Digital options and distribution recovery, financial engineering for quants lesson

Suppose European calls on one stock trade at every strike for a single maturity. A contract that pays one unit of currency if the stock finishes above 100 is not among them, and neither is a claim that pays when the stock finishes between 99 and 101, yet with deterministic discounting the call prices already fix what both are worth. The slope of the call price in the strike gives the digital price, and its curvature gives the discounted pricing density of the terminal stock price. This lesson derives both relations, checks them on the Black-Scholes formula, and shows what changes when the terminal law has atoms, when only a few strikes are quoted and when discounting is random.

Key terms

Cash-or-nothing digital option

Definition Fix the current time $t$ , a maturity $T > t$ and $\tau = T - t$ in years. Assume a constant continuously compounded rate $r$ , so that the money-market account is $B_u = e^{ru}$ and the discount factor from $T$ back to $t$ is the deterministic number $e^{-r\tau}$ , and let $\mathbb{Q}$ be a pricing measure for this numeraire. Write $\mathbb{E}^{\mathbb{Q}}_t[\,\cdot\,] = \mathbb{E}^{\mathbb{Q}}[\,\cdot \mid \mathcal{F}_t]$ and $\mathbb{Q}_t$ for the conditional expectation and the conditional probability given the information $\mathcal{F}_t$ at time $t$ . A unit cash-or-nothing digital call with strike $K > 0$ pays one unit of currency at $T$ if $S_T > K$ and nothing otherwise. Its payoff is $\mathbf{1}_{\{S_T > K\}}$ and its value at $t$ is

\begin{aligned} D(K) &= e^{-r\tau}\,\mathbb{E}^{\mathbb{Q}}_t\bigl[\mathbf{1}_{\{S_T > K\}}\bigr] \\[0.4em] &= e^{-r\tau}\,\mathbb{Q}_t(S_T > K). \end{aligned}

A digital that pays $A$ units is worth $A\,D(K)$ . Some contracts pay on $\{S_T \ge K\}$ instead, and the two prices differ by $e^{-r\tau}\,\mathbb{Q}_t(S_T = K)$ , which is zero unless the terminal law puts mass on $K$ . The digital put pays on the complementary event $\{S_T \le K\}$ , so the digital call and the digital put together pay one unit for sure and their values add up to $e^{-r\tau}$ . In the Black-Scholes model with $\sigma > 0$ and $\tau > 0$ , given $S_t = S$ ,

D(K) = e^{-r\tau}\,\Phi(d_2),

where

\begin{aligned} d_2 &= \frac{\ln(S/K) + (r - q)\tau}{\sigma\sqrt{\tau}} \\[0.4em] &\quad - \tfrac{1}{2}\sigma\sqrt{\tau}. \end{aligned}

At expiry the value is the payoff $\mathbf{1}_{\{S_T > K\}}$ itself. Because the payoff jumps at $K$ , the Black-Scholes delta of the digital, $e^{-r\tau}\varphi(d_2)/(S\sigma\sqrt{\tau})$ , grows without bound as $\tau \to 0$ when $S = K$ , while the delta $e^{-q\tau}\Phi(d_1)$ of a call stays between $0$ and $1$ .

In plain words A digital is a bet on one event that pays a fixed amount. Its price is the discounted pricing probability of that event. That probability is a weight implied by prices under $\mathbb{Q}$ , not a forecast of how often the event happens.

Example Take the constructed inputs $S = K = 100$ , $r = q = 0$ , $\sigma = 0.2$ and $\tau = 1$ . Then $d_2 = -0.1$ and

D(100) = \Phi(-0.1) \approx 0.460172.

The digital put is worth $\Phi(0.1) \approx 0.539828$ , and the two add up to $1 = e^{-r\tau}$ . The pricing probability of finishing above 100 is below one half even though the forward price $S e^{(r - q)\tau}$ equals 100, because under $\mathbb{Q}$ the median of $S_T$ is $100\,e^{-\sigma^2\tau/2} = 100\,e^{-0.02} \approx 98.02$ .

Strike derivative

Definition Write $C(K)$ for the time $t$ value of the European call with strike $K$ and maturity $T$ , viewed as a function of $K > 0$ with $t$ , $T$ , the current state and the model held fixed,

C(K) = e^{-r\tau}\,\mathbb{E}^{\mathbb{Q}}_t\bigl[(S_T - K)^{+}\bigr].

Assume $S_T \ge 0$ and $\mathbb{E}^{\mathbb{Q}}_t[S_T] < \infty$ , so that every call value is finite. The strike derivative is $C_K = \partial C/\partial K$ . For every $K > 0$ both one-sided derivatives exist, and

\begin{aligned} C_{K,+}(K) &= -e^{-r\tau}\,\mathbb{Q}_t(S_T > K), \\[0.4em] C_{K,-}(K) &= -e^{-r\tau}\,\mathbb{Q}_t(S_T \ge K). \end{aligned}

For the right derivative take $\varepsilon > 0$ and write the normalized call spread as

\begin{aligned} &\frac{C(K) - C(K + \varepsilon)}{\varepsilon} \\[0.4em] &\quad = e^{-r\tau}\,\mathbb{E}^{\mathbb{Q}}_t\bigl[f_{\varepsilon}(S_T)\bigr], \end{aligned}

where $f_{\varepsilon}(s) = \bigl((s - K)^{+} - (s - K - \varepsilon)^{+}\bigr)/\varepsilon$ is $0$ for $s \le K$ , rises linearly from $0$ to $1$ on $[K, K + \varepsilon]$ and equals $1$ for $s \ge K + \varepsilon$ . As $\varepsilon \downarrow 0$ it converges to $\mathbf{1}_{\{s > K\}}$ at every $s$ , and $0 \le f_{\varepsilon} \le 1$ , so dominated convergence moves the limit inside the expectation. The discount factor is a deterministic constant and comes out of the expectation. The left derivative follows in the same way from $\bigl((s - K + \varepsilon)^{+} - (s - K)^{+}\bigr)/\varepsilon$ , whose limit is $\mathbf{1}_{\{s \ge K\}}$ . The two derivatives agree, and $C$ is differentiable at $K$ , exactly when $\mathbb{Q}_t(S_T = K) = 0$ , and then

D(K) = -C_K(K).

The strike derivative is dimensionless, currency per currency unit of strike, and lies in $[-e^{-r\tau}, 0]$ . It should not be confused with the stock delta $C_S = \partial C/\partial S$ , which differentiates the same price function in a different variable.

In plain words Raising the strike from $K$ to $K + \varepsilon$ lowers the call payoff by $\varepsilon$ on every path that finishes above $K + \varepsilon$ , by less on paths that finish between $K$ and $K + \varepsilon$ , and not at all on the others. Per unit of strike, the call therefore loses value at the rate of the discounted pricing probability of finishing above the strike, and that is the digital price.

Example With the inputs of the first card the call is worth $C(100) \approx 7.965567$ . One-sided and centered call spreads give

width $h$	one-sided $(C(100) - C(100 + h))/h$	centered $(C(100 - h) - C(100 + h))/(2h)$
$1$	$0.450300$	$0.460222$
$0.01$	$0.460073$	$0.460172$

Both approach $D(100) \approx 0.460172$ . The centered spread gets there much faster, because for a smooth $C$ its error is of order $h^2$ , while the error of the one-sided spread is of order $h$ . The stock delta at the same point is $e^{-q\tau}\Phi(d_1) = \Phi(0.1) \approx 0.539828$ , a different number.

Breeden-Litzenberger formula

Definition Keep the deterministic discount factor $e^{-r\tau}$ , and suppose that given $\mathcal{F}_t$ the terminal price $S_T$ is strictly positive, has a finite mean and has a $\mathbb{Q}$ density $p_{\mathbb{Q}}$ on $(0, \infty)$ that is continuous. Then $C$ is twice continuously differentiable on $(0, \infty)$ , its slope is

C_K(K) = -e^{-r\tau}\,\mathbb{Q}_t(S_T > K),

and its second derivative is

C_{KK}(K) = e^{-r\tau}\,p_{\mathbb{Q}}(K),

so that

p_{\mathbb{Q}}(K) = e^{r\tau}\,C_{KK}(K).

This is the Breeden-Litzenberger formula. To derive it, write the call as

C(K) = e^{-r\tau}\!\int_K^{\infty} (s - K)\,p_{\mathbb{Q}}(s)\,ds.

A density puts no mass on single points, so the previous card makes $C$ differentiable with

C_K(K) = -e^{-r\tau}\!\int_K^{\infty} p_{\mathbb{Q}}(s)\,ds.

Since $p_{\mathbb{Q}}$ is continuous, the fundamental theorem of calculus gives the tail integral the derivative $-p_{\mathbb{Q}}(K)$ , and differentiating once more gives the formula. Without continuity of the density, $C_K$ is still absolutely continuous and the formula holds for almost every $K$ . The formula recovers the conditional law of $S_T$ under $\mathbb{Q}$ at the one maturity $T$ . It says nothing about the physical law under $\mathbb{P}$ , and nothing about the joint law of the price at several dates, so two models with the same terminal law can still give different prices to a barrier option.

In plain words The slope of the call price in the strike is minus the discounted probability of finishing above the strike. The rate at which that slope changes is the discounted probability per unit of price of finishing near the strike. Reading these two derivatives off call prices recovers the pricing distribution of the terminal price.

Example In the Black-Scholes model,

\begin{aligned} C(K) &= S e^{-q\tau}\Phi(d_1) \\[0.4em] &\quad - K e^{-r\tau}\Phi(d_2), \end{aligned}

with $d_1 = d_2 + \sigma\sqrt{\tau}$ . Both $d_1$ and $d_2$ have the strike derivative $-1/(K\sigma\sqrt{\tau})$ , and the identity $S e^{-q\tau}\varphi(d_1) = K e^{-r\tau}\varphi(d_2)$ from Greeks and differential identities cancels the two terms that contain the normal density, leaving $C_K = -e^{-r\tau}\Phi(d_2) = -D(K)$ . Differentiating once more gives

C_{KK}(K) = \frac{e^{-r\tau}\varphi(d_2)}{K\sigma\sqrt{\tau}}.

This is $e^{-r\tau}$ times the lognormal density of $S_T$ at $K$ , because $\ln S_T$ is normal with mean $\ln S + (r - q - \sigma^2/2)\tau$ and variance $\sigma^2\tau$ , and the factor $1/K$ comes from the change of variable $s = e^{x}$ . The density computed from the payoff expectation and the one computed from the closed-form price agree. With the constructed inputs, $\varphi(-0.1) \approx 0.396953$ and

\begin{aligned} p_{\mathbb{Q}}(100) &= \frac{\varphi(-0.1)}{100 \cdot 0.2} \\[0.4em] &\approx 0.019848 \end{aligned}

per currency unit, which also equals $C_{KK}(100)$ because $r = 0$ .

State-price density

Definition Under the hypotheses of the Breeden-Litzenberger formula, the state-price density at time $t$ for maturity $T$ is the function

\begin{aligned} \pi_t(s) &= e^{-r\tau}\,p_{\mathbb{Q}}(s) \\[0.4em] &= C_{KK}(s), \qquad s > 0, \end{aligned}

measured in units of inverse currency. For $0 < a < b$ the claim that pays $\mathbf{1}_{\{a < S_T \le b\}}$ at $T$ is worth

\begin{aligned} \int_a^b \pi_t(s)\,ds &= C_K(b) - C_K(a) \\[0.4em] &= D(a) - D(b), \end{aligned}

and a payoff $h(S_T)$ with $h \ge 0$ measurable, or with $\mathbb{E}^{\mathbb{Q}}_t[\lvert h(S_T) \rvert] < \infty$ , is worth $\int_0^{\infty} h(s)\,\pi_t(s)\,ds$ . The density integrates to $e^{-r\tau}$ , the price of one unit paid for sure. Some texts give the name state-price density to the pricing kernel of One period replication and state prices, which is a density of state prices with respect to $\mathbb{P}$ . Here it is a density in the terminal price level. When interest rates are random, the call value is

\begin{aligned} C(K) &= \mathbb{E}^{\mathbb{Q}}_t\Bigl[e^{-\int_t^T r_u\,du} \\[0.4em] &\qquad \times (S_T - K)^{+}\Bigr], \end{aligned}

and the same strike derivatives recover the discounted state prices $\mathbb{E}^{\mathbb{Q}}_t\bigl[e^{-\int_t^T r_u\,du}\,\mathbf{1}_{\{S_T \in A\}}\bigr]$ of sets $A$ of terminal prices. Their total mass is the zero-coupon bond price $P(t,T)$ , and after division by $P(t,T)$ they give the law of $S_T$ under the $T$ -forward measure $\mathbb{Q}^{T}$ , the pricing measure for the bond numeraire, not its law under $\mathbb{Q}$ . The two laws agree when, given $\mathcal{F}_t$ , the discount factor and $S_T$ are independent under $\mathbb{Q}$ , which is the case with deterministic rates.

In plain words The state-price density spreads the price $e^{-r\tau}$ of a sure unit over the possible terminal prices. Multiplying it by the width of a small price interval gives approximately what a claim paying one unit on that interval costs today. Its value at a single point is a rate per unit of price, not the price of that point, which is zero.

Example With the constructed inputs, $\pi_t(100) \approx 0.019848$ per currency unit. The claim that pays one unit if $99 < S_T \le 101$ is then worth about $2 \cdot 0.019848 \approx 0.039695$ , while its exact lognormal value, $\mathbb{Q}_t(99 < S_T \le 101)$ since $r = 0$ , is about $0.039681$ . The small gap is the error of treating the density as constant over an interval of width 2. A claim that pays one unit only if $S_T$ equals 100 exactly is worth nothing.

Call price curvature

Definition The call price curvature is the second strike derivative $C_{KK}$ where it exists. For every terminal law with $S_T \ge 0$ and $\mathbb{E}^{\mathbb{Q}}_t[S_T] < \infty$ , with or without a density, the call value is convex and nonincreasing in $K$ , because $(s - K)^{+}$ is convex and nonincreasing in $K$ for each $s$ . The right derivative of the strike derivative card gives, for $0 < a < b$ ,

\begin{aligned} &C_{K,+}(b) - C_{K,+}(a) \\[0.4em] &\quad = e^{-r\tau}\,\mathbb{Q}_t(a < S_T \le b), \end{aligned}

so the increase of the slope over a strike interval is always the discounted pricing probability of that interval, whether the law has a density, atoms or both. A butterfly spread with wing width $0 < h < K$ is long one call at $K - h$ , short two calls at $K$ and long one call at $K + h$ . It pays the tent $(h - \lvert S_T - K \rvert)^{+} \ge 0$ and costs

\begin{aligned} \mathrm{BF}_h(K) &= C(K - h) - 2C(K) \\[0.4em] &\quad + C(K + h). \end{aligned}

The tent has area $h^2$ . If $C$ is twice continuously differentiable near $K$ , then $\mathrm{BF}_h(K)/h^2 \to C_{KK}(K)$ as $h \to 0$ , and if it is four times continuously differentiable, the error is $h^2 C^{(4)}(\xi)/12$ for some $\xi$ between $K - h$ and $K + h$ . Without arbitrage $\mathrm{BF}_h(K) \ge 0$ . If a butterfly had a negative price, buying it and putting the credit in the money-market account would cost nothing and would pay the nonnegative tent plus a positive amount of cash in every state. So if a density recovered from quotes turns negative at some strike, the quotes or the interpolation between strikes are inconsistent, and the negative value cannot be read as a probability.

In plain words Call prices bend upward as the strike rises, and the amount of bending over a strike interval is the price of a claim that pays one unit when the stock finishes in that interval. A butterfly spread buys this bend directly, so in a market without arbitrage it never has a negative price.

Example With the constructed inputs the calls with strikes 95, 100 and 105 are worth about 10.519541, 7.965567 and 5.905593. The butterfly with $h = 5$ costs about 0.494000, and

\frac{\mathrm{BF}_5(100)}{25} \approx 0.019760,

just below the exact curvature 0.019848. With $h = 1$ the error is about $-3.5 \cdot 10^{-6}$ and with $h = 0.1$ about $-3.5 \cdot 10^{-8}$ , so dividing $h$ by 10 divides the error by about 100, as the $h^2$ term predicts. Shrinking $h$ further eventually makes the result worse in floating-point arithmetic. With $h = 10^{-6}$ the second-order part of each price difference, $\tfrac12 C_{KK} h^2 \approx 10^{-14}$ , is comparable to the rounding error of a price near 8 in double precision, so the computed butterfly says nothing about the curvature.

Probability atom

Definition A level $a \ge 0$ is a probability atom of the terminal law if $\mathbb{Q}_t(S_T = a) > 0$ . At an atom $a > 0$ the one-sided strike derivatives differ by

\begin{aligned} &C_{K,+}(a) - C_{K,-}(a) \\[0.4em] &\quad = e^{-r\tau}\,\mathbb{Q}_t(S_T = a), \end{aligned}

so the call price has a corner at $K = a$ and no second derivative there, and in place of a density its curvature carries a point mass of size $e^{-r\tau}\,\mathbb{Q}_t(S_T = a)$ . The digital that pays on $\{S_T > a\}$ is worth $-C_{K,+}(a)$ , the digital that pays on $\{S_T \ge a\}$ is worth $-C_{K,-}(a)$ , and centered call spreads converge to their average, which is the price of neither contract. An atom at zero, such as a default to zero, is invisible to the curvature at positive strikes. Since $C_{K,+}(b) \to 0$ as $b \to \infty$ , letting $a \downarrow 0$ and $b \to \infty$ in the slope identity of the previous card gives

\begin{aligned} &-\lim_{a \downarrow 0} C_{K,+}(a) \\[0.4em] &\quad = e^{-r\tau}\,\mathbb{Q}_t(S_T > 0), \end{aligned}

which falls short of $e^{-r\tau}$ by $e^{-r\tau}\,\mathbb{Q}_t(S_T = 0)$ . The missing mass appears in puts with small strikes. The European put $P(K) = e^{-r\tau}\,\mathbb{E}^{\mathbb{Q}}_t[(K - S_T)^{+}]$ satisfies $P(K)/K \to e^{-r\tau}\,\mathbb{Q}_t(S_T = 0)$ as $K \downarrow 0$ , by dominated convergence applied to $(1 - S_T/K)^{+}$ .

In plain words When the terminal price lands exactly on one level with positive pricing probability, the call price curve has a corner there instead of a smooth bend. The size of the corner is the discounted probability of that level, and whether a digital pays when the price lands exactly on the strike changes its price.

Example In a constructed terminal law with $r = q = 0$ and $S_t = 100$ , let $S_T$ equal 80, 100 or 120 with $\mathbb{Q}$ probabilities $\tfrac14$ , $\tfrac12$ and $\tfrac14$ , so that $\mathbb{E}^{\mathbb{Q}}_t[S_T] = 100$ . For $80 \le K \le 100$ the call is worth $80 - \tfrac34 K$ , and for $100 \le K \le 120$ it is worth $30 - \tfrac14 K$ , both equal to 5 at $K = 100$ . The right slope at 100 is $-\tfrac14 = -\mathbb{Q}_t(S_T > 100)$ and the left slope is $-\tfrac34 = -\mathbb{Q}_t(S_T \ge 100)$ . So the digital that pays only above 100 costs 0.25, the digital that also pays at 100 costs 0.75, and the corner has size $\tfrac12$ , the atom. The centered spread $(C(99) - C(101))/2 = (5.75 - 4.75)/2 = 0.5$ prices neither digital. The butterfly costs $\mathrm{BF}_h(100) = \tfrac12 h$ for $0 < h \le 20$ , so $\mathrm{BF}_h(100)/h^2 = 1/(2h)$ grows without bound as $h \to 0$ instead of converging to a density.

Check yourself

Keep $S = K = 100$ , $q = 0$ , $\sigma = 0.2$ and $\tau = 1$ , but set $r = 0.05$ . Price a digital that pays 10 currency units if $S_T > 100$ , give the $\mathbb{Q}$ probability of that event, and find the pricing density and the state-price density at 100.

Show the answer

Now $d_2 = (0 + 0.05)/0.2 - 0.1 = 0.15$ , so $\mathbb{Q}_t(S_T > 100) = \Phi(0.15) \approx 0.559618$ . The discount factor is $e^{-0.05} \approx 0.951229$ , so the unit digital is worth about 0.532325 and the digital paying 10 units is worth about 5.323248. As a check, the digital put is worth $e^{-0.05}\Phi(-0.15) \approx 0.418905$ , and the two add up to 0.951229. The pricing density is

\begin{aligned} p_{\mathbb{Q}}(100) &= \frac{\varphi(0.15)}{100 \cdot 0.2} \\[0.4em] &\approx 0.019724 \end{aligned}

per currency unit, and the state-price density is $C_{KK}(100) = e^{-0.05}\,p_{\mathbb{Q}}(100) \approx 0.018762$ . The pricing probability rose from 0.460172 to 0.559618 because the drift of the stock under $\mathbb{Q}$ is $r - q$ . This says nothing about the physical probability under $\mathbb{P}$ , which depends on the physical drift.

In the constructed example with $r = q = 0$ , a classmate prices the unit digital with strike 100 at 0.539828, the call delta, arguing that delta is the probability of finishing in the money. What is wrong, and which identity links the call price, its delta and its strike derivative?

Show the answer

The digital is minus the strike derivative, $D(100) = -C_K(100) = e^{-r\tau}\Phi(d_2) = \Phi(-0.1) \approx 0.460172$ . The delta is the derivative in the current stock price, $C_S = e^{-q\tau}\Phi(d_1) = \Phi(0.1) \approx 0.539828$ . The number $\Phi(d_1)$ is a probability of $\{S_T > K\}$ under a different measure, the one that uses the stock as numeraire (here $q = 0$ ), not under $\mathbb{Q}$ . In the Black-Scholes model the law of $S_T/S$ does not depend on $S$ , so the call price is homogeneous of degree one in the pair $(S, K)$ , which means $C(\lambda S, \lambda K) = \lambda\,C(S, K)$ for $\lambda > 0$ . Differentiating in $\lambda$ at $\lambda = 1$ gives Euler’s identity

C = S\,C_S + K\,C_K.

Here $100 \cdot 0.539828 - 100 \cdot 0.460172 = 7.965567$ , which is the call price. At $S = K$ the identity says that the delta minus the digital price equals $C/K \approx 0.079656$ , so the two numbers differ whenever the call has positive value.

In a constructed model with $r = q = 0$ and $S_t = 100$ , the call prices are $C(K) = 100 - 0.8K$ for $0 < K \le 100$ , $C(K) = 0.008\,(150 - K)^2$ for $100 \le K \le 150$ , and $C(K) = 0$ for $K \ge 150$ . Recover the pricing density where you can, integrate it over positive strikes and explain where the rest of the probability went. At which strikes does the Breeden-Litzenberger formula fail to hold pointwise?

Show the answer

The second derivative is $0$ on $(0, 100)$ , $0.016$ on $(100, 150)$ and $0$ above 150, so the pricing density is $0.016$ per currency unit on $(100, 150)$ and $0$ elsewhere in $(0, \infty)$ , a uniform density of total mass $0.016 \cdot 50 = 0.8$ . The two formulas meet at 100 with value 20 and slope $-0.8$ , so $C$ is continuously differentiable, but $C_{KK}$ jumps at 100 and at 150 because the density is discontinuous there. The formula holds at every positive strike except these two, and almost everywhere as the general statement says. The integral of the curvature is 0.8, not 1, and the slope for small strikes is $-0.8 = -\mathbb{Q}_t(S_T > 0)$ , so $\mathbb{Q}_t(S_T = 0) = 0.2$ . The missing mass is an atom at zero, which no curvature at positive strikes can see. The hypothesis of a strictly positive terminal price fails. Put-call parity with zero rates and dividends gives the put $P(K) = C(K) - 100 + K = 0.2K$ for $0 < K \le 100$ , and $P(K)/K = 0.2$ is the pricing probability of the zero state. As a check, the law with an atom of 0.2 at zero and mass 0.8 spread uniformly over $[100, 150]$ has mean $0.8 \cdot 125 = 100 = S_t$ , as the martingale property under $\mathbb{Q}$ requires when $r = q = 0$ , and it reproduces

\begin{aligned} C(120) &= 0.8 \int_{120}^{150} \frac{s - 120}{50}\,ds \\[0.4em] &= 7.2 = 0.008 \cdot 30^2. \end{aligned}

Digital options and distribution recovery

Key terms

Cash-or-nothing digital option

Strike derivative

Breeden-Litzenberger formula

State-price density

Call price curvature

Probability atom

Check yourself

How this lesson connects

Builds on

Used later in