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For the quadratic Ornstein-Uhlenbeck state-price density, apply the Itô's formula under the reference probability measure. Since $f^{\prime }(x)=x$ and $f^{\prime \prime }(x)=1$,The stochastic integral is a true martingale on finite horizons by the finite moments of the Ornstein-Uhlenbeck process. A uniform sufficient condition is $\boxed{\alpha \ge {\sigma }^2/(2a)}$, making the finite-variation drift nonpositive. This is also the condition for nonpositive drift at every state.

Use the positive state-price density $\zeta$ to price a terminal payoff $H$ by ${\Pi }_t(H)=\mathbb{E}[{\zeta }_{T}H\mid {\mathcal{F}}_t]/{\zeta }_t$. The riskless account must cancel the drift of $\zeta$, giving the short rateThis pricing system is consistent with a money-market risk-neutral measure: with $M_t=e^{{\int }_0^t{r}_{u}du}$, the process ${\zeta }_t{M}_t/{\zeta }_0$ is a stochastic exponential whose diffusion coefficient $\sigma X_t/f(X_t)$ is bounded. The Novikov condition makes it a true density martingale. The original Brownian motion is for the reference measure, not automatically for this new pricing measure.

For $X_0=x_0$, the OU second moment is $\mathbb{E}{X}_T^2=x_0^2{e}^{-2\lambda T}+{\sigma }^2(1-e^{-2\lambda T})/(2\lambda )$. Thus the zero-coupon bond price isConditionally at time $t$, replace $T$ by $T-t$ and $x_0$ by $X_t$.

The short rate is nonnegative under the condition above, increases with $X_t^2$, and lies between $\alpha -{\sigma }^2/(2a)$ and $\alpha +2\lambda$. The model is tractable but imposes fixed rate bounds, identical rates for opposite values of $X$, and a restricted one-factor term structure. Its long-maturity yield is $\alpha$. These structural restrictions, including exclusion of negative rates under the supermartingale condition, may be unsuitable for a general interest-rate fit.

Pricing kernel and replication. In the nondegenerate Black-Scholes model, put $\kappa =(\mu -r)/\sigma$ and normalize the state-price density by ${\zeta }_0=1$. The process isThe density $e^{rt}{\zeta }_t$ changes probability to the risk-neutral measure. Under that measure $W_t+\kappa t$ is a Brownian motion and the stock drift is $r$. Thus an integrable contingent claim $H$ has time-$t$ priceIn the usual augmented natural Brownian filtration, the Brownian martingale representation theorem supplies a replicating strategy; this is the complete market assumption. For a nonnegative admissible trading strategy without intermediate consumption, the state-price budget constraint is $\mathbb{E}[{\zeta }_T{w}_T]\le w_0$, with equality for a fully invested replicated claim. In particular

Feasibility and the largest slope. First take the intended regime $r>0$, $T>0$, $w_0,S_0>0$, and the usual nonnegative portfolio wealth constraint. The terminal wealth floor isFor any feasible claim the state-price budget constraint impliesTherefore $\alpha \le w_0/S_0$. Conversely, when $0\le \alpha \le w_0/S_0$, hold $\alpha$ shares and put the remaining $w_0-\alpha S_0$ in the continuous-time bank account. Its terminal portfolio wealth isHenceAt equality, $\xi =\bar{\alpha }{S}_T$ has cost exactly $w_0$. Positivity of the state-price density forces $w_T=\bar{\alpha }{S}_T$ almost surely: any strict improvement would cost more. Invest all initial portfolio wealth in $w_0/S_0$ shares and hold them until $T$.

Optimal payoff below the feasibility limit. For $\alpha <\bar{\alpha }$ the floor is strictly positive, and its price is strictly less than $w_0$. Assume the utility function is increasing, differentiable and strictly concave, satisfies the Inada conditions, and has the integrability needed for the finite-budget optimization. These are the usual hypotheses implicit in using inverse marginal utility. For each positive multiplier $\lambda$, maximizeseparately in every state. Its derivative decreases through zero at $I(\lambda {\zeta }_T)$, so the floored marginal utility optimizer isThe multiplier is characterized byUnder the stated integrability hypotheses the left side is continuous and decreasing, tends to the floor cost as $\lambda \uparrow \infty$, and tends to infinity as $\lambda \downarrow 0$. It is strictly decreasing wherever it exceeds the floor cost: on the event where the inverse-marginal-utility payoff exceeds the floor, a larger multiplier strictly reduces that payoff. Therefore the budget determines a unique finite multiplier. For CRRA utility the inverse marginal utility is $I(y)=y^{-1/R}$; lognormal moments provide the needed integrability.

For completeness, pointwise maximality gives, for any feasible competing terminal portfolio wealth $Y$,Taking expectations and using $\mathbb{E}[{\zeta }_{T}Y]\le w_0=\mathbb{E}[{\zeta }_T{w}_T^{\ast }]$ proves optimality. Strict concavity gives uniqueness of the terminal claim. Its price processis nonnegative and starts from $w_0$; claim replication therefore turns the payoff optimizer into an admissible portfolio.

What the missing interest-rate hypothesis changes. The PDF does not explicitly assume $r>0$ or give the utility and admissibility hypotheses above. These omissions matter. At $r=0$, under nonnegative admissibility, the largest feasible slope remains $w_0/S_0$: for larger slopes the positive-part floor costs strictly more than $w_0$, since $S_T$ has support $(0,\infty )$. But every $\alpha \le w_0/S_0$ already gives floor cost exactly $w_0$, so the only feasible terminal claim is $\xi$. There is no spare budget for an inverse-marginal-utility improvement. For example $\alpha =0$, $\kappa \ne 0$ and CRRA utility make $\mathbb{E}[{\zeta }_T\max \{w_0,I(\lambda {\zeta }_T)\}]>w_0$ for every finite $\lambda$. The prescribed positive finite multiplier then does not exist.

For negative $r$, nonnegative admissibility requires the effective floor ${\xi }_{+}=\max \{\xi ,0\}$. Define its costThis is continuous and convex. For $\alpha \le w_0/S_0$ it equals $w_0{e}^{-rT}+\alpha S_0(1-e^{-rT})$, exceeding $w_0$ below the endpoint. At ${\alpha }_0=w_0/S_0$, $G({\alpha }_0)=w_0$ and $G^{\prime }({\alpha }_0)=S_0(1-e^{-rT})<0$. Above that endpoint the lognormal stock gives strict convexity, and $G(\alpha )\to \infty$. Thus there is a unique second root ${\alpha }_1>{\alpha }_0$ of $G({\alpha }_1)=w_0$, the feasible slopes are $[{\alpha }_0,{\alpha }_1]$, and the largest is ${\alpha }_1$. Equivalently, for $\alpha >{\alpha }_0$ the floor price is $\alpha$ times the European call option price with strike $S_0-w_0/\alpha$. At the upper endpoint replicate ${\xi }_{+}$; in the interval with strict budget slack the same floored marginal utility optimizer applies with ${\xi }_{+}$.

If instead portfolio wealth may be negative and utility is defined on all real portfolio wealth, the floor itself has its affine replication cost: at $r=0$ every slope is feasible, and at negative $r$ every $\alpha \ge w_0/S_0$ is feasible. There is then no largest finite slope. The intended stock-only endpoint and strict-slack optimizer use positive interest and standard nonnegative admissibility.

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.

To see minimality, any other nonnegative self-financing portfolio replicating the payoff has discounted wealth a nonnegative local martingale under $\mathbb{Q}$, hence a supermartingale. Its initial capital $x$ must satisfy $x\ge {\mathbb{E}}^{\mathbb{Q}}[e^{-rT}(S_T-K{)}^{+}]=C(0,S_0)$. The strategy constructed above attains equality. ThusThe additive physical diffusion may take negative stock values; the formula and nonnegative replicating wealth remain valid. Replacing it by a multiplicative Black–Scholes diffusion would give the wrong price and hedge.

Differentiate the Gaussian price with respect to $s$. The terms involving derivatives of $d$ cancel because ${\varphi }^{\prime }(d)=-d\varphi (d)$ and $s-K{e}^{-r\tau }=\nu d$. Thus the delta hedge isFor every $t<T$, $\nu >0$ and $S_t$ is finite, so $0<{\pi }_t<1$. At maturity its limiting value is the payoff derivative $1_{\{S_T>K\}}$ except at the kink, an event of probability zero under the equivalent Gaussian law. Consequently the stock holding is always nonnegative and never exceeds one. The initial drift $\mu$ does not enter this hedge; it is removed by the change to the risk-neutral measure.

Normalize $B_0=1$ and put ${\tilde{S}}_t=e^{-rt}{S}_t$. Its dynamics are $d{\tilde{S}}_t={\tilde{S}}_t\sigma (t,S_t)d{W}_t$. Bounded volatility makes this stochastic exponential a true martingale on each finite horizon, by the Novikov condition. It also gives a finite second moment: stopping the Itô formula for $S^2$ and applying the Gronwall inequality yields $\mathbb{E}{S}_T^2\le S_0^2{e}^{(2r+L^2)T}$ when $\sigma \le L$.

The discounted payoff $\xi =e^{-rT}(S_T-K{)}^{+}$ is thus square-integrable. Define the nonnegative martingaleThe Brownian martingale representation theorem says that every square-integrable martingale in the Brownian filtration has a representation $M_t=M_0+{\int }_0^t{h}_{u}d{W}_u$ with predictable $h$ and $\mathbb{E}{\int }_0^T{h}_u^{2}du<\infty$. Since $\tilde{S}>0$ and $\sigma >0$, chooseThen discounted gains satisfy $d{M}_t={\pi }_{t}d{\tilde{S}}_t$. Consequently $X_t=B_t{M}_t={\varphi }_t{B}_t+{\pi }_t{S}_t$ is self-financing, nonnegative, and hence an admissible trading strategy. At maturity $X_T=(S_T-K{)}^{+}$, proving claim replication at cost $C(T,K)$.

For minimality, discounted wealth of any admissible self-financing portfolio is a local martingale bounded below, hence a supermartingale by localization and the conditional Fatou lemma. Thus any such replication with initial wealth $x$ obeysTogether with the constructed portfolio, this provesThis is Brownian representation replication in a local volatility market. The given drift $r$ means that the original probability measure already serves as the risk-neutral measure.

Expand the payoff as $(S_T-2K+K^2/S_T)1_{\{S_T>K\}}$. The same risk-neutral measure applies. For any real $p$, completing the square gives the truncated lognormal momentUsing $p=1,0,-1$ and discounting by $e^{-rT}$ givesThe inverse-price moment explains both the extra volatility factor and the shifted normal argument in the last term.

In the Black-Scholes model, absence of arbitrage and completeness give the price as the discounted expectation under the risk-neutral measure. Under that measure,The physical drift $\mu$ therefore does not enter the option price. PutFor the standard normal distribution function $\Phi$, $\mathbb{P}(S_T>K)=\Phi (d_2)$. Completing the square in the Gaussian density also gives $\mathbb{E}[S_T{1}_{\{S_T>K\}}]=S_0{e}^{rT}\Phi (d_1)$. Expanding the call payoff into these two truncated moments yields

Take the original measure as the risk-neutral measure for this zero-interest model, or interpret $S$ as a discounted stock price. ${\xi }_t$ is the arbitrage-free value of the European contingent claim with payoff $g(S_T)$, and ${\pi }_t$ is its option delta and replicating stock holding. Part (b) supplies the self-financing portfolio:With the cash asset identically one, its value is ${\xi }_t$ and its gains satisfy $d{\xi }_t={\pi }_{t}d{S}_t$, so it exactly replicates $g(S_T)$. This delta hedge is admissible since the assumed nonnegative $V$ gives nonnegative wealth. In this continuous-path setting the holdings can be taken predictable.

The auxiliary change of measure in part (c) expresses the option delta as an expectation of terminal payoff sensitivity. It changes the stock drift to $a{a}^{\prime }$ and is generally not the original risk-neutral measure for the zero-interest stock market. Its role is to value sensitivity under the tilted state dynamics, while the actual replicating strategy trades under the original dynamics. No general claim of market completeness is needed, including when $a$ can vanish.

For constant coefficients, the density process $Y_t{B}_t/B_0$ is a true exponential martingale and defines the risk-neutral measure $Q$. Under $Q$,The minimal value process is thereforeThe Markov property and the lognormal transition law make this a deterministic function of $(t,s)$, and

The one-period spot interest rate is defined byand the bank account byA probability measure $Q$ equivalent to the physical measure is a risk-neutral measure when every discounted zero-coupon bond priceis a $Q$-martingale. Equivalently,

If $T\mapsto P_t^T$ is nonincreasing, thenso $1+r_t=(P_t^{t+1}{)}^{-1}\ge 1$ and $r_t\ge 0$.

Conversely, if every spot rate is nonnegative, then $B_{T+1}=B_T(1+r_T)\ge B_T$. Under a risk-neutral measure,Hence

In the Black-Scholes model, take $B_t=e^{-\rho (T-t)}$ as a zero-coupon bond numéraire. Holding the fixed stock value $\theta B_t$ requires $g_t=\theta B_t/S_t$ shares. The self-financing conditions for smooth stock and bond holdings giveunits of the bond. Since $V_t=g_t{S}_t+h_t{B}_t$, this yields the displayed wealth. Under the risk-neutral measure, $V_t/B_t=w_0/B_0+\theta \sigma W_t^Q$: the discounted gain is a constant multiple of Brownian motion.

A risk-neutral measure $Q$ is a probability measure equivalent to the physical probability measure $P$, meaning they agree on which events have zero probability. For assets without intermediate payments, prices expressed in the chosen numéraire must be martingales under $Q$. For a positive bank account $B$ and a stock without dividends, this requires $S/B$ to be a martingale; with dividends, use discounted dividend gains. In an unrestricted finite-state one-period financial market, existence of such a measure is equivalent to absence of arbitrage, by the fundamental theorem of asset pricing.

Physical probability $p=P(\text{up})$ describes the modeled frequency or belief about an up move. The risk-neutral probability $q=Q(\text{up})$ is constrained by traded prices. In the worked binomial market, $q=5/8$ even if $p=1/2$ or $p=4/5$. Calling $Q$ risk-neutral does not assume that investors have no risk aversion; it changes the weights used for risk-neutral pricing.

The martingale condition for the discounted stock is $qu{S}_0+(1-q)d{S}_0=R{S}_0$. Therefore the up risk-neutral probability isWith down factor $d=1+a$, up factor $u=1+b$ and $R=1+r$, this is equivalently $(r-a)/(b-a)$. For $d=0.8$, $u=1.2$, $R=1.05$, it gives $q=0.25/0.4=0.625$. Under the risk-neutral measure, the terminal stock price has expected value $(5/8)120+(3/8)80=105$, whose discounted value is 100. The physical probability $p$ need only satisfy $0<p<1$; it does not enter this calculation.