FX Digital Call Pricing and Delta Blow-Up
You are pricing a European digital (binary) call option in the FX market. The setup:
- Spot exchange rate $S_0$ is quoted as USD per EUR.
- Domestic (USD) continuously compounded rate is $r_d$.
- Foreign (EUR) continuously compounded rate is $r_f$.
- Maturity is $T$, strike is $K$, and volatility is $\sigma$ (constant).
- The option pays $\$ USD at time $T$ if $S_T > K$, and nothing otherwise.
Assume the Garman--Kohlhagen model (Black--Scholes extended to FX).
- Write the time-0 price of this digital call in closed form.
- Compute the digital's delta, $\partial V / \partial S_0$.
- Explain why the digital's delta (and other Greeks) can become extremely large near the strike as $T \to 0$.
Hints
- Under Garman--Kohlhagen, the spot follows GBM with drift $r_d - r_f$. The digital price is just the discounted risk-neutral probability that $S_T > K$.
- Express $P(S_T > K)$ in terms of $\Phi(d_2)$ where $d_2$ uses the same log-moneyness formula as Black--Scholes but with $r_d - r_f$ replacing the drift.
- For the delta, note that $\partial d_2/\partial S_0 = 1/(S_0 \sigma \sqrt{T})$. As $T \to 0$ near the strike, this factor causes delta to diverge like /\sqrt{T}$.
Worked Solution
How to Think About It: A digital call is the simplest discontinuous payoff -- it is a step function at $K$. Under Garman--Kohlhagen, the risk-neutral dynamics of the spot rate $S_t$ are geometric Brownian motion with drift $r_d - r_f$ instead of $r_d$. The price of any European claim is just the discounted risk-neutral expectation. For a digital call, that expectation reduces to $P(S_T > K)$ under the risk-neutral measure, discounted at the domestic rate. Before writing anything, think about limiting cases: if $S_0 \gg K$, the digital is almost certain to pay off, so price $\approx e^{-r_d T}$. If $S_0 \ll K$, price $\approx 0$. At expiry ($T \to 0$), the price collapses to a step function -- either $e^{-r_d T} \approx 1$ if $S_0 > K$, or $0$ if $S_0 < K$. That step function is the source of the Greek blow-up.
Quick Sanity Checks:
- Price should be bounded between 0 and $e^{-r_d T}$ (the discounted payoff).
- Delta should be positive (higher spot means more likely to finish ITM).
- Delta should peak near the strike and decay to zero far away on either side.
- As $T \to 0$, delta at the strike should diverge -- the payoff is a Heaviside step function, and its "slope" is a Dirac delta.
Derivation:
*Part 1: Closed-form price.*
Under Garman--Kohlhagen, the spot evolves as:
$S_T = S_0 \exp\!\left[\left(r_d - r_f - \tfrac{1}{2}\sigma^2\right)T + \sigma \sqrt{T}\, Z\right]$
where $Z \sim N(0,1)$ under the domestic risk-neutral measure. The digital pays $\
$ when $S_T > K$, so:$V_0 = e^{-r_d T}\, P(S_T > K)$
Taking logs, $S_T > K$ is equivalent to:
$\sigma \sqrt{T}\, Z > \ln(K/S_0) - (r_d - r_f - \tfrac{1}{2}\sigma^2)T$
$Z > -\frac{\ln(S_0/K) + (r_d - r_f - \tfrac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}$
Define:
$d_2 = \frac{\ln(S_0/K) + (r_d - r_f - \tfrac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}$
Since $P(Z > -d_2) = \Phi(d_2)$, the price is:
$\boxed{V_0 = e^{-r_d T}\,\Phi(d_2)}$
This is exactly the $\Phi(d_2)$ term from the Garman--Kohlhagen formula for a vanilla call, discounted at the domestic rate. In fact, a digital call is the derivative of the vanilla call price with respect to the strike (up to a sign and discounting), which is why $d_2$ appears naturally.
*Part 2: Delta.*
Differentiate with respect to $S_0$:
$\Delta = \frac{\partial V_0}{\partial S_0} = e^{-r_d T}\,\phi(d_2)\,\frac{\partial d_2}{\partial S_0}$
where $\phi$ is the standard normal PDF. Computing the partial derivative:
$\frac{\partial d_2}{\partial S_0} = \frac{1}{S_0 \,\sigma\sqrt{T}}$
Therefore:
$\boxed{\Delta = \frac{e^{-r_d T}\,\phi(d_2)}{S_0\,\sigma\sqrt{T}}}$
*Part 3: Why Greeks blow up near the strike as $T \to 0$.*
As $T \to 0$ with $S_0 \approx K$ (near ATM), we have $d_2 \to 0$, so $\phi(d_2) \to \phi(0) = 1/\sqrt{2\pi} \approx 0.399$. The numerator stays finite, but the denominator contains $\sigma\sqrt{T}$, which vanishes. The delta therefore scales as:
$\Delta \sim \frac{e^{-r_d T}\,\phi(0)}{S_0\,\sigma\sqrt{T}} \propto \frac{1}{\sqrt{T}} \to \infty$
Physically, the digital payoff is a step function at $K$. The "derivative" of a step function is a Dirac delta. With finite time remaining, diffusion smooths this discontinuity -- the price function is a smoothed step, with the smoothing width proportional to $\sigma\sqrt{T}$. As $T \to 0$, the smoothing disappears, the price curve steepens into a vertical cliff at $K$, and delta (the slope) diverges.
This has severe practical consequences: a digital that is near ATM and close to expiry has enormous delta (and gamma). Hedging it with the underlying requires huge position changes from tiny spot moves. In practice, traders hedge digitals with tight call spreads (a call spread whose width approximates the digital), which caps the Greeks at the cost of a small model error.
Practical Interpretation: The digital call formula $e^{-r_d T}\Phi(d_2)$ looks deceptively simple, but the real interview insight is the Greek blow-up. On a desk, nobody trades naked digitals near expiry -- the hedging costs become prohibitive because every tick near the strike requires massive rebalancing. The standard trick is to approximate the digital with a call spread: buy a call at $K - \epsilon$ and sell a call at $K + \epsilon$, scaled by
/(2\epsilon)$. This replicates the digital payoff in the limit $\epsilon \to 0$ and keeps Greeks bounded for any finite $\epsilon$. Understanding this spread replication is often the real point of the question.Answer: The digital call price is $V_0 = e^{-r_d T}\,\Phi(d_2)$ where $d_2 = [\ln(S_0/K) + (r_d - r_f - \frac{1}{2}\sigma^2)T]/(\sigma\sqrt{T})$. Its delta is $\Delta = e^{-r_d T}\,\phi(d_2)/(S_0\,\sigma\sqrt{T})$. Near the strike as $T \to 0$, delta diverges as
/\sqrt{T}$ because the payoff discontinuity is smoothed only by diffusion of width $\sigma\sqrt{T}$, which vanishes at expiry.Intuition
The digital call is the purest example of a discontinuous payoff, and discontinuous payoffs create hedging nightmares. The price formula itself is just $\Phi(d_2)$ discounted -- the same $d_2$ that appears in the vanilla Black--Scholes formula. In fact, a digital call is the partial derivative of a vanilla call with respect to the strike (times $-e^{-r_d T}$), which is a useful mental model: anything you know about how vanilla prices change with $K$ immediately tells you about digitals.
The deeper lesson is about the tension between model and reality near discontinuities. In the model, delta goes to infinity as $T \to 0$ near the strike. In practice, this means your hedge ratio oscillates wildly with tiny moves in spot, making continuous delta-hedging impossible. This is why traders never hedge a digital with the underlying alone near expiry -- they use call spread replication to smooth the payoff. The width of the spread is a direct trade-off: narrower gives a better payoff approximation but worse Greeks, wider gives smoother Greeks but introduces payoff slippage. This spread-vs-accuracy trade-off shows up everywhere in exotic derivatives hedging, not just digitals.