

This is a martingale under the same measure used in the stipulated conditional expectation. To interpret it as futures pricing, that measure must be a suitable pricing measure. For a non-dividend-paying traded stock at a deterministic interest rate, the absence of arbitrage instead gives a forward contract delivery price $S_t{e}^{r(T-t)}$. A physical-measure expectation under the given mean-reverting drift is not automatically that price; the question's expectation convention is sufficient for the calculation above, but not for this additional identification.

Fix a finite horizon $N$, a filtration $({\mathcal{F}}_n)$, a positive bank account $B_n$ with predictable one-step growth, ex-dividend risky prices $S_n$, and asset dividends $D_n$ paid at date $n$. Put $s_n=S_n/B_n$ and $d_n=D_n/B_n$. An equivalent martingale measure $Q$ makes the discounted cum-dividend gainsmartingales; equivalently ${\mathbb{E}}_Q[s_n-s_{n-1}+d_n\mid {\mathcal{F}}_{n-1}]=0$. It is these gains, not necessarily the discounted ex-dividend prices alone, which enter the definition.

Let ${\theta }_n$ be risky holdings chosen at date $n-1$ and carried to date $n$. Let $C_n$ be the portfolio's net cash distribution to its owner at date $n$, and $V_n$ its remaining value after that distribution and rebalancing. A self-financing portfolio with these distributions has the budget identityTo see this, value the previous stock and bank holdings at the new prices, add the asset dividends, subtract the owner distribution, and observe that the discounted bank holding is unchanged. Rebalancing then alters holdings but not value. An asset dividend reinvested in the portfolio is already in the gain term; it is not also an owner distribution.

For a predictable strategy whose gains are integrable under $Q$ (bounded holdings in a finite-state market suffice), conditioning the gain increment gives zero. Therefore discounted remaining wealth plus cumulative discounted owner payments is a martingale. Conditioning its terminal value proves the multi-period dividend pricing identityThe relation holds under every equivalent martingale measure for which these integrability conditions hold. Each payment is discounted from its own payment date, not from one common terminal date. The expression inside the expectation is in bank account units; multiplication by $B_n$ returns current currency units.

When the portfolio is liquidated and all its final value is paid out, $V_N=0$ after liquidation and that value is included in $C_N$. The displayed identity then involves only future discounted distributions. Without liquidation, the terminal value must remain. A bank account holding with no coupons before redemption has positive value despite having no intervening dividends; omitting its redemption or terminal holding value would plainly give the wrong result. In an infinite horizon, deleting the terminal term instead requires a dividend-price transversality condition, namely convergence of its conditional discounted expectation to zero.

For a traded asset the same proof gives its discounted expected dividends plus terminal sale value. For a replicated claim paid only at $N$, it gives $V_n=B_n{\mathbb{E}}_Q[H/B_N\mid {\mathcal{F}}_n]$. In a complete market this prices every admissible claim uniquely. In an incomplete market only the portfolio's attainable payout stream has this common value under all measures; an arbitrary unattainable payoff can have different expectations under different measures. For a merely dominated measure the conditional identity holds $Q$-almost surely; it makes no assertion on physical states assigned zero mass by $Q$. External capital injections must also be counted, as negative owner distributions. Thus the valuation principle follows from the gain martingale and the trading budget, together with the terminal convention and integrability, rather than from unconditional averaging of arbitrary future cash flows.

Assume $f\in C^{1,2}$ on positive prices before maturity and the usual local trading integrability. If its value is held in stock units ${\delta }_t$ and bank units ${\beta }_t$, thenfor a self-financing portfolio. The Itô formula diffusion coefficient is $\sigma S_t{f}_x$, so $\sigma S_t>0$ forces ${\delta }_t=f_x(S_t,t)$ and ${\beta }_t{B}_t=f-S_t{f}_x$. Equating drifts yieldsThe physical drift cancels, giving the Black-Scholes equationConversely, if the smooth function satisfies this equation, set $\delta =f_x$ and $\beta =(f-x{f}_x)/B$. Substituting into Itô formula gives exactly $df=\delta dS+\beta dB$, so these holdings are self-financing. This checks the holdings as well as the value equation; a partial differential equation solution does not justify arbitrary prescribed holdings. Appropriate admissibility and growth conditions are imposed when using it as an economic price.

For the logarithmic stock payoff, let $\tau =T-t$ and condition on $S_t=x$. Under the risk-neutral measure,The normal exponential moment $\mathbb{E}{e}^{uZ}=e^{u^2/2}$ and its derivative $\mathbb{E}[Z{e}^{uZ}]=u{e}^{u^2/2}$ giveThereforeThe option delta is the displayed stock holding. The bank account value is $f-S_t{f}_x=-S_t$, so ${\beta }_t=-S_t/B_t$. Directly, $f_{xx}=1/x$ and $f_t=-x(\rho +{\sigma }^2/2)$ verify the partial differential equation, while $f(x,T)=x\log x$ verifies the payoff. Lognormal moments give the required pricing integrability. Since $x\log x\ge -1/e$, the conditional price is bounded below over this finite horizon, so the resulting claim replication has the usual admissibility property.

Expected utility of terminal wealth. Let initial wealth be $x_0>0$, maturity $T=t_0$, and stock-dollar investment ${\pi }_t$. A self-financing strategy in the Black-Scholes model has wealthChoose admissible strategies with nonnegative wealth, and an increasing strictly concave differentiable utility function $U$ satisfying the Inada conditions. Assume the required expectations are finite and the optimization is well posed; the resulting budget equation below must have a solution. Write $\lambda =(\alpha -\rho )/\sigma$ andThis state-price density is the discounted density of the unique equivalent martingale measure. Nonnegative discounted wealth is a supermartingale under that measure, so every admissible terminal wealth $Y$ obeys the state-price budget constraint $\mathbb{E}[H_{T}Y]\le x_0$.

Conversely a nonnegative terminal claim with finite such cost is financed by its conditional martingale price, which remains nonnegative; Brownian Martingale representation theorem constructs its holdings. Thus market completeness turns dynamic optimization into a terminal-payoff problem. Let $I=(U^{\prime }{)}^{-1}$ be the inverse marginal utility, and choose $y>0$ from $\mathbb{E}[H_{T}I(y{H}_T)]=x_0$. Then the complete-market terminal utility optimizer isFor any feasible $Y$, concavity gives the pointwise tangent inequalityTaking expectations and using the budget proves optimality; strict concavity gives uniqueness up to null events. The optimal wealth process isIf its discounted martingale has representation $d(e^{-\rho t}{X}_t^{\ast })={\zeta }_{t}d{W}_t^Q$, hold ${\delta }_t=e^{\rho t}{\zeta }_t/(\sigma S_t)$ stock units and invest the remainder in the bank account. This derives the strategy, not only the terminal first-order condition.

The equivalent dynamic approach is the Bellman equation for terminal-wealth utility. For a smooth concave value $v(t,x)$, dynamic programming and Itô formula giveFor $v_{xx}<0$, maximizing the concave quadratic gives ${\pi }^{\ast }=-(\alpha -\rho )v_x/({\sigma }^2{v}_{xx})$. For an arbitrary admissible control the Itô drift of $v(t,X_t)$ is nonpositive; localization and integrability therefore bound its expected terminal utility by $v(0,x_0)$. The maximizing control makes the drift zero, giving equality when the verification expectations are valid.

For CRRA utility $U(x)=x^{1-\gamma }/(1-\gamma )$, $\gamma >0$, $\gamma \ne 1$, substitution givesThe optimal stock fraction is constant. Its geometric wealth dynamics stay positive, so it is feasible. For logarithmic utility the fraction is $(\alpha -\rho )/{\sigma }^2$ and $v(t,x)=\log x+(\rho +{\lambda }^2/2)(T-t)$. Risk aversion controls the risky exposure, while the market price of risk controls its reward.

Pricing claims depending on the path. For a path-dependent contingent claim $C=F((S_u{)}_{0\le u\le T})$, the risk-neutral pricing process isFor square-integrable discounted $C$, the Brownian Martingale representation theorem gives $d(e^{-\rho t}{V}_t)={\phi }_{t}d{W}_t^Q$. Since $d(e^{-\rho t}{S}_t)=\sigma e^{-\rho t}{S}_{t}d{W}_t^Q$, the replicating holding is ${\delta }_t={\phi }_t/(\sigma e^{-\rho t}{S}_t)$, with bank units $(V_t-{\delta }_t{S}_t)/B_t$. Thus path dependence does not destroy market completeness; it changes the information needed to determine the price and hedge.

For an arithmetic Asian option, introduce $A_t={\int }_0^t{S}_{u}du$ and seek $V_t=v(t,S_t,A_t)$. Because $d{A}_t=S_{t}dt$, Itô formula givesThe stock holding is $v_s$. The state $a$ records the already observed average; current stock price alone cannot recover it.

A Geometric Asian option admits an explicit conditional calculation. Let $I_t={\int }_0^t\log S_{u}du$ and $G_T=e^{I_T/T}$. Conditional on time $t$, the normal law of $\log G_T$ has mean and varianceIndeed its random part is $\sigma T^{-1}{\int }_t^T(T-u)d{W}_u^Q$, by stochastic Fubini; the Itô isometry gives $q_t$. Completing the square in a normal exponential integral proves the conditional geometric-average Asian option formulafor $K>0,t<T$. This explicitly includes the known past log-average. Holding that past integral fixed when differentiating the price gives the stock holdingThe normal-density derivative terms cancel because $e^{m_t+q_t/2}\phi (d_1)=K\phi (d_2)$.

For a lookback option use $M_t={\max }_{u\le t}{S}_u$. A floating-strike put pays $M_T-S_T$. Its price $v(t,s,m)$ obeys the ordinary Black-Scholes equation in $0<s<m$, with terminal value $m-s$. The extra Itô term is $v_{m}d{M}_t$; since $d{M}_t$ is carried by $S_t=M_t$, self-financing requires the running-maximum boundary for a lookback option $v_m(t,m,m)=0$ before maturity. Equivalently its price follows from the upper-hitting probabilities in Question 3:This uses $\mathbb{E}\max (m,Z)=m+{\int }_m^{\infty }P(Z\ge z)dz$ and ${\mathbb{E}}_Q{S}_T=s{e}^{\rho (T-t)}$. For a barrier option, the state must additionally record whether the barrier has already been hit; a no-rebate knockout price has an absorbing zero boundary. These augmentations give tractable PDEs or conditional-expectation computations while retaining the same martingale pricing and claim replication principle.

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.

An American option allows exercise at a stopping time $\tau \in [t,T]$. For its exercise financial payoff $H$, the pricing problem isThe discounted value is the Snell envelope of discounted exercise payoffs. Define the pricing operatorIn the continuation region the Black-Scholes equation holds, $\mathcal{L}V=0$. In the exercise region $V=H$, while the ability to wait and the supermartingale property give $\mathcal{L}V\le 0$. Together these conditions are the obstacle problemThus an unknown exercise boundary must be found alongside the value. At a regular boundary in the nondegenerate Itô diffusion, value matching is $V=H$, and smooth pasting is $V_S=H^{\prime }$ where the financial payoff is differentiable. A finite difference method or a binomial options pricing model steps backward taking the larger of continuation and immediate exercise values at each node.

For a non-dividend-paying American call option with $r\ge 0$, no early exercise of a call without dividends follows directly from the European lower boundAt every candidate exercise time, keeping the European call has value at least the immediate exercise financial payoff, so early exercise cannot improve the value; the American and European prices coincide. The first inequality follows from Jensen inequality or put-call parity and nonnegative put value. With $r>0$ an American put option can benefit from early receipt of its strike at sufficiently low stock prices, so its exercise boundary is generally nontrivial and its value is at least the European put value. Dividends can make call exercise worthwhile, and negative interest invalidates the stated no-early-exercise call argument.

Let $Y=h(U)$ have finite variance $v$ and mean $I$. An independent-sample Monte Carlo estimator has variance $v/N$, so its root-mean-square random error is $\sqrt{v/N}$. Three techniques reduce the variance coefficient rather than changing this ordinary $N^{-1/2}$ rate.

For antithetic variates, couple $Y=h(U)$ with $Y^{\prime }=h(1-U)$ and average the pair. With $M$ independent pairs, the variance is $(v+\mathrm{Cov}(Y,Y^{\prime }))/(2M)$, compared with $v/(2M)$ for $2M$ independent function evaluations. If $h$ is monotone in a scalar uniform, the covariance is nonpositive: for independent uniforms $U,V$,Opposite fluctuations therefore cancel. The reduction factor at equal evaluation count is $1+\rho$, where $\rho$ is the paired correlation coefficient; it can approach zero, but positive covariance would make the method worse. This is appropriate for monotone or otherwise negatively coupled payoffs, with cheap complementary paths.

For control variates, use a jointly simulated $C$ with known mean $m_C$ and average $Y-\beta (C-m_C)$. It remains an unbiased estimator for fixed $\beta$, andCompleting the square gives ${\beta }_{\ast }=\mathrm{Cov}(Y,C)/\mathrm{Var}C$ and minimal variance $v(1-{\rho }^2)$. For contingent claim simulation, an analytically priced related financial payoff, such as a simpler option or discounted terminal stock, can be a useful control. It works best when correlation has large absolute value and simulation of the control adds little cost. A coefficient fixed from an independent pilot keeps the elementary unbiasedness argument valid; estimating it from the same sample can introduce finite-sample bias.

For stratified sampling, partition the sampling space into strata $A_h$ of probabilities $p_h$, and independently simulate $N_h$ samples conditional on each stratum. The estimator ${\sum }_h{p}_h{\bar{Y}}_h$ is unbiased and has varianceWith proportional allocation $N_h=N{p}_h$, ignoring integer rounding, this equals ${\sum }_h{p}_h{v}_h/N$. The law of total variance shows the reduction from ordinary sampling is $\mathrm{Var}(\mathbb{E}[Y\mid A_h])/N$: fixed representation removes randomness in the stratum counts. At equal per-draw cost, minimizing the displayed variance gives $N_h\propto p_h\sqrt{v_h}$ and minimum $({\sum }_h{p}_h\sqrt{v_h}{)}^2/N$ for positive within-stratum variances. This follows by Cauchy-Schwarz inequality and is useful when a low-dimensional coordinate strongly predicts the financial payoff and conditional simulation is easy.

There is no universal ranking. At comparable cost the variance factors depend respectively on paired covariance, squared control correlation, and within-stratum variation. A nearly perfect control can outperform a weak antithetic pair; an appropriate stratification can be almost exact for a nearly stratum-constant financial payoff. Extra construction and sampling costs must be included in a fair accuracy comparison, and the methods can be combined.

Take bank-account values $B_k=R^k$ with $R>0$, and let the stock multiply independently by $u$ or $d$ at each step, where $0<d<R<u$. Under the physical probability measure $P$, the up probability is $p\in (0,1)$; the filtration records the successive moves. These assumptions specify the standard binomial market. If $R$ lies outside the strict interval $(d,u)$, a long or short stock position financed through the bank account gives an arbitrage.

The risk-neutral probability in a binomial market is fixed by $qu+(1-q)d=R$, givingFor successor claim values $V_u,V_d$ at a node with current stock price $S$, solve the two replication equations. The current stock holding and current cash value areThe cost of this replicating portfolio in a binomial market isApply this backward option pricing recursion from any terminal contingent claim $H$, including claims depending on the whole sequence of moves. It constructs a self-financing portfolio and proves market completeness by induction on the remaining number of periods. Taking successive conditional expectations under the measure $Q$ with independent up probability $q$ gives the unique risk-neutral pricing valueAny other price would give an arbitrage by buying the cheaper of the claim and its replica and selling the dearer.

If $K$ is the number of up moves, a particular path has probabilities $p^K(1-p{)}^{n-K}$ under $P$ and $q^K(1-q{)}^{n-K}$ under $Q$. Therefore the binomial-market probability density, namely the Radon-Nikodym derivative, isThis last expression makes $Z$ a function of the terminal stock price. Summing over paths gives ${\mathbb{E}}_{P}Z=1$, and strict positivity proves equivalence of the two probability measures.

For unrestricted terminal portfolio wealth $H$, its initial replication cost is $R^{-n}{\mathbb{E}}_{Q}H$. Hence feasible wealth satisfies the state-price budget constraint ${\mathbb{E}}_{Q}H=R^n{w}_0$. For the stated exponential utility, $v^{\prime }(x)=e^{-ax}$ and $v^{\prime \prime }(x)=-a{e}^{-ax}<0$. The binomial exponential-utility terminal wealth iswhereIt satisfies the budget, and its marginal utility is $v^{\prime }(H^{\ast })=\lambda Z$ with $\lambda =\exp (-a{R}^n{w}_0-{\mathbb{E}}_Q\log Z)>0$. The concave supporting-tangent inequality givesTaking expectations under $P$ makes the last term zero by the budget. Strict concavity makes equality possible only when $H=H^{\ast }$ in every positive-probability state. The already constructed complete market supplies its replicating strategy, so this proves global optimality and uniqueness. The formula allows negative final wealth, as permitted by the unrestricted trading problem.

Let ${\varphi }_T(y)=(2\pi T{)}^{-1/2}{e}^{-y^2/(2T)}$ be the normal distribution density. The Brownian reflection principle gives the killed endpoint densityIndeed, reflection after the first hit of $a$ maps paths ending below $a$ that have hit the barrier to paths ending at $2a-y$; symmetry gives ${\varphi }_T(2a-y)={\varphi }_T(y-2a)$.

By the Girsanov theorem, weighting the path law by $\exp (\nu W_T-{\nu }^{2}T/2)$ makes the coordinate process a Brownian motion with drift $\nu$. Completing the two squares yieldsIntegration up to $x\le a$ proves the joint endpoint and maximum law for drifted Brownian motion:Here $\Phi$ is the standard normal distribution function. Setting $x=a$ also gives a continuous barrier distribution, so equality at the barrier has zero probability.

For the up-and-in option, use the risk-neutral measure of the Black-Scholes model with interest rate $\rho$ and spot volatility $\sigma >0$. ThenAbove $a$, the barrier has certainly been reached. Below $a$, subtracting the killed density from the unrestricted endpoint density leaves $e^{2a\nu }{\varphi }_T(y-2a-\nu T)$. Thus the discounted expected value of its payoff isIn the second integral put $z=y-2a$, and defineThen $S_0{e}^{\sigma y}=S_0{e}^{\sigma z}/\kappa$, $z\le -a$ means $S_0{e}^{\sigma z}\le \kappa b$, and $e^{2a\nu }={\kappa }^{-\nu /\sigma }$. Both integrals now use the unrestricted terminal stock distribution. This proves that the reflected European payoff for an up-and-in option isTheir initial risk-neutral pricing values agree for payoffs with finite absolute expectations in these integrals. The identity concerns prices; the two terminal payoffs need not agree along individual paths.

Write $T=t_0$, $B_t=e^{-\rho (T-t)}$, and let $V_t=f(S_t,t)$ with $f(x,t)=xg(x,t)+B_{t}h(x,t)$. The portfolio holds $g(S_t,t)$ shares of stock and $h(S_t,t)$ units of the zero-coupon bond. It is self-financing when all changes of holdings are funded within it, which is the gain identityUnder the physical probability measure, write $d{S}_t=\mu S_{t}dt+\sigma S_{t}d{W}_t$, where $\sigma >0$ and $d{B}_t=\rho B_{t}dt$. The Itô formula givesMatching the Brownian motion coefficients forces $f_x=g$. Since $f_x=g+x{g}_x+B_t{h}_x$, this is preciselyDifferentiating $f_x=g$ gives $f_{xx}=g_x$. Since $f_t=x{g}_t+\rho B_{t}h+B_t{h}_t$, matching the drift coefficients now givesThese are the self-financing conditions for smooth stock and bond holdings. Conversely, substituting both identities into the Itô formula gives the gain identity, so they are sufficient as well. Smoothness and the positive lognormal distribution density of $S_t$ extend the identities along the price process to every $x>0$, with boundary times obtained by continuity.

The portfolio with constant stock value in bond units must have $xg(x,t)=\theta B_t$, so $g(x,t)=\theta B_t/x$. The first condition then gives $h_x=\theta /x$, henceThe second condition reduces to $c^{\prime }(t)=\theta ({\sigma }^2/2-\rho )$. Initial portfolio wealth requires $h(S_0,0)=w_0/B_0-\theta =w_0{e}^{\rho T}-\theta$. Therefore the holdings areThe corresponding portfolio wealth isUnder the risk-neutral measure, $\log (S_t/S_0)=(\rho -{\sigma }^2/2)t+\sigma W_t^Q$, so its value in zero-coupon bond units is $V_t/B_t=w_0{e}^{\rho T}+\theta \sigma W_t^Q$. This also verifies the self-financing gain equation after using the bond as numéraire.