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This part's conditional-expectation representation is its own hypothesis; it does not need the incorrect unrestricted claim in part (b). Fix $T$, and writeThe assumed representation makes $M(T)$ a martingale. The stochastic Fubini theorem givesApply the Itô formula to $e^{-I}$ and use the equation for $U$ from part (a), including the cross-variation. The result isUniqueness of the continuous semimartingale decomposition makes the drift vanish. Initially this is a $dtd\mathbb{P}$ statement; the assumed continuity in time and maturity extends it to the continuous versions simultaneously. Let $T\downarrow t$ to obtainSubstitute back and differentiate the maturity integrals using their continuous integrands:This is the forward drift restriction for square-root stock claims. The $-{\sigma }_t/2$ term comes from the product cross-variation and must be retained. For the uninformative zero-stock case, the representation does not identify $f$; as usual a positive initial stock price is understood.

Use a telescoping replication of a stock-price sum. Hold $T-t+1$ shares during interval $(t-1,t]$; at time $t$, sell one share and keep its proceeds in the bond. Start with $T$ shares and no cash, costing $T{S}_0$.

After the time-$t$ rebalance, the stock holdings are $T-t$ and the cash holdings are ${\sum }_{u=1}^t{S}_u$, so wealth isThe sale of one share exactly funds the cash increase, making the strategy self-financing. Equivalently,At time $T$ there are no remaining shares, and the cash equals the claim. All stock positions over trading intervals are predictable.

For every integer $m\in \{0,\dots ,N\}$,Thus the static replication on a finite terminal support consists of one share and two calls of every listed strike, held to maturity:The strike-$N$ call contributes zero at maturity but is harmless in this identity. The portfolio costs $S_0+2{\sum }_{K=1}^N{C}_0(K)$. No distributional assumption on the stock is needed beyond its stated terminal support.

The pathwise identity $(S_t-S_{t-1}{)}^2=S_t^2-S_{t-1}^2-2S_{t-1}(S_t-S_{t-1})$ telescopes toThis is the discrete realized-variance replication identity. Replicate $S_T^2$ using part (b). Add a self-financing portfolio with initial wealth $-S_0^2$ and stock holdings $-2S_{t-1}$ during $(t-1,t]$.

To give the cash positions explicitly, letThe added portfolio holds $G_{t-1}+2S_{t-1}^2$ units of the bond over that interval. Its starting value is $G_{t-1}$ and its change is exactly $-2S_{t-1}(S_t-S_{t-1})$. The combined holdings are thereforeThe terminal value equals the squared-increment sum by the telescoping identity. Its initial cost isThe dynamic stock hedge uses only prices known before each interval, while the calls remain static.

Differentiating the European call option price in strike gives, for $T>0$,The strike derivative uses dominated convergence; the second uses the continuous density. Also,Differentiate the supplied time-integral identity and the discount factor. Continuity of the density supplies the diffusion-term derivative. For the tail first moment, continuity in time follows from continuous stock paths, locally uniformly bounded second moments, and the absence of an atom at $K$. ThereforeHence the Dupire equation isIts initial condition is $C(0,K)=(S_0-K{)}^{+}$; natural strike boundaries are $C(T,0)=S_0$ and $C(T,K)\to 0$ as $K\to \infty$. These are consistent with the discounted stock martingale and integrable tails. Where $C_{KK}>0$, the same identity gives local volatility recovery from call prices:The equation evolves in maturity and strike, unlike the backward option-value equation in calendar time and spot.

The Brownian martingale representation theorem argument in part (a) also replicates the bounded put payoff. The discounted stock is a true martingale, so the elementary terminal payoff identity yields put-call paritySet $H(T,K)=-S_0+K{e}^{-rT}$. Then $H_T=-rK{e}^{-rT}$, $H_K=e^{-rT}$, and $H_{KK}=0$. ThusLinearity of the Dupire equation and $P=C+H$ giveThe initial payoff is $P(0,K)=(K-S_0{)}^{+}$, and $P(T,0)=0$. Therefore calls and puts obey the same maturity-strike differential equation, with their respective initial and boundary data.

The natural filtration of $S$ is the natural Brownian filtration: since $\sigma >0$, the equation $W_t=(\log (S_t/S_0)-\mu t)/\sigma$ recovers the entire Brownian motion history from the stock history. For $s<t$, independent increments and the moment-generating function of a normal distribution giveAll these expectations are finite. Since $S_s>0$, the martingale identity holds for every $s<t$ exactly when the last exponential equals one. The required logarithmic drift isThis is the distinction between the drift of $\log S$ and the drift of $S$: here the latter is zero.

First work with the filtration ${\mathcal{G}}_t={\mathcal{F}}_t^W\vee \sigma (1_{\{u\le \tau \}}:0\le u\le t)$, which records the Brownian motion and whether default has occurred. On $\{s>\tau \}$ all subsequent defaultable stock prices vanish. On $\{s\le \tau \}$, the memoryless property of the exponential distribution gives survival from $s$ to $t$ with conditional probability $e^{-\lambda (t-s)}$. The residual survival event is independent of the future independent increments of the Brownian motion. Thus, for deterministic $s<t$,Also $\mathbb{E}{\hat{S}}_t=e^{\lambda t}{e}^{-\lambda t}{S}_0=S_0$, so this is a true martingale, with no localization needed. Its natural filtration is contained in $\mathcal{G}$; applying the tower property of conditional expectation to the displayed identity yields the requested natural-filtration martingale property:The printed convention keeps the stock alive at $t=\tau$. It is a left-continuous convention at that one random time. Replacing it by $1_{\{t<\tau \}}$ gives the usual càdlàg zero-recovery default model; since $\mathbb{P}(\tau =t)=0$ at each fixed $t$, the fixed-time martingale calculations above are unchanged.

Let $h$ be the stock holding and let $c$ be the initial cost, so the cash holding is $b=c-10h$. Terminal portfolio wealth at stock price $s$ is $c+h(s-10)$. Dominating the European call option payoff at the three possible prices givesAdding the two endpoint inequalities yields $2c\ge 1$. Equality is attained by $h=1/2$ and $c=1/2$, which imply $b=-9/2$. The corresponding terminal wealth is $0,1/2,1$ at $s=9,10,11$, respectively, compared with the required payoff $0,0,1$. The cheapest super-replication strategy isThe middle-state excess shows why this is superhedging rather than exact claim replication. The endpoint bound proves global minimality, without relying on the physical state probabilities.

A complete market can replicate every contingent claim in its specified payoff class: for every terminal ${\mathcal{F}}_T$-measurable payoff $X$, there is an admissible self-financing portfolio whose terminal wealth equals $X$ almost surely. Equivalently, in an arbitrage-free finite-state discrete-time model the equivalent martingale measure is unique. In notation,Market completeness is a replication property. The following price comparisons also use the usual absence of arbitrage, which the source leaves implicit: without it the initial cost of claim replication need not be unique. For example, in a deterministic one-period model with cash worth one at both dates and a stock worth one initially but zero finally, all terminal claims are replicable using cash. Adding any stock holding changes the initial cost without changing the terminal payoff. This is a complete model with arbitrage, so it has no well-defined unique replication price.

Let $b$ and $h$ be the numbers of units of the numéraire and the stock in a replicating strategy. The two terminal payoffs of the European call option are $2$ and $0$, soSolving these simultaneous equations gives $b=-6/35$ and $h=8/35$. The initial replication cost isAs an independent pricing check, let $q$ be the risk-neutral probability of the state $(15,20)$ under the numéraire measure. The discounted stock must have initial value one and terminal mean one:Risk-neutral valuation then gives $10(3/7)(2/15)=4/7$. The physical probabilities $1/2,1/2$ are not the numéraire-measure probabilities.

Apply the Itô formula to ${\xi }_t=V(t,S_t)$. The backward equation cancels its drift, leavingBounded $a$ and $V_S$ make this Itô integral square-integrable on the finite horizon; in addition $V$ itself is bounded. Hence $\xi$ is a true martingale, not just a local martingale, and ${\xi }_T=g(S_T)$. Taking conditional expectations givesThe stock is understood to be adapted to the stated Brownian filtration, with its initial value fixed there. This is the Feynman-Kac formula in the zero-potential, zero-drift case.

Differentiate the backward equation for $V$ with respect to the stock state, and put $w=V_S$. The product derivative of $a^2/2$ is $a{a}^{\prime }$, givingThe assumed smoothness and bounded derivatives put $w$ in the uniqueness class for the equation defining $U$. Therefore $U=V_S$, and ${\pi }_t=V_S(t,S_t)$ is the option delta. Substituting in the Itô integral from part (a) and using $d{S}_t=a(S_t)d{W}_t$ gives the self-financing representationThe boundedness hypotheses justify both stochastic integrals. This derivation does not divide by $a$ and remains valid even at states with zero diffusion amplitude.

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.

Write $q_{15},q_{12},q_9$ for the state probabilities under an equivalent martingale measure. Since cash is constant and the European put option pays only in the lowest state, the pricing equations areTheir unique solution is $q_9={\xi }_0/2$, $q_{15}={\xi }_0/2-2/3$, $q_{12}=5/3-{\xi }_0$. Every physical state has positive probability, so equivalence requires all three values to be strictly positive. ThusFor each price in this open interval the pricing kernel takes values $3q_{15},3q_{12},3q_9$ in the three equally likely states, proving absence of arbitrage.

The necessity, including the exclusion of endpoints, can also be checked directly. The three terminal payoff vectors of cash, the stock and the European put option form the invertible matrixEach unit state payoff therefore has a replicating strategy, whose initial cost is the corresponding $q$. A zero or negative $q$ supplies an arbitrage, so the endpoints are genuinely excluded.

Write $(a,b,c)$ for holdings of cash, the stock and the European put option, and let $\gamma =-V_0$ be initial consumption. At the given price, $a=-10b-c-\gamma$, and the three terminal payoffs, in descending order of the stock price, areNonnegative terminal payoffs require $c+\gamma \le 2b$ and $c-\gamma \ge b$. Together with $\gamma \ge 0$, these imply $b\ge 2\gamma \ge 0$. If $b=0$ they force $\gamma =c=a=0$, which is not an arbitrage. If $b>0$, the highest-state payoff is at least $3b>0$, so every such choice is an arbitrage. Consequently the complete set isThe pure-investment arbitrages are exactly the choices $\gamma =0$:For example, $(a,b,c)=(-11,1,1)$ costs zero and pays $(4,1,0)$.

Assume the usual positive initial stock price and strike, so the logarithm is defined, and interpret $U$ as a classical solution of the displayed partial differential equation. For a fixed complex $z$, put $F_t(z)=S_t^{z}U(t,{\sigma }_t,z)$, defining $S_t^z=\exp (z\log S_t)$. The Itô formula for the two Brownian motions with correlation coefficient $\rho$ givesThe partial differential equation cancels the entire drift. By the permission to treat the resulting local martingales as true martingales, and the terminal condition, $F_t(z)=\mathbb{E}[S_T^z\mid {\mathcal{F}}_t]$.

The bounded spot volatility ensures $\mathbb{E}{S}_T^{x_0}<\infty$: the Itô formula for a real power and localization bound its moment by $S_0^{x_0}\exp (x_0(x_0-1)\Vert \sigma {\Vert }_{\infty }^{2}T/2)$. Rewrite the proposed integrand asNow $|F_t(x_0+iy)|\le \mathbb{E}[S_T^{x_0}\mid {\mathcal{F}}_t]$, an integrable bound independent of $y$. Conditional Fubini's theorem and part (b) therefore implyThis route justifies the contour exchange without assuming bounds on $U$ uniform in all complex $z$. Cash is constant, $S$ is a true martingale under the stated allowance, and the displayed $C$ is a true martingale. The original measure is thus an equivalent martingale measure for all three assets. The fundamental theorem of asset pricing gives the market has no arbitrage under the usual admissible trading convention.

Let ${\theta }_t$ and ${\eta }_t$ be the predictable holdings of the stock and the bank account. With no consumption, the self-financing portfolio has $X=\theta S+\eta B$ and $dX=\theta dS+\eta dB$. The Itô product rule, together with the dynamics from part (a), givesThe drift vanishes because $X=\theta S+\eta B$ and $\sigma \lambda =\mu -r$. Thus $XY$ is a local martingale. It is nonnegative by the assumed nonnegative wealth and strict positivity of $Y$, and henceFor example, this last fact follows directly from Conditional Fatou lemma applied to a localizing sequence; the initial capital is assumed finite. As usual, holdings must be integrable against the asset semimartingales for the self-financing portfolio equation to be defined.

With constant coefficients, $\lambda =(\mu -r)/\sigma$ is constant, so the stochastic exponential $Z_t=e^{rt}{Y}_t$ is a true martingale. Under its equivalent martingale measure $\mathbb{Q}$, the stock follows the Black-Scholes model $d{S}_t=S_t(rdt+\sigma d{W}_t^{\mathbb{Q}})$. The Gaussian exponential moment gives the value of the power optionThe option delta is ${\partial }_{s}V=V/(2s)$, so the replicating strategy holdsHalf the wealth value is in the stock and half in the bank account, with continuous rebalancing. To check the self-financing portfolio property under the original measure, the Itô formula gives $d{V}_t=\frac{1}{2}(r+\mu )V_{t}dt+\frac{1}{2}\sigma V_{t}d{W}_t={\theta }_{t}d{S}_t+{\eta }_{t}d{B}_t$. Also $V_T=\sqrt{S_T}$ and $V_t>0$.

The payoff is unbounded, so the bounded-payoff restriction of part (c) is not invoked automatically; the Black-Scholes model has the necessary finite Gaussian moments, and the explicit strategy attains $V_0=\mathbb{E}[Y_T\sqrt{S_T}]$. The supermartingale bound from part (b) therefore proves its minimality. The cost is independent of the physical drift $\mu$.

For risk-neutral stock dynamics $dS=(r-\delta )Sdt+\sigma SdW$, with $r>0$ and $\sigma >0$, let ${\beta }_{-}<0$ solve ${\sigma }^2\beta (\beta -1)/2+(r-\delta )\beta -r=0$. A fixed downward trigger $L\in (0,X)$ gives continuation value $(X-L)(S/L{)}^{{\beta }_{-}}$ for $S>L$. Optimizing over the trigger, or imposing smooth fit, gives the displayed boundary. The optimal value equals $X-S$ below $S^{\ast }$ and the continuation expression above. A verification argument uses its payoff majorization and nonpositive discounted generator. At $r=0$ and $\delta \ge 0$, the perpetual value is $X$ for positive spot, approached by triggers tending to zero rather than attained at a positive finite trigger.