Small-Volatility Asymptotics¶
Consider \(\sigma\downarrow 0\). Randomness vanishes and prices approach deterministic limits.
Deterministic limit¶
As \(\sigma\to 0\),
so \(S_T\to S e^{r\tau}\).
OTM becomes exponentially small¶
If \(K>S e^{r\tau}\) (OTM call in the deterministic limit),
Rate function identification. The constant \(c\) is the squared geodesic distance in the log-moneyness metric:
This represents the squared "distance" the asset must travel (in log-space) from its deterministic trajectory to reach the strike.
Derivation. For small \(\sigma\), using \(d_2 = \frac{\ln(S/K) + (r - \frac12\sigma^2)\tau}{\sigma\sqrt{\tau}}\):
For OTM calls, \(\ln(S/K) + r\tau < 0\), so \(d_2 \to -\infty\) as \(\sigma \to 0\).
Using the Gaussian tail: [ N(d_2) \approx \frac{1}{|d_2|\sqrt{2\pi}}e^{-d_2^2/2} \approx \exp!\left(-\frac{(\ln(K/S) - r\tau)^2}{2\sigma^2\tau}\right) ]
Large deviations interpretation¶
The small-\(\sigma\) asymptotics connect to Varadhan's lemma: for the diffusion \(S_t\),
where \(I(s)\) is the rate function (action functional):
This is the cost of deviating from the deterministic path.
ITM call in small volatility¶
For ITM calls (\(K < Se^{r\tau}\)):
The call converges to its intrinsic value plus carry.
ATM call in small volatility¶
For ATM forward (\(K = Se^{r\tau}\)):
The price is linear in \(\sigma\) at leading order.
Greeks in small volatility¶
Vega dominates ATM risk: [ \nu = S\sqrt{\tau}N'(d_1) \approx \frac{S\sqrt{\tau}}{\sqrt{2\pi}} \quad \text{(nearly constant)} ]
Gamma for ATM: [ \Gamma = \frac{N'(d_1)}{S\sigma\sqrt{\tau}} \approx \frac{1}{S\sigma\sqrt{2\pi\tau}} \to \infty \quad \text{as } \sigma \to 0 ]
Delta transition: As \(\sigma \to 0\), delta becomes a step function: [ \Delta \to \begin{cases} 1 & S > Ke^{-r\tau} \ \frac{1}{2} & S = Ke^{-r\tau} \ 0 & S < Ke^{-r\tau} \end{cases} ]
Connection to heat kernel expansion¶
The Black–Scholes equation in log-coordinates is a heat equation. The small-\(\sigma^2\) expansion corresponds to the small-time expansion of the heat kernel:
The leading exponential term gives the rate function; corrections appear as polynomial prefactors.
Laplace principle¶
The small-\(\sigma\) limit can be analyzed via Laplace's method:
where \(p(s)\) concentrates near \(Se^{r\tau}\). For OTM options, the integral is over the tail, giving exponential decay.
What to remember¶
- OTM prices shrink exponentially in \(1/\sigma^2\) with rate \(c = (\ln(K/S) - r\tau)^2/\tau\).
- The rate function is the squared geodesic distance from the deterministic trajectory.
- These asymptotics connect to large deviations and Laplace principles.
- Delta becomes step-like; gamma diverges as \(\sigma \to 0\) for ATM options.
- Heat kernel small-time expansion provides the mathematical framework.
Exercises¶
Exercise 1. For an OTM call with \(K = 110\), \(S = 100\), \(r = 0.03\), \(\tau = 1\) year, compute the rate function \(c = (\ln(K/S) - r\tau)^2/\tau\) and the exponential decay factor \(\exp(-c/(2\sigma^2))\) for \(\sigma = 0.10\) and \(\sigma = 0.05\). By how many orders of magnitude does the price decrease when volatility halves?
Solution to Exercise 1
First compute \(\ln(K/S) = \ln(110/100) = \ln(1.1) = 0.09531\) and \(r\tau = 0.03 \times 1 = 0.03\).
The rate function is
For \(\sigma = 0.10\):
For \(\sigma = 0.05\):
The ratio of decay factors is \(0.4261 / 0.8082 = 0.5272\). In log terms:
So the price decreases by about \(0.28\) orders of magnitude (roughly a factor of 1.9) when volatility halves. The key insight is that the decay scales as \(1/\sigma^2\), so halving \(\sigma\) quadruples the exponent. The effect becomes much more dramatic for larger values of \(c\) (further OTM options).
Exercise 2. For the ATM forward strike \(K = Se^{r\tau}\), verify that \(d_1 = \sigma\sqrt{\tau}/2\) and \(d_2 = -\sigma\sqrt{\tau}/2\). Show that \(C_{\text{ATM}} \approx \frac{S\sigma\sqrt{\tau}}{\sqrt{2\pi}}\) to leading order in \(\sigma\), confirming linearity in \(\sigma\).
Solution to Exercise 2
For the ATM forward strike \(K = Se^{r\tau}\), we have \(\ln(S/K) = -r\tau\). Substituting into the \(d_1\) formula:
Now the call price is \(C = SN(d_1) - Ke^{-r\tau}N(d_2)\). Since \(Ke^{-r\tau} = S\):
For small \(\sigma\), \(d_1 = \sigma\sqrt{\tau}/2\) is small, so we expand \(N(z) \approx \frac{1}{2} + \frac{z}{\sqrt{2\pi}}\):
This confirms \(C_{\text{ATM}} \approx \frac{S\sigma\sqrt{\tau}}{\sqrt{2\pi}}\), which is linear in \(\sigma\) at leading order. Higher-order corrections are \(\mathcal{O}(\sigma^3)\).
Exercise 3. As \(\sigma \to 0\), delta becomes the step function \(\Delta \to \mathbf{1}_{S > Ke^{-r\tau}}\). For \(K = 100\), \(r = 0.05\), \(\tau = 0.5\), compute the critical spot level \(S^* = Ke^{-r\tau}\) where the step occurs. Plot (or sketch) delta as a function of \(S\) for \(\sigma = 0.30\), \(0.10\), and \(0.01\) to illustrate the convergence to a step function.
Solution to Exercise 3
The critical spot level where the step occurs is
The delta of a call is \(\Delta = N(d_1)\) where \(d_1 = \frac{\ln(S/K) + (r + \frac{1}{2}\sigma^2)\tau}{\sigma\sqrt{\tau}}\).
At \(S = S^* = 97.53\): \(\ln(S/K) = -r\tau = -0.025\), so \(d_1 = \frac{\sigma^2\tau/2}{\sigma\sqrt{\tau}} = \frac{\sigma\sqrt{\tau}}{2}\).
- For \(\sigma = 0.30\): \(d_1 = 0.30\sqrt{0.5}/2 = 0.1061\), \(\Delta = N(0.1061) = 0.5423\)
- For \(\sigma = 0.10\): \(d_1 = 0.10\sqrt{0.5}/2 = 0.0354\), \(\Delta = N(0.0354) = 0.5141\)
- For \(\sigma = 0.01\): \(d_1 = 0.01\sqrt{0.5}/2 = 0.00354\), \(\Delta = N(0.00354) = 0.5014\)
As \(\sigma \to 0\), delta at \(S^*\) converges to \(1/2\) (the midpoint of the step). Away from \(S^*\), the convergence to 0 or 1 is rapid. For example, at \(S = 95\) (below \(S^*\)):
- \(\sigma = 0.30\): \(d_1 \approx -0.63\), \(\Delta \approx 0.26\)
- \(\sigma = 0.01\): \(d_1 \approx -36.4\), \(\Delta \approx 0\)
The transition from \(\Delta \approx 0\) to \(\Delta \approx 1\) becomes infinitely sharp as \(\sigma \to 0\), confirming convergence to the step function \(\mathbf{1}_{S > S^*}\).
Exercise 4. The gamma formula \(\Gamma = N'(d_1)/(S\sigma\sqrt{\tau})\) diverges as \(\sigma \to 0\) for ATM options. Show that \(\Gamma \to \infty\) at rate \(1/\sigma\) when \(S = Ke^{-r\tau}\). What physical interpretation does this have --- why is the option's convexity infinite in the zero-volatility limit?
Solution to Exercise 4
At \(S = Ke^{-r\tau}\) (ATM forward), we have \(d_1 = \sigma\sqrt{\tau}/2\) and
For small \(\sigma\), \(\exp(-\sigma^2\tau/8) \to 1\), so
This diverges as \(1/\sigma\) when \(\sigma \to 0\), confirming the claimed rate.
Physical interpretation. In the zero-volatility limit, the call payoff \((S_T - K)^+\) as a function of \(S_0\) becomes a kink: zero for \(S_0 < Ke^{-r\tau}\) and \(S_0 - Ke^{-r\tau}\) for \(S_0 > Ke^{-r\tau}\). The second derivative of this payoff (with respect to \(S_0\)) is a Dirac delta at \(S_0 = Ke^{-r\tau}\).
For any finite \(\sigma > 0\), diffusion smooths this kink into a curve, but the curvature near the kink grows without bound as \(\sigma \to 0\). Gamma measures this curvature, so it diverges. Practically, this means delta-hedging becomes infinitely sensitive to small spot moves near ATM when volatility is very low --- the hedger must adjust the position dramatically for tiny price changes.
Exercise 5. The large deviations rate function \(I(s) = \frac{1}{2\tau}(\ln(s/S) - r\tau)^2\) describes the cost of the diffusion reaching level \(s\) from \(S\). Compute \(I(s)\) for \(S = 100\), \(r = 0.03\), \(\tau = 1\), at \(s = 80, 100, 120, 150\). Which of these levels is "cheapest" to reach?
Solution to Exercise 5
With \(S = 100\), \(r = 0.03\), \(\tau = 1\), we have \(r\tau = 0.03\):
At \(s = 80\): \(\ln(80/100) = -0.2231\)
At \(s = 100\): \(\ln(100/100) = 0\)
At \(s = 120\): \(\ln(120/100) = 0.1823\)
At \(s = 150\): \(\ln(150/100) = 0.4055\)
The minimum occurs at \(s = Se^{r\tau} = 100 \times e^{0.03} = 103.05\), where \(I = 0\). Among the given levels, \(s = 100\) is cheapest (\(I = 0.00045\)), being closest to the deterministic forward price. The level \(s = 150\) is the most expensive to reach, consistent with it being the farthest from the drift trajectory.
Exercise 6. The heat kernel expansion \(p(x,y;t) \sim \frac{1}{\sqrt{2\pi t}}\exp(-\frac{(y-x)^2}{2t}) \sum_{n=0}^\infty a_n(x,y)t^n\) provides systematic corrections to the leading-order Gaussian. In the Black--Scholes log-coordinate, what is the leading correction \(a_1(x,y)\), and how does it relate to the drift term \((r - \frac{1}{2}\sigma^2)\)?
Solution to Exercise 6
In log-coordinates \(X_t = \ln S_t\), the Black--Scholes SDE becomes
Rescaling to unit diffusion via \(Y = X/\sigma\), the transition density satisfies a heat equation. The exact transition density in log-coordinates is
Comparing with the heat kernel expansion \(p(x,y;t) \sim \frac{1}{\sqrt{2\pi t}}\exp(-\frac{(y-x)^2}{2t})\sum_n a_n t^n\) where \(t = \sigma^2\tau\):
The leading Gaussian has center \(y - x\) with no drift. The drift \(\mu = r - \frac{1}{2}\sigma^2\) enters through rewriting
Expanding the correction factor in powers of \(t = \sigma^2\tau\):
The leading correction coefficient is therefore
More precisely, collecting terms at order \(t = \sigma^2\tau\): the \(a_1\) correction encodes how the drift \(\mu = r - \frac{1}{2}\sigma^2\) shifts the center of the Gaussian away from \(y - x\). In the Black--Scholes case with constant coefficients, the expansion terminates because the exact density is Gaussian. For models with state-dependent volatility, \(a_1\) involves curvature of the metric and is genuinely non-trivial.