Static arbitrage and volatility surfaces

Definition Consider a frictionless market in which the stock, zero-coupon bonds and a set of European options on the stock trade at their quoted prices, with no limits on short sales. A static arbitrage is a portfolio of these instruments that is chosen at time $0$ and then held without trading until the options expire, such that its cost is at most zero, its payoff is nonnegative in every state, and either its cost is negative or its payoff is positive with positive probability. A set of quotes is free of static arbitrage when no such portfolio exists. Throughout, $D(T) = P(0, T) > 0$ is the time-$0$ price of a zero-coupon bond paying $1$ at $T$, which the strike tests use as a traded price and which equals the discount factor to $T$ when rates are deterministic, and $C(K, T)$ is the time-$0$ price of a European call with strike $K \ge 0$ and maturity $T$, all calls having the same underlying, settlement and exercise style. If rates are deterministic and there is a pricing measure $\mathbb{Q}$, equivalent to $\mathbb{P}$, with $\mathbb{E}^{\mathbb{Q}}[S_T] < \infty$ and

for every strike, then every portfolio of these calls and the bond of maturity $T$ costs $D(T)$ times the $\mathbb{Q}$-expectation of its payoff, and none of them is a static arbitrage. The tests in the next cards are necessary conditions for this. They concern executable prices of identical contracts, so with bid and ask quotes a test uses the ask for every option bought and the bid for every option sold.

Definition For strikes $K_1 < K_2$ with the same maturity $T$, the call spread that is long the call at $K_1$ and short the call at $K_2$ pays

which lies between $0$ and $K_2 - K_1$. In a market free of static arbitrage this gives strike monotonicity together with a slope bound,

If the left inequality failed, buying the spread would bring in cash and leave a nonnegative payoff. If the right one failed, selling the spread and buying $K_2 - K_1$ bonds would do the same, since the bonds pay $K_2 - K_1$, at least what the spread pays. Dividing by $K_2 - K_1$, every difference quotient of $C(\cdot, T)$ lies in $[-D(T), 0]$, and where the price is differentiable in $K$,

Under the pricing representation of the first card the right derivative is $C_K(K+, T) = -D(T)\,\mathbb{Q}(S_T > K)$, minus the price of a digital call. The inequalities need not be strict. If $\mathbb{Q}(S_T > K_1) = 0$, both calls are worth zero and the spread costs nothing.

Definition For strikes $K_1 < K_2 < K_3$ of the same maturity let

so that $K_2 = \lambda K_1 + (1 - \lambda) K_3$. The butterfly spread is long $\lambda$ calls at $K_1$, short one call at $K_2$ and long $1 - \lambda$ calls at $K_3$. For each $s \ge 0$ the payoff $(s - K)^{+}$ is a convex function of $K$, so the butterfly pays a nonnegative amount. Its payoff is zero for $S_T \le K_1$ and for $S_T \ge K_3$, rises linearly to $\lambda (K_2 - K_1)$ at $S_T = K_2$ and falls linearly back to zero. In a market free of static arbitrage its price is nonnegative, which is strike convexity,

For equally spaced strikes $K - h$, $K$, $K + h$ the weights are $\tfrac12$, $-1$, $\tfrac12$, and the condition reads $C(K - h, T) - 2C(K, T) + C(K + h, T) \ge 0$. Equivalently, the slopes between adjacent strikes are nondecreasing. Where $C$ is twice differentiable in $K$, $C_{KK}(K, T) \ge 0$. If $S_T$ has a density $p_T$ under $\mathbb{Q}$ in the representation of the first card, then $C_{KK}(K, T) = D(T)\,p_T(K)$, so negative curvature would be a negative state-price density. With atoms in the law of $S_T$ the price has kinks and the inequality holds as a statement about slopes.

Definition Assume a constant rate $r$ and a constant dividend yield $q$, so that $D(T) = e^{-rT}$ and the forward is $F(0, T) = S_0 e^{(r - q)T}$. Assume that call prices are given by a pricing measure $\mathbb{Q}$ as in the first card, under which $S_t \ge 0$ and the value of the stock with dividends reinvested, discounted by the money-market account, $e^{qt} S_t/B_t$ with $B_t = e^{rt}$, is a true martingale with respect to the filtration $(\mathcal{F}_t)$. Then $Y_t = S_t/F(0, t)$ is a nonnegative $\mathbb{Q}$-martingale with $Y_0 = 1$. For the normalized strike $k = K/F(0, T) \ge 0$ define the normalized call price

For $T_1 \le T_2$ and every $k \ge 0$, the conditional Jensen inequality for the convex function $y \mapsto (y - k)^{+}$ and the martingale property give

and taking expectations proves calendar consistency,

The comparison is at the same normalized strike, so the cash strikes are $K_1 = kF(0, T_1)$ and $K_2 = kF(0, T_2)$, and in cash terms it reads $e^{q(T_2 - T_1)} C(K_2, T_2) \ge C(K_1, T_1)$. The raw price at a fixed cash strike need not increase with maturity when $q > 0$. When the inequality fails in a frictionless market, selling the call at $(K_1, T_1)$ and buying $e^{q(T_2 - T_1)}$ calls at $(K_2, T_2)$ brings in cash. If the sold call ends in the money, shorting one share at $T_1$ to pay its payoff $S_{T_1} - K_1$, paying the dividends on the short by shorting more, and investing the remaining $K_1$ until $T_2$ leaves the payoff $e^{q(T_2 - T_1)}(K_2 - S_{T_2})^{+} \ge 0$ at $T_2$, and otherwise the bought calls are simply held. This calendar spread trades the stock once, at $T_1$. A surface is usually called free of static arbitrage when it passes the strike tests and this calendar test. Stochastic rates, discrete dividends or different settlement conventions need their own numeraire and carry treatment, which this lesson does not cover.

Definition Fix $S_0$, a constant rate $r$ and dividend yield $q$ as in the calendar card, and the contract conventions. For each strike $K$ and maturity $T$ whose call price lies strictly inside the arbitrage bounds $D(T)(F(0, T) - K)^{+} < C(K, T) < D(T)F(0, T)$, the implied volatility $\sigma_{\mathrm{imp}}(K, T)$ is the unique Black-Scholes volatility that reproduces $C(K, T)$. The function $(K, T) \mapsto \sigma_{\mathrm{imp}}(K, T)$ on the quoted points is the volatility surface. It is often written in the normalized strike $k = K/F(0, T)$ together with the total implied variance

The normalized Black-Scholes call price $c_{\mathrm{BS}}(k, w) = \Phi(d_1) - k\,\Phi(d_2)$, with $d_1 = (-\ln k + w/2)/\sqrt{w}$ and $d_2 = d_1 - \sqrt{w}$, is strictly increasing in $w$, so the calendar test of the previous card is equivalent to $w(k, T)$ being nondecreasing in $T$ at each fixed $k$. At each point the surface and the price carry the same information, but the conditions of this lesson are linear in prices and nonlinear in volatilities. A flat surface comes from one Black-Scholes model and is free of static arbitrage, while a smooth-looking surface that is not flat may still violate a butterfly or calendar test. The surface is a set of coordinates for European prices. It is not a model of how the stock moves, and it does not say how the surface itself will move.

Definition Let calls of one maturity be quoted at strikes $K_1 < \dots < K_n$, with prices $C_j = C(K_j, T)$ and adjacent slopes $m_j = (C_{j+1} - C_j)/(K_{j+1} - K_j)$ that pass the strike tests,

An interpolation $C(K)$ between the quoted strikes is arbitrage-free in strike when it is convex and has every slope in $[-D(T), 0]$. The piecewise linear interpolation has exactly the slopes $m_j$, so it qualifies. Its curvature is zero between the grid points, so as a state-price distribution it puts the $\mathbb{Q}$-mass $(m_j - m_{j-1})/D(T)$ at each interior strike $K_j$ and none in between. These are atoms, not a smooth density. A smooth interpolation must keep $C_{KK} \ge 0$ and $-D(T) \le C_K \le 0$ everywhere, and a smooth curve of implied volatilities guarantees neither, because the map from volatility to price is nonlinear and depends on the strike. Extrapolation beyond the grid must stay consistent with total probability one and with the forward, since $C(0, T) = D(T)F(0, T)$ and $C(K, T) \to 0$ as $K \to \infty$ when $\mathbb{E}^{\mathbb{Q}}[S_T] < \infty$. Across maturities, under the assumptions of the calendar card, linear interpolation of normalized prices in $T$ at fixed $k$ keeps all the tests, because a convex combination of $c(\cdot, T_1)$ and $c(\cdot, T_2)$ is the normalized call price of a mixture of the two laws of $Y$, which still has mean one, and it moves monotonically from one slice to the other. When bid and ask quotes allow several consistent fits, choosing one of them is a choice of regularization.