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The joint generator. With $\theta$ again measured in dollars, portfolio wealth obeysIts noise and the factor noise are driven by the same Brownian motion. Their quadratic covariation is $\sigma (x)\theta \alpha (x)dt$, so the diffusion generator has a cross derivative. Dynamic programming and the Itô formula giveAll coefficient functions in this formula are evaluated at $x$. The cross derivative is essential: it gives intertemporal hedging demand. Normalizing CRRA utility as $U(c)=c^p/p$, where $p=1-R$, the two optimizations give, at nonzero volatility,and henceHere $p(-1/R)=1-1/R$.

Homogeneity and the reduced equation. Scaling initial portfolio wealth and both controls preserves the portfolio wealth constraint and multiplies the objective by the positive number $b^p$. ThusThis expression is valid for both signs of $p$: $V$ is negative when $R>1$, but its portfolio wealth derivative is positive. Its derivatives areSubstitution yieldsThe resulting feedback isThe first portfolio term is myopic, and the second is intertemporal hedging demand.

Constant market price of risk. If $\mu -r=\sigma \kappa$ and volatility is nonzero, the portfolio term becomes $p(\kappa f+\alpha f^{\prime }{)}^2/(2Rf)$, so the magnitude of stock volatility disappears. Applying the power transformation of a complete-market investment equation $f=g^R$ cancels the squared-gradient terms and gives the further reductionFor ${\gamma }_M>0$ its economic solution is $g=1/{\gamma }_M$. ThusTo see why this solves the investment-consumption problem, optimize directly over Brownian portfolio exposures, writing $y=\sigma (x)\theta$. The portfolio wealth equation becomes $dw=ydW+(rw+\kappa y-c)dt$, which no longer contains $X$. With nonzero volatility the same admissible exposure processes are available for every factor state, so the attainable wealth-consumption pairs, and therefore the value, are exactly those of the Merton consumption-investment problem. This also excludes extraneous solutions of the linear equation without imposing artificial factor boundary data.

The printed boundedness assumptions do not ensure nonzero volatility or a finite value. The unsimplified HJB equation remains the correct control equation at a zero of $\sigma$. There the hedge term vanishes; if $\mu \ne r$, the riskless excess return gives arbitrage with unrestricted holdings. Under $\mu -r=\sigma \kappa$, a zero-volatility state offers only the bank exposure at that instant. For example $\sigma \equiv 0$, $\mu \equiv r$ satisfies this relation for any chosen $\kappa$, but its value uses ${\gamma }_0=[\rho -pr]/R$, not a fictitious nonzero risk premium. The constant-value formula using $\kappa$ presupposes access to the Brownian exposure, with sufficient integrability for the corresponding holdings. Additive utility constants again only shift $V$ by a constant divided by $\rho$.

Buy one lower-strike European call option and sell one higher-strike European call option. The initial cost is $C(K_i)-C(K_{i+1})<0$, so the strategy releases strictly positive cash. Its terminal payoff isfor every stock price. Thus it is an arbitrage: a positive initial receipt accompanies a nonnegative terminal obligation. One may consume the receipt immediately, or hold it in cash to make a zero-initial-capital strategy with strictly positive terminal wealth. This is the vertical-spread arbitrage for increasing call prices.

Buy half a call at each neighboring strike and sell one call at the middle strike. Its cost isFor fixed terminal stock price $s$, the function $K\mapsto (s-K{)}^{+}$ is convex. Since $K_i$ is the midpoint, the payoffis nonnegative for every $s$. More explicitly, it is zero outside $[K_{i-1},K_{i+1}]$, equals $(s-K_{i-1})/2$ on $[K_{i-1},K_i]$, and equals $(K_{i+1}-s)/2$ on $[K_i,K_{i+1}]$. The negative cost and nonnegative payoff produce an arbitrage. This is the butterfly-spread arbitrage for nonconvex call prices.

Use positive strikes, the natural real-power domain of this price curve. For any such $K$, the stock-minus-call payoff isalmost surely, since $S_1>0$. Therefore the call price must be strictly below the stock price $S_0=1$: if $C(K)\ge 1$, buying stock and selling the call has nonpositive initial cost and strictly positive terminal payoff. Also a negative call price is an immediate arbitrage by buying the call.

For $p<0$, $1+K^p>K^p$ and raising to the negative power $1/p$ reverses the inequality, giving $(1+K^p{)}^{1/p}<K$ and $C(K)<0$. The expression is undefined at $p=0$. For $0<p<1$, strict concavity of the power implies $(1+K{)}^p<1+K^p$, whenceAt $p=1$, $C(K)=1$. Every defined case with $p\le 1$ therefore violates the necessary no-arbitrage bounds. ConsequentlyThis argument does not require a dense family of strikes; even one positive-strike call gives the contradiction. The strict stock-minus-call payoff explains why the borderline $p=1$ is also excluded.

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.

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.

Under the usual no-arbitrage interpretation from part (a), let $Q^N$ be an equivalent martingale measure for the strictly positive numéraire $N$. Then $M_t=S_t/N_t$ is a martingale. Define the discounted European call option payoff $Y_t=(S_t-K{)}^{+}/N_t=(M_t-K/N_t{)}^{+}$. Because $N_{t+1}\ge N_t$ and $K>0$, we have $K/N_{t+1}\le K/N_t$. The conditional Jensen inequality therefore givesFor the middle step, the strike $K/N_t$ is fixed conditionally on ${\mathcal{F}}_t$; alternatively use $\mathbb{E}[X^{+}\mid {\mathcal{F}}_t]\ge \max \{\mathbb{E}[X\mid {\mathcal{F}}_t],0\}$. Thus $Y$ is a submartingale. Risk-neutral valuation in units of $N$ gives $C(T,K)=N_0{\mathbb{E}}^{Q^N}{Y}_T$, with the usual fixed initial prices. ConsequentlyThe source's word “increasing” means nondecreasing: a European call option can have zero payoff at successive maturities, so strict increase is not guaranteed. Integrability is understood in the claim class for which the stated replication prices exist.

The positive-part function is convex, so pathwise Jensen inequality givesTo compare the Asian option price with European call option prices at different dates, carry each earlier payoff forward using the numéraire. Purchase $1/T$ of each replicating strategy for maturity $t$, and, when its payoff is received, reinvest it in $N$ until $T$. This is a self-financing portfolio, with terminal wealthThe first inequality uses $N_T\ge N_t$ and nonnegative payoffs. Since the Asian option is replicable in the complete market, absence of arbitrage makes its replication cost no larger than this superhedging cost. ThereforeEquivalently, divide the first pathwise bound by $N_T$, use $(S_t-K{)}^{+}/N_T\le (S_t-K{)}^{+}/N_t$, and take expectations under the numéraire equivalent martingale measure. Merely averaging earlier payoffs without reinvestment would miss the difference in payment dates.

For holdings $h\in {\mathbb{R}}^n$, let $V_0=h\cdot P_0$ and $V_1=h\cdot P_1$. Under the investment-consumption arbitrage convention used here, an arbitrage satisfiesThe initial consumption is $-V_0$, and the terminal consumption is $V_1$: there is a possible gain with no external funding and no negative consumption. A pure-investment arbitrage additionally requires $V_0=0$. This convention is important because the last part distinguishes initial consumption from pure investment.

For any holdings $h$, the pricing kernel identity implies $V_0=\mathbb{E}[R{V}_1]$. If $V_1\ge 0$ almost surely, then $V_0\ge 0$, so an arbitrage cannot have $V_0<0$. If also $V_0=0$, then the nonnegative integrable random variable $R{V}_1$ has zero expectation and hence is zero almost surely. Since $R>0$ almost surely, $V_1=0$ almost surely, ruling out a positive terminal gain. Therefore a pricing kernel rules out arbitrage.

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.