Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 38 / 4

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.