Boundary and Terminal Behavior of Greeks¶
Greeks exhibit characteristic behavior near \(S\to 0\), \(S\to\infty\), and \(t\uparrow T\). This matters for hedging and numerics.
Far-field behavior (vanilla calls)¶
As \(S\to\infty\), a call behaves like
so
Quantitative refinement. More precisely, for large \(S\):
Since \(d_1 \sim \frac{\ln(S/K)}{\sigma\sqrt{\tau}}\) for large \(S\), the decay is faster than any polynomial.
As \(S\to 0\),
Quantitative refinement. For small \(S\):
Near maturity: quantitative asymptotics¶
At maturity \(V(T,S)=\Phi(S)\). For payoffs with kinks, derivatives at \(T\) are not classical; gamma concentrates near the kink.
ATM behavior as \(\tau \to 0\). For \(S = K\) (at-the-money):
Transition layer width. The region where delta transitions from 0 to 1 has width
in spot space, or equivalently \(\delta x \sim \sigma\sqrt{\tau}\) in log-moneyness \(x = \ln(S/K)\).
OTM and ITM behavior near expiry¶
Out-of-the-money call (\(S < K\)): As \(\tau \to 0\),
In-the-money call (\(S > K\)): As \(\tau \to 0\),
Numerical boundary conditions¶
For finite-difference PDE solvers:
| Boundary | Condition for Call |
|---|---|
| \(S = 0\) | \(V = 0\), \(\Delta = 0\), \(\Gamma = 0\) |
| \(S = S_{\max}\) | \(V \approx S - Ke^{-r\tau}\), \(\Delta \approx 1\), \(\Gamma \approx 0\) |
| \(t = T\) | \(V = (S-K)^+\), \(\Delta = \mathbf{1}_{S>K}\) (step), \(\Gamma = \delta(S-K)\) |
What to remember¶
- Boundary conditions for PDE numerics are guided by far-field Greek limits.
- Near maturity, gamma spikes near the strike with \(\Gamma \sim \tau^{-1/2}\); delta becomes step-like.
- Transition layer has width \(\mathcal{O}(\sigma\sqrt{\tau})\) in log-moneyness.
- Far from the money, Greeks decay exponentially with rate \(\sim 1/\tau\).
Exercises¶
Exercise 1. For a European call as \(S \to \infty\), the option behaves like \(V \approx S - Ke^{-r\tau}\). Using this, verify that \(\Delta \to 1\), \(\Gamma \to 0\), and \(\Theta \to -rKe^{-r\tau}\) in this limit. Interpret these results financially.
Solution to Exercise 1
As \(S \to \infty\), the European call behaves like \(V \approx S - Ke^{-r\tau}\), since the probability of finishing in-the-money approaches 1. Computing Greeks in this limit:
Delta: Differentiating \(V \approx S - Ke^{-r\tau}\) with respect to \(S\):
This means a deep ITM call moves dollar-for-dollar with the stock. The holder is virtually certain to exercise, so the option behaves like the stock minus a fixed payment.
Gamma: Differentiating again:
With delta locked at 1, there is no curvature in the price function. No rebalancing of the hedge is needed.
Theta: From the PDE, \(\Theta = -\frac{1}{2}\sigma^2 S^2 \Gamma - rS\Delta + rV\). Substituting \(\Gamma \to 0\), \(\Delta \to 1\), and \(V \to S - Ke^{-r\tau}\):
Financial interpretation. The theta converges to \(-rKe^{-r\tau}\), which is negative. This represents the time value gained from the strike payment being discounted: as time passes, the present value \(Ke^{-r\tau}\) of the strike increases (approaches \(K\)), reducing the option value. The deep ITM call has no optionality value left -- its only time dependence comes from the discounting of the strike. For \(r > 0\), the option loses value over time at rate \(rKe^{-r\tau}\).
Exercise 2. The ATM gamma diverges as \(\Gamma_{\text{ATM}} \sim (K\sigma\sqrt{2\pi\tau})^{-1}\) when \(\tau \to 0\). For \(K = 100\) and \(\sigma = 0.25\), compute the gamma values at \(\tau = 30\), \(7\), \(1\), and \(0.1\) trading days (use \(\tau = \text{days}/252\)). Plot or tabulate the results.
Solution to Exercise 2
The ATM gamma formula is \(\Gamma_{\text{ATM}} = \frac{1}{K\sigma\sqrt{2\pi\tau}}\). With \(K = 100\) and \(\sigma = 0.25\):
Converting trading days to years using \(\tau = \text{days}/252\):
30 trading days (\(\tau = 30/252 = 0.1190\)):
7 trading days (\(\tau = 7/252 = 0.02778\)):
1 trading day (\(\tau = 1/252 = 0.003968\)):
0.1 trading days (\(\tau = 0.1/252 = 0.0003968\)):
| Days to expiry | \(\tau\) | \(\Gamma_{\text{ATM}}\) |
|---|---|---|
| 30 | 0.1190 | 0.046 |
| 7 | 0.0278 | 0.096 |
| 1 | 0.00397 | 0.253 |
| 0.1 | 0.000397 | 0.801 |
The \(\tau^{-1/2}\) divergence is clearly visible: gamma roughly doubles each time \(\tau\) decreases by a factor of 4. This explosive growth near expiry makes delta-hedging extremely costly for short-dated ATM options.
Exercise 3. The transition layer where delta moves from 0 to 1 has width \(\delta S \sim S \cdot \sigma\sqrt{\tau}\) in spot space. For \(S = K = 100\), \(\sigma = 0.20\), compute the transition width for \(\tau = 1\) year, \(1\) month, and \(1\) day. How does this affect a hedger's rebalancing requirements?
Solution to Exercise 3
The transition width in spot space is \(\delta S \sim S\sigma\sqrt{\tau}\). With \(S = K = 100\) and \(\sigma = 0.20\):
1 year (\(\tau = 1\)):
Delta transitions from near 0 to near 1 over a range of roughly \(\$80\) to \(\$120\) (i.e., \(\pm \$20\) around the strike). The hedge ratio changes gradually, and infrequent rebalancing suffices.
1 month (\(\tau = 1/12 = 0.0833\)):
The transition zone narrows to roughly \(\$94\) to \(\$106\). A hedger must rebalance more frequently, since moderate price moves can cause significant changes in delta.
1 day (\(\tau = 1/252 = 0.003968\)):
Delta transitions over just \(\pm \$1.26\) around the strike. Even small price movements cause large delta swings, requiring near-continuous rebalancing.
Impact on hedging. As the transition width shrinks with \(\sqrt{\tau}\), the gamma (which scales as \(1/\delta S\)) increases correspondingly. A hedger must rebalance more frequently to track the rapidly changing delta. The rebalancing frequency scales as \(1/(\delta S)^2 \sim 1/\tau\), meaning costs grow dramatically near expiry. In practice, this makes precise delta-hedging of short-dated ATM options prohibitively expensive, motivating wider hedging bands or the use of option spreads.
Exercise 4. For a deep OTM call with \(S < K\), delta decays as \(\Delta \approx \exp\left(-\frac{(\ln(K/S))^2}{2\sigma^2 \tau}\right)\) as \(\tau \to 0\). Compute this exponential decay rate for \(S = 95\), \(K = 100\), \(\sigma = 0.20\) at \(\tau = 5\) days. At what point does delta become effectively zero (say, less than \(10^{-6}\))?
Solution to Exercise 4
For a deep OTM call with \(S < K\), the exponential decay rate is:
With \(S = 95\), \(K = 100\), \(\sigma = 0.20\):
At \(\tau = 5\) days \(= 5/252 = 0.01984\) years:
So delta is approximately 0.19 -- still meaningful.
For delta to fall below \(10^{-6}\), we need:
Converting: \(0.002381 \times 252 = 0.600\) trading days, or approximately 14.4 trading hours. So a 5% OTM call becomes effectively zero-delta less than one trading day before expiry. This extremely rapid decay illustrates why OTM options near expiry have negligible delta exposure and essentially no hedging value.
Exercise 5. A finite-difference PDE solver requires boundary conditions at \(S = 0\) and \(S = S_{\max}\). For a European put, write down the analogous boundary conditions to those given for the call, including the values of \(V\), \(\Delta\), and \(\Gamma\) at each boundary.
Solution to Exercise 5
For a European put with payoff \(\Phi(S) = (K - S)^+\), the boundary conditions are:
At \(S = 0\): The put is deep in-the-money. The price approaches the discounted intrinsic value:
The delta is:
(put delta is \(N(d_1) - 1 \to -1\) as \(S \to 0\)). The gamma is:
(the price is asymptotically linear in \(S\) at \(S = 0\)).
At \(S = S_{\max}\): The put is deep out-of-the-money. All values approach zero:
More precisely, \(V = Ke^{-r\tau}N(-d_2) - SN(-d_1) \to 0\) exponentially as \(S \to \infty\) since \(N(-d_1), N(-d_2) \to 0\).
At \(t = T\): The terminal values are:
Note that the terminal gamma is the same Dirac delta as for the call (same kink location, same second derivative).
| Boundary | \(V\) | \(\Delta\) | \(\Gamma\) |
|---|---|---|---|
| \(S = 0\) | \(Ke^{-r\tau}\) | \(-1\) | \(0\) |
| \(S = S_{\max}\) | \(\approx 0\) | \(\approx 0\) | \(\approx 0\) |
| \(t = T\) | \((K-S)^+\) | \(-\mathbf{1}_{S<K}\) (step) | \(\delta(S-K)\) |
Exercise 6. Near maturity, the vega of an ATM option satisfies \(\nu_{\text{ATM}} \sim K\sqrt{\tau}/\sqrt{2\pi}\), which vanishes as \(\tau \to 0\). Explain why this is consistent with the fact that an option about to expire is insensitive to volatility. What does this imply about hedging vega risk for short-dated options?
Solution to Exercise 6
An option about to expire has its payoff nearly determined by the current spot price. If \(\tau\) is very small, the stock price \(S_T\) will be very close to \(S_t\) regardless of \(\sigma\), because the diffusion has almost no time to act: the standard deviation of \(\ln(S_T/S_t)\) is \(\sigma\sqrt{\tau}\), which vanishes as \(\tau \to 0\).
Quantitatively, for an ATM option:
The vega vanishes because: (i) the option payoff depends on where \(S_T\) ends up, (ii) as \(\tau \to 0\), the distribution of \(S_T\) collapses to a point mass at \(S_t\) regardless of \(\sigma\), so changing \(\sigma\) has negligible effect on the expected payoff. The \(\sqrt{\tau}\) rate of decay reflects the diffusion scaling -- the standard deviation of the stock's log-return is proportional to \(\sqrt{\tau}\).
Implications for vega hedging. For short-dated options, the small vega means:
-
Low vega exposure per option: Each short-dated option contributes little vega to a portfolio. To achieve a given vega target, a trader needs many more short-dated options than long-dated ones.
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Unstable vega hedge ratios: While total vega is small, the vega per unit of gamma can be expressed as \(\nu/\Gamma = \sigma S^2 \tau\), which also shrinks with \(\tau\). This means that the "vega-to-gamma" ratio deteriorates, making it difficult to hedge vega without taking on excessive gamma.
-
Volatility uncertainty less relevant: Since the option is nearly insensitive to \(\sigma\), model risk from volatility misspecification is small for short-dated options. The dominant risk is gamma (spot movement), not vega (volatility movement).
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Practical consequence: Traders typically hedge vega using longer-dated options and manage gamma with short-dated options, exploiting the natural separation of risk horizons.