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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.

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.

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.

The difference of the terminal European call option and European put option payoffs is $S_T^1-K$. Replicating that difference by one share and a borrowing position gives conditional put-call parityThus the call is more valuable exactly when the right side is positive. With a put selected at equality, the chooser option payoff isAt equality both choices have the same price, even though their eventual payoffs can differ. Put-call parity assumes no dividends or other intermediate stock cash flows, as in the stated asset-price model.

Put $K^{\prime }=K(1+r{)}^{t_0-T}$. At the choice time,Buy a European call option of strike $K$ and maturity $T$, together with a European put option of strike $K^{\prime }$ and maturity $t_0$. If the call is selected, the short-maturity put expires worthless. If the put is selected, sell the call and combine its value with the short-maturity put's cash payment; put-call parity makes this exactly enough to buy the maturity-$T$ put. This self-financing replication provesIt requires tradability of the priced options at the choice time, but not completeness of every other market claim. The identity also covers $t_0=0$ and $t_0=T$.

Let $p_K=C(T,K+1)-2C(T,K)+C(T,K-1)>0$ and $D_K=C(T+1,K)-C(T,K)$. Given $S_T=K$, the next value is one of $K-1,K,K+1$. The martingale property makes its upward and downward conditional probabilities equal; their sum is the conditional variance of the increment, ${\sigma }^2(T,K)=2D_K/p_K$. ThusThe discrete Dupire equation therefore recovers all positive-probability one-step conditional transitions from the European call option price surface. No Markov property is required: these are conditional probabilities given the current value, not necessarily given the whole past.

Recovery is impossible at a zero-probability state. For example, $S_t\equiv S_0$ satisfies every stated assumption, but at $T\ge 1$ a different integer $K$ can satisfy $|K-S_0|\le T$ while $p_K=0$. Its conditional transitions can be specified arbitrarily without altering $C$. Thus the literal all-states claim needs the positive-probability qualification.

By the bound in part (a), Fubini's theorem applies. Conditional on $S$,where the Characteristic function of the Cauchy distribution was used. Since $\mathbb{E}S=1$,The formula expresses a European call option value through complex moments of $S$. In an affine stochastic-volatility model such as the Heston model, those moments are available from an explicit transform, so call prices reduce to a one-dimensional Fourier expectation or integral.

For a European call option, the pointwise inequality $(S_T-K{)}^{+}\ge S_T-K$ and risk-neutral pricing give

At time $t$, immediate exercise is worth $(S_t-K{)}^{+}$, while waiting is worth the European call option price with remaining maturity $\tau =T-t$. The conditional version of part (iii) givesUnder the standard Black-Scholes assumption $r\ge 0$, this is at least $S_t-K$, while a call value is nonnegative. Hence $C(t,S_t)\ge (S_t-K{)}^{+}$, and the investor should wait until $T$; with $\sigma >0$ and $\tau >0$, waiting has strictly positive time value.

If negative interest rates are allowed, the unconditional instruction “wait” need not be correct. The exact decision is to exercise whenand otherwise to continue.

For the European call option, risk-neutral valuation givesDefine the stock-numeraire measure in a binomial market byIt is a probability measure because the risk-neutral martingale property gives ${\mathbb{E}}_Q{S}_N=S_0(1+r{)}^N$. Consequently

Tilting one up move by its stock factor changes its probability toSimilarly $1-\hat{q}=(1-q)(1+a)/(1+r)$. Thus under $\hat{Q}$ the number of up moves is binomial with parameter $\hat{q}$, and

The European call option with strike $5$ pays $2,0,0$ at terminal stock prices $7,5,3$. Its lower-node value is zero, while its upper-node value isA second step of backward option pricing givesBy the fundamental theorem of asset pricing, this discounted expectation under the unique risk-neutral probabilities is the unique no-arbitrage price.

A European call option with maturity $N$ and strike $K$ paysat time $N$. Let $q$ be the risk-neutral conditional probability of the factor $1+b$. The discounted stock must be a martingale, soand thereforeBoth probabilities are strictly positive. At every node these are the unique probabilities satisfying the martingale condition, so the fundamental theorem of asset pricing gives a unique no-arbitrage price.

If exactly $n$ of the $N$ moves use the factor $1+a$, thenand there are $\left(\genfrac{}{}{0ex}{}{N}{n}\right)$ such paths. Discounted risk-neutral expectation now giveswhere