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Construct the strictly positive rolling one-period bond account using successive one-period zero-coupon bonds:At time $t-1$, invest the entire account value in the bond maturing at $t$. This gives a positive self-financing portfolio usable as a numéraire.

We use the finite-discrete-time fundamental theorem of asset pricing: in a frictionless market with finitely many adapted assets and trading dates, no arbitrage is equivalent to existence of an equivalent martingale measure for asset prices, including dividends, expressed in a positive traded numéraire. Maturing bond payoffs are reinvested in that numéraire. Let $\mathbb{Q}$ be such a measure and $L_t=\mathbb{E}[d\mathbb{Q}/d\mathbb{P}\mid {\mathcal{F}}_t]$ its positive density process. The martingale pricing relation for a unit zero-coupon bond isDefine $Z_t=L_t/{\beta }_t$. The Bayes formula for conditional expectation then givesThe positive expectations are finite because these are the traded finite bond prices; in particular $\mathbb{E}{Z}_T=Z_0{P}_0(T)$ when initial information is trivial. Normalize $Z_0=1$ without changing any ratio. With nontrivial initial information the same identities are conditional on that information. The state-price density need not be unique when the bond market is incomplete; existence is sufficient.

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

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

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

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

Use zero-interest cash as the one-period numéraire, consistent with the stated expectation-price formula. For $p>1$, differentiate the proposed call-price curve twice. The first derivative and the candidate density areThis is the power call-curve pricing density. It is strictly positive. Its integral is $C^{\prime }(\infty )-C^{\prime }(0+)=1$, and its survival function is $-C^{\prime }(u)$. Since $C(0+)=1$ and $C(\infty )=0$,Likewise, integrating the survival function from $K$ onwards givesThus a market whose terminal stock has this law under an equivalent martingale measure prices the stock at one, every proposed call at $C(K)$, and any integrable claim $g(S_1)$ at $\int g(u)f_p(u)du$. The finite-market fundamental theorem of asset pricing says that an equivalent measure pricing every traded discounted payoff by expectation excludes arbitrage. This proves the intended conclusion when such an equivalent pricing law is part of the model. For example, take the canonical terminal state space $(0,\infty )$ with stock equal to its coordinate and physical law equivalent to the positive density $f_p$.

There is, however, a genuine insufficiency in the literal finite-strike formulation: a finite list of call prices and no-arbitrage alone do not force this pricing law, nor even a continuous terminal distribution. Here is an explicit counterexample. Take $p=2$, one strike $K=1$, and two terminal stock valuesGive the lower stock value the remaining strictly positive probability and take this as the physical measure too. Direct calculation gives $\mathbb{E}{S}_1=1$ andso the stock/cash/call market is arbitrage-free. Now letThis bounded nonnegative function is zero at both actual stock values, so $g(S_1)=0$ almost surely. Yet $\int g(u)f_2(u)du>0$. Charging that positive amount for the identically zero payoff creates an arbitrage by selling it. In fact no Lebesgue probability density can price every claim correctly on this two-state market.

Therefore the displayed $f_p$ is the intended continuous pricing density, but the promised no-arbitrage extension requires an equivalent pricing measure with this terminal law; a full call curve identifies that law if such a measure exists, but it does not follow from the printed finite-strike hypotheses alone. This is the finite-strike nonidentification of a pricing density. The counterexample and the corrected sufficient hypothesis account for the literal and intended readings separately.

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.