Delta of a Digital Call Option

Worked Solution

How to Think About It: A digital call is a discounted bet on the event $\{S_T > K\}$, so its price is $e^{-rT}$ times the risk-neutral probability of that event, $N(d_2)$. Delta asks: how fast does that probability move when the stock moves? The heuristic is a chain-rule + dimensional check. The only place $S$ enters is through $d_2$, and $d_2$ shifts by $\partial d_2/\partial S = 1/(S\sigma\sqrt{T})$ per dollar. The probability responds to that shift at the rate of the normal *density* $\phi(d_2)$. So $\Delta = e^{-rT}\phi(d_2)\cdot\tfrac{1}{S\sigma\sqrt{T}}$ almost writes itself. The trap is to reach for the vanilla-call delta $N(d_1)$ — that is the *derivative of a payoff that grows linearly past $K$*, not of an indicator. A digital's payoff is flat except for a jump, so its sensitivity lives entirely in the density at the boundary.

Quick Estimate: The interview payoff here is the *scaling law*, and you can get it dimensionally without a single BSM table lookup. At the money $d_2\approx 0$, so $\phi(d_2)\approx\phi(0)=1/\sqrt{2\pi}\approx 0.4$. Everything else in $\Delta=\tfrac{e^{-rT}\phi(d_2)}{S\sigma\sqrt{T}}$ is order-1 except the lone $\sqrt{T}$ in the denominator, so $\Delta$ scales like $1/\sqrt{T}$. Put numbers on it: take $S=K=100$, $\sigma=0.2$, $r=0$. At $T=1$: $\Delta\approx \tfrac{0.4}{100\cdot 0.2\cdot 1}=\tfrac{0.4}{20}=0.02$. Now shrink the horizon to $T=0.01$ (about 2–3 trading days): $\sqrt{T}$ drops from $1$ to $0.1$, a factor of $10$, so $\Delta\approx 0.02\times 10 = 0.2$. Shrink again to $T=10^{-4}$: another factor of $10$, $\Delta\approx 2$. The number has no ceiling — halving the time to expiry multiplies delta by $\sqrt2$ forever. That runaway *is* the answer to part 2: the hedge ratio diverges.

Approach: Chain rule through $d_2$; then read off the $T\to 0$ behaviour from the $1/\sqrt{T}$ factor with $\phi(d_2)\to\phi(0)$.

Formal Solution:

(a) Delta. With $V=e^{-rT}N(d_2)$ and $d_2=\dfrac{\ln(S/K)+(r-\sigma^2/2)T}{\sigma\sqrt{T}}$, only $\ln S$ carries the $S$-dependence, so $$\frac{\partial d_2}{\partial S}=\frac{1}{S\sigma\sqrt{T}}.$$ By the chain rule, using $N'(x)=\phi(x)$, $$\boxed{\;\Delta=\frac{\partial V}{\partial S}=e^{-rT}\phi(d_2)\,\frac{\partial d_2}{\partial S}=\frac{e^{-rT}\,\phi(d_2)}{S\sigma\sqrt{T}}\;}$$ Sanity: $\Delta>0$ always (higher $S$ raises the ITM probability). Deep ITM ($d_2\to+\infty$) or deep OTM ($d_2\to-\infty$) both send $\phi(d_2)\to 0$, so $\Delta\to 0$ in the tails; the sensitivity peaks near ATM.

(b) The limit $T\to 0$ at the money. Set $S=K$, so $\ln(S/K)=0$ and $d_2=\dfrac{(r-\sigma^2/2)\sqrt{T}}{\sigma}\to 0$; hence $\phi(d_2)\to\phi(0)=\tfrac{1}{\sqrt{2\pi}}$ and $e^{-rT}\to 1$. The denominator $S\sigma\sqrt{T}\to 0$, so $$\Delta\sim\frac{1}{S\sigma\sqrt{2\pi}}\cdot\frac{1}{\sqrt{T}}\;\longrightarrow\;+\infty.$$ Delta blows up without bound — it concentrates into a Dirac spike at $S=K$ as the payoff sharpens into a step. A tiny stock move flips the whole \$1 payoff, so no finite position in the underlying can hedge it; the required share count explodes.

Practical note: desks never hedge a naked digital near expiry with stock. They replicate it with a tight call spread — long $C(K)$, short $C(K+\epsilon)$, scaled by $1/\epsilon$ — whose delta stays bounded. Tighter $\epsilon$ means better replication but more gamma to carry.

Answer: $\Delta=\dfrac{e^{-rT}\,\phi(d_2)}{S\sigma\sqrt{T}}$. At the money it grows like $1/\sqrt{T}$ and diverges as $T\to 0$, so the underlying alone essentially cannot hedge the digital right near expiry.