Digital Option Pricing via Girsanov's Theorem¶

The digital is the cleanest possible payoff: it pays $\$1$ if the stock finishes above $K$, and nothing otherwise. Stripped of the "how far in the money" question, its time-$0$ price is literally the discounted risk-neutral exercise probability — exactly the second term of the Black–Scholes call formula divided by $K$. This pages explores three consequences of that observation: (i) the digital is the strike-derivative of the standard call, (ii) differentiating once more recovers the implied risk-neutral density (Breeden–Litzenberger), and (iii) the diverging hedge ratio near expiry forces practitioners to replace the digital by a tight call spread.

Where this fits

Roadmap row(s): Strike-space; also Probability through $\mathcal{N}(d_2)$.
Builds on: The Black-Scholes formula (the call price differentiated in $K$) and Properties and bounds (strike monotonicity).
Feeds into: Computational examples — Figure 3 visualises the implied-density extraction discussed here.

Payoff Structure¶

Section goal: specifying the cash-or-nothing payoff and how it differs from the standard call.

A digital (cash-or-nothing) call option pays a fixed amount $1$ if the underlying asset exceeds the strike price $K$ at maturity $T$:

\[ \text{Payoff at } T: \quad \mathbf{1}_{\{S_T > K\}} \]

Unlike a standard European call, which pays the excess $(S_T - K)^+$, the digital call pays a flat amount regardless of how far in the money it finishes.

Risk-Neutral Pricing¶

Recall (see § Risk-Neutral Valuation Principle and § Probabilistic Interpretation): $D_0 = e^{-rT}\mathbb{E}^{\mathbb{Q}}[\mathbf{1}_{\{S_T>K\}}] = e^{-rT}\mathbb{Q}(S_T>K) = e^{-rT}\mathcal{N}(d_2)$, the fair price of a digital call paying $1$ at maturity if $S_T > K$.

Connection to Black–Scholes Formula¶

Section goal: the strike-derivative identity $D_0 = -\partial C/\partial K$ and the Breeden-Litzenberger bridge to implied densities.

The Black–Scholes formula for a European call can be written as:

\[ C_0 = S_0\,\mathcal{N}(d_1) - K e^{-rT}\,\mathcal{N}(d_2) \]

The second term $K e^{-rT}\,\mathcal{N}(d_2)$ is exactly $K$ times the digital call price. The digital is the cleanest concrete object that pins down what $\mathcal{N}(d_2)$ means: the time-$0$ value of receiving $\$1$ if and only if $S_T > K$. (The full measure-theoretic interpretation of $\mathcal{N}(d_2) = \mathbb{Q}(S_T > K)$ and $\mathcal{N}(d_1) = \mathbb{Q}^S(S_T > K)$ is developed in probabilistic interpretation; here we use only that $\mathcal{N}(d_2)$ is the discounted exercise probability.)

Equivalently, the digital call price is the negative derivative of the standard call price with respect to the strike:

\[ D_0 = -\frac{\partial C_0}{\partial K} \]

At first glance this identity looks suspicious: $d_1$ and $d_2$ both depend on $K$, so differentiating $C_0 = S_0\mathcal{N}(d_1) - Ke^{-rT}\mathcal{N}(d_2)$ in $K$ should produce extra terms from $\partial d_1/\partial K$ and $\partial d_2/\partial K$ via the chain rule. Those terms do appear, but they cancel exactly because of the Black–Scholes density identity

\[ S_0\,\phi(d_1) = Ke^{-rT}\,\phi(d_2) \]

which forces $S_0\phi(d_1)\,\partial_K d_1$ and $Ke^{-rT}\phi(d_2)\,\partial_K d_2$ to coincide (since $\partial_K d_1 = \partial_K d_2 = -1/(K\sigma\sqrt{T})$). What survives is the single explicit-$K$ term $-e^{-rT}\mathcal{N}(d_2)$, giving the clean identity above. The cancellation is one of the structural elegances of the Black–Scholes formula; Exercise 2 carries out the differentiation in detail.

This reveals that the digital call captures the density of the call price with respect to strike.

The next step is conceptually larger than it looks: differentiating once more in $K$ does not just give another Greek—it inverts the very integral that produced $C_0$ in the first place, recovering the risk-neutral density itself from prices alone. This matters because liquid markets quote calls across a near-continuum of strikes, so the second strike-derivative is something one can actually estimate from market data, turning a snapshot of the volatility surface into an implied distribution. The non-obviousness lies in this reversal: pricing integrates the payoff against an unknown density, yet differentiating the resulting price function in the strike undoes that integration and exposes the density. Carried out at each maturity, this construction yields one implied risk-neutral density per maturity slice of the volatility surface—the Breeden-Litzenberger identity that follows.

Differentiating once more gives the Breeden-Litzenberger result: the risk-neutral density of $S_T$ can be extracted from observed call prices via

\[ f^{\mathbb{Q}}(K) = e^{rT}\frac{\partial^2 C_0}{\partial K^2} \]

Why this works. Start from the risk-neutral pricing integral

\[ C_0(K) = e^{-rT}\int_K^{\infty} (s - K)\,f^{\mathbb{Q}}_{S_T}(s)\,ds \]

and differentiate twice in $K$ using Leibniz's rule. For the first derivative, the boundary term $(s - K)\big|_{s=K} = 0$, so the lower-limit contribution vanishes and only the integrand derivative survives:

\[ \frac{\partial C_0}{\partial K} = -e^{-rT}\int_K^{\infty} f^{\mathbb{Q}}_{S_T}(s)\,ds = -e^{-rT}\,\mathbb{Q}(S_T > K) \]

This is the model-free version of the earlier identity $D_0 = -\partial C_0/\partial K$: it holds for any underlying with a density, not just lognormal — the Black–Scholes cancellation $S_0\phi(d_1) = Ke^{-rT}\phi(d_2)$ shown above is a special case. Differentiating once more, Leibniz on the tail integral now produces a nonzero boundary contribution at the lower limit:

\[ \frac{\partial^2 C_0}{\partial K^2} = -e^{-rT}\,\frac{\partial}{\partial K}\int_K^{\infty} f^{\mathbb{Q}}_{S_T}(s)\,ds = e^{-rT}\,f^{\mathbb{Q}}_{S_T}(K) \]

Rearranging gives the formula above. The structure is now transparent: $C_0(K)$ is a double integral of the density (an integral of the survival function, itself an integral of $f$, weighted by the payoff ramp), so two strike-derivatives strip away both integrations and expose the density directly. In summary:

first strike-derivative $\;\to\;$ tail probability $\mathbb{Q}(S_T > K)$,
second strike-derivative $\;\to\;$ the density $f^{\mathbb{Q}}_{S_T}(K)$ itself.

This is one of the deepest connections between option markets and probability theory—it shows that a complete set of call prices implicitly encodes the entire risk-neutral distribution.

Digital Put Option¶

Section goal: pricing the cash-or-nothing put.

By symmetry, a digital put paying $1$ if $S_T < K$ has price:

\[ D_0^{\text{put}} = e^{-rT}\,\mathcal{N}(-d_2) = e^{-rT}\left[1 - \mathcal{N}(d_2)\right] \]

The sum of digital call and digital put prices equals the price of a zero-coupon bond:

\[ D_0^{\text{call}} + D_0^{\text{put}} = e^{-rT} \]

Hedging Considerations¶

Section goal: why digitals are pathological to hedge and how traders use call spreads as approximations.

The delta of a digital call is:

\[ \Delta_{\text{digital}} = \frac{\partial D_0}{\partial S_0} = e^{-rT} \frac{\phi(d_2)}{S_0 \sigma \sqrt{T}} \]

where $\phi$ is the standard normal density. Near expiry ($T \to 0$) and near the strike ($S_0 \approx K$), this delta becomes extremely large, making digital options notoriously difficult to hedge in practice. This discontinuity in the payoff creates significant gamma risk near maturity.

Call-Spread Approximation¶

A practical way to reason about (and bound the cost of hedging) a digital is to approximate it by a tight call spread. Buy one call at $K - \tfrac{\epsilon}{2}$ and sell one call at $K + \tfrac{\epsilon}{2}$, then divide the position by $\epsilon$:

\[ D_0 \;\approx\; \frac{C(K - \tfrac{\epsilon}{2}) - C(K + \tfrac{\epsilon}{2})}{\epsilon} \;\xrightarrow{\;\epsilon \to 0\;}\; -\frac{\partial C_0}{\partial K} = e^{-rT}\mathcal{N}(d_2) \]

The intermediate ratio is the discrete version of $-\partial C_0/\partial K$; the limit recovers the digital exactly. The pre-limit call spread is itself a tradable, well-behaved instrument: bounded gamma, linear payoff, replicable from vanilla calls. Practitioners therefore quote a digital as the limit of call spreads of decreasing width and hedge it with a call spread of finite width, accepting basis risk in the strike interval $(K - \tfrac{\epsilon}{2}, K + \tfrac{\epsilon}{2})$ in exchange for stable Greeks. This is the standard market resolution of the unhedgeable digital. (Exercise 6 works through the convergence; Exercise 7 quantifies how the digital delta diverges as $T \to 0$.)

Exercises¶

Exercise 1. Derive the price of a digital call option that pays an amount $Q$ (instead of $1$) at maturity if $S_T > K$. Express the result in terms of $d_2$ and verify that your formula reduces to the standard result when $Q = 1$.

Solution to Exercise 1

The digital call paying $1$ at maturity has price $D_0 = e^{-rT}\mathcal{N}(d_2)$. By linearity of expectation, a digital call paying $Q$ has price:

\[ D_0^{(Q)} = Q \cdot e^{-rT}\mathcal{N}(d_2) \]

This follows because the payoff is $Q \cdot \mathbf{1}_{\{S_T > K\}}$, so:

\[ D_0^{(Q)} = e^{-rT}\mathbb{E}^{\mathbb{Q}}[Q \cdot \mathbf{1}_{\{S_T > K\}}] = Q \cdot e^{-rT}\mathbb{Q}(S_T > K) = Q \cdot e^{-rT}\mathcal{N}(d_2) \]

When $Q = 1$: $D_0^{(1)} = e^{-rT}\mathcal{N}(d_2)$, which is the standard result. ✓

Exercise 2. Show that the digital call price $D_0 = e^{-rT}\mathcal{N}(d_2)$ can be recovered by differentiating the Black-Scholes call price with respect to the strike:

\[ D_0 = -\frac{\partial C_0}{\partial K} \]

Carry out the differentiation explicitly, using the relationship $\frac{\partial d_1}{\partial K} = \frac{\partial d_2}{\partial K} = -\frac{1}{K \sigma \sqrt{T}}$ and the identity $S_0 \phi(d_1) = K e^{-rT} \phi(d_2)$.

Solution to Exercise 2

The Black-Scholes call price is $C_0 = S_0\mathcal{N}(d_1) - Ke^{-rT}\mathcal{N}(d_2)$.

Differentiating with respect to $K$:

\[ \frac{\partial C_0}{\partial K} = S_0 \phi(d_1)\frac{\partial d_1}{\partial K} - e^{-rT}\mathcal{N}(d_2) - Ke^{-rT}\phi(d_2)\frac{\partial d_2}{\partial K} \]

Using $\frac{\partial d_1}{\partial K} = \frac{\partial d_2}{\partial K} = -\frac{1}{K\sigma\sqrt{T}}$:

\[ \frac{\partial C_0}{\partial K} = -\frac{S_0\phi(d_1)}{K\sigma\sqrt{T}} - e^{-rT}\mathcal{N}(d_2) + \frac{Ke^{-rT}\phi(d_2)}{K\sigma\sqrt{T}} \]

\[ = -e^{-rT}\mathcal{N}(d_2) + \frac{1}{\sigma\sqrt{T}}\left[-S_0\phi(d_1)/K + e^{-rT}\phi(d_2)\right] \]

Now we use the identity $S_0\phi(d_1) = Ke^{-rT}\phi(d_2)$. This identity follows from:

\[ \frac{\phi(d_1)}{\phi(d_2)} = \exp\left(-\frac{d_1^2 - d_2^2}{2}\right) = \exp\left(-\frac{(d_1-d_2)(d_1+d_2)}{2}\right) \]

Since $d_1 - d_2 = \sigma\sqrt{T}$ and $d_1 + d_2 = \frac{2\ln(S_0/K) + 2rT}{\sigma\sqrt{T}} - \sigma\sqrt{T} + \sigma\sqrt{T}$... more directly, one can verify $S_0\phi(d_1) = Ke^{-rT}\phi(d_2)$ by showing $\ln(S_0/K) + rT = \frac{1}{2}(d_1^2 - d_2^2)$.

With this identity, $S_0\phi(d_1)/K = e^{-rT}\phi(d_2)$, so the bracketed term vanishes:

\[ \frac{\partial C_0}{\partial K} = -e^{-rT}\mathcal{N}(d_2) \]

Therefore:

\[ D_0 = e^{-rT}\mathcal{N}(d_2) = -\frac{\partial C_0}{\partial K} \]

Exercise 3. A digital call and a digital put on the same underlying with the same strike and maturity have prices $D_0^{\text{call}}$ and $D_0^{\text{put}}$. Prove that their sum equals $e^{-rT}$. What is the financial interpretation of this identity?

Solution to Exercise 3

The digital call price is $D_0^{\text{call}} = e^{-rT}\mathcal{N}(d_2)$ and the digital put price is $D_0^{\text{put}} = e^{-rT}\mathcal{N}(-d_2)$.

Their sum:

\[ D_0^{\text{call}} + D_0^{\text{put}} = e^{-rT}\mathcal{N}(d_2) + e^{-rT}\mathcal{N}(-d_2) = e^{-rT}[\mathcal{N}(d_2) + \mathcal{N}(-d_2)] \]

Using the symmetry property $\mathcal{N}(x) + \mathcal{N}(-x) = 1$:

\[ D_0^{\text{call}} + D_0^{\text{put}} = e^{-rT} \cdot 1 = e^{-rT} \]

Financial interpretation: Holding both a digital call and a digital put with the same strike guarantees a payment of $1$ at maturity regardless of where $S_T$ ends up (either $S_T > K$ or $S_T \leq K$ must hold). This is equivalent to holding a zero-coupon bond that pays $1$ at time $T$, whose present value is $e^{-rT}$.

Exercise 4. Compute the gamma of a digital call option. Show that the gamma is:

\[ \Gamma_{\text{digital}} = -e^{-rT} \frac{\phi(d_2) \, d_1}{S_0^2 \sigma^2 T} \]

At what value of $S_0$ (in terms of $K$, $r$, $\sigma$, $T$) does the digital call gamma equal zero?

Solution to Exercise 4

Start from the digital call delta:

\[ \Delta = \frac{\partial D_0}{\partial S_0} = \frac{e^{-rT}\,\phi(d_2)}{S_0\,\sigma\sqrt{T}} \]

Take logs to make the differentiation cleaner:

\[ \ln \Delta = -rT + \ln\phi(d_2) - \ln S_0 - \ln(\sigma\sqrt{T}) \]

Differentiating with respect to $S_0$ and using $\frac{\partial d_2}{\partial S_0} = \frac{1}{S_0\sigma\sqrt{T}}$ together with $\frac{d}{dx}\ln\phi(x) = -x$:

\[ \frac{1}{\Delta}\,\frac{\partial \Delta}{\partial S_0} = -d_2 \cdot \frac{1}{S_0\sigma\sqrt{T}} - \frac{1}{S_0} = -\frac{1}{S_0}\!\left(\frac{d_2}{\sigma\sqrt{T}} + 1\right) = -\frac{d_2 + \sigma\sqrt{T}}{S_0\,\sigma\sqrt{T}} = -\frac{d_1}{S_0\,\sigma\sqrt{T}} \]

where the last equality uses $d_1 = d_2 + \sigma\sqrt{T}$. Therefore:

\[ \Gamma_{\text{digital}} = \Delta \cdot \left(-\frac{d_1}{S_0\,\sigma\sqrt{T}}\right) = -\,e^{-rT}\,\frac{\phi(d_2)\,d_1}{S_0^{\,2}\,\sigma^{2}\,T} \]

Gamma equals zero when $d_1 = 0$, i.e., when $\ln(S_0/K) + (r + \tfrac{1}{2}\sigma^2)T = 0$, equivalently:

\[ S_0 = K\exp\!\left[-\left(r + \tfrac{1}{2}\sigma^2\right)T\right] \]

At this stock price, the digital call delta achieves its maximum, and gamma changes sign. For $S_0$ below this level, $\Gamma > 0$ (delta increasing in $S_0$); above it, $\Gamma < 0$ (delta decreasing).

Exercise 5. Consider a double digital option (also called a digital range option) that pays $1$ at maturity if $K_1 < S_T < K_2$ for $K_1 < K_2$. Express the price of this option in terms of the standard normal CDF $\mathcal{N}$. Compute its price when $S_0 = 100$, $K_1 = 95$, $K_2 = 105$, $r = 5\%$, $\sigma = 20\%$, and $T = 0.5$.

Solution to Exercise 5

The double digital pays $1$ if $K_1 < S_T < K_2$. This can be decomposed as:

\[ \mathbf{1}_{\{K_1 < S_T < K_2\}} = \mathbf{1}_{\{S_T > K_1\}} - \mathbf{1}_{\{S_T > K_2\}} \]

(a digital call at $K_1$ minus a digital call at $K_2$). Therefore:

\[ D_0^{\text{range}} = e^{-rT}[\mathcal{N}(d_2(K_1)) - \mathcal{N}(d_2(K_2))] \]

where $d_2(K) = \frac{\ln(S_0/K) + (r - \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}$.

Numerical computation with $S_0 = 100$, $K_1 = 95$, $K_2 = 105$, $r = 0.05$, $\sigma = 0.20$, $T = 0.5$:

For $K_1 = 95$:

\[ d_2(95) = \frac{\ln(100/95) + (0.05 - 0.02) \times 0.5}{0.20\sqrt{0.5}} = \frac{0.05129 + 0.015}{0.14142} = \frac{0.06629}{0.14142} = 0.4688 \]

For $K_2 = 105$:

\[ d_2(105) = \frac{\ln(100/105) + 0.015}{0.14142} = \frac{-0.04879 + 0.015}{0.14142} = \frac{-0.03379}{0.14142} = -0.2390 \]

\[ \mathcal{N}(0.4688) = 0.6804, \quad \mathcal{N}(-0.2390) = 0.4055 \]

\[ D_0^{\text{range}} = e^{-0.05 \times 0.5}(0.6804 - 0.4055) = 0.9753 \times 0.2749 = 0.2681 \]

The double digital option is worth approximately $\$0.27$.

Exercise 6. A trader replicates a digital call by constructing a call spread: buying a call at strike $K - \epsilon$ and selling a call at strike $K$, then scaling by $1/\epsilon$. Show that in the limit $\epsilon \to 0$, this replicating portfolio converges to the digital call payoff. What practical difficulties arise for small but nonzero $\epsilon$?

Solution to Exercise 6

The call spread replicating portfolio has payoff:

\[ \Pi_T = \frac{1}{\epsilon}\left[(S_T - (K-\epsilon))^+ - (S_T - K)^+\right] \]

Case 1: $S_T \leq K - \epsilon$: Both calls expire worthless. $\Pi_T = 0$.

Case 2: $K - \epsilon < S_T \leq K$: Only the first call is ITM. $\Pi_T = \frac{S_T - K + \epsilon}{\epsilon} \in (0, 1]$.

Case 3: $S_T > K$: Both calls are ITM. $\Pi_T = \frac{(S_T - K + \epsilon) - (S_T - K)}{\epsilon} = 1$.

As $\epsilon \to 0$, in Case 2 the region $(K-\epsilon, K]$ shrinks to the single point $\{K\}$ (measure zero under continuous distributions), so:

\[ \lim_{\epsilon \to 0} \Pi_T = \begin{cases} 0 & \text{if } S_T < K \\ 1 & \text{if } S_T > K \end{cases} = \mathbf{1}_{\{S_T > K\}} \]

This is exactly the digital call payoff.

Practical difficulties for small but nonzero $\epsilon$:

The position involves $1/\epsilon$ units of each call, so the notional size becomes very large.
The hedge requires frequent rebalancing near the strike, with large position sizes amplifying transaction costs.
The payoff is not exactly $0$ or $1$ in the region $(K-\epsilon, K]$, creating basis risk.
Near expiry, the gamma of the call spread becomes extremely large, making delta-hedging unstable.

Exercise 7. The digital call delta $\Delta_{\text{digital}} = e^{-rT} \frac{\phi(d_2)}{S_0 \sigma \sqrt{T}}$ diverges as $T \to 0$ when $S_0 = K$. Compute the delta explicitly for $S_0 = K = 100$, $\sigma = 25\%$, $r = 3\%$, and $T \in \{1, 0.1, 0.01, 0.001\}$. Discuss why this behavior makes near-expiry digital options extremely difficult to delta-hedge in practice.

Solution to Exercise 7

With $S_0 = K = 100$, $\sigma = 0.25$, $r = 0.03$:

\[ d_2 = \frac{\ln(1) + (0.03 - 0.03125)T}{0.25\sqrt{T}} = \frac{-0.00125\,T}{0.25\sqrt{T}} = \frac{-0.005\sqrt{T}}{1} = -0.005\sqrt{T} \]

For small $T$, $d_2 \approx 0$, so $\phi(d_2) \approx \phi(0) = \frac{1}{\sqrt{2\pi}} \approx 0.3989$.

\[ \Delta_{\text{digital}} = e^{-rT}\frac{0.3989}{100 \times 0.25 \times \sqrt{T}} = \frac{e^{-rT} \times 0.3989}{25\sqrt{T}} \]

$T$	$\sqrt{T}$	$e^{-rT}$	$\Delta_{\text{digital}}$
1.000	1.0000	0.9704	0.01548
0.100	0.3162	0.9970	0.05029
0.010	0.1000	0.9997	0.15956
0.001	0.03162	1.0000	0.50451

As $T \to 0$, $\Delta \propto 1/\sqrt{T} \to \infty$.

Why this makes hedging extremely difficult: The diverging delta means that to maintain a delta-neutral hedge, the trader must hold an increasingly large number of shares as expiry approaches. A tiny move in the stock price around $K$ flips the digital payoff from $0$ to $1$ (or vice versa), requiring the hedge to swing from $0$ shares to a very large position. In practice, this means enormous transaction costs, as the hedge must be rebalanced constantly with large position changes. Any discrete rebalancing interval results in significant hedging error, and bid-ask spreads make the rebalancing prohibitively expensive. This is the fundamental reason why digital options near expiry are among the most difficult derivatives to manage.