Large-Time Behavior and Ergodicity¶

Large-time limits depend on whether the model admits a stationary distribution. The behavior of option prices and Greeks as \(T \to \infty\) differs fundamentally between ergodic and non-ergodic models.

Non-ergodic Black–Scholes¶

Geometric Brownian motion has no stationary distribution in \(S\). Under the risk-neutral measure,

\[ S_T = S_0 \exp\!\left[\left(r - \frac{\sigma^2}{2}\right)T + \sigma W_T\right] \]

As \(T \to \infty\):

\(\log S_T\) has variance \(\sigma^2 T \to \infty\), so the distribution of \(S_T\) spreads without bound.
The call price grows like \(S_0\) (bounded below by intrinsic value) while the put price converges to \(Ke^{-rT} \to 0\).

More precisely, for a European call:

\[ C(S_0, K, T) \to S_0 \quad \text{as } T \to \infty \]

since \(N(d_1) \to 1\) and \(Ke^{-rT}N(d_2) \to 0\).

Large-time behavior of Greeks in Black–Scholes¶

As \(T \to \infty\) (equivalently \(\tau \to \infty\)):

\[ \Delta_{\text{call}} = N(d_1) \to 1, \quad \Gamma \to 0, \quad \nu \to 0 \]

The option behaves increasingly like the underlying itself. This is intuitive: with enough time, any OTM call becomes ATM in expectation, and the option's optionality premium vanishes relative to the forward price.

For theta:

\[ \Theta_{\text{call}} \to rKe^{-rT}N(d_2) \to 0 \]

So time decay vanishes for very long-dated options — they are dominated by their delta exposure.

Ergodic factors in multi-factor models¶

In multi-factor models, mean-reverting factors (e.g., variance in Heston-type models) may be ergodic with invariant measure \(\pi\). For a CIR-type variance process \(v_t\) satisfying

\[ dv_t = \kappa(\bar{v} - v_t)\,dt + \xi\sqrt{v_t}\,dW_t \]

the process is ergodic when the Feller condition \(2\kappa\bar{v} > \xi^2\) holds, and for suitable functions \(f\):

\[ \frac{1}{T}\int_0^T f(v_s)\,\mathrm{d}s \xrightarrow{a.s.} \int f\,\mathrm{d}\pi \]

where \(\pi\) is the Gamma distribution with shape \(\alpha = 2\kappa\bar{v}/\xi^2\) and rate \(\beta = 2\kappa/\xi^2\).

Implications for long-dated option pricing¶

In stochastic volatility models with ergodic variance:

Effective volatility convergence. The time-averaged variance converges to the long-run mean:

\[ \frac{1}{T}\int_0^T v_s\,ds \xrightarrow{a.s.} \bar{v} \quad \text{as } T \to \infty \]

This suggests that long-dated options are approximately priced by Black–Scholes with \(\sigma = \sqrt{\bar{v}}\), plus corrections that decay with maturity.

Implied volatility term structure. As \(T \to \infty\), implied volatility for ATM options converges to

\[ \sigma_{\text{implied}}(T) \to \sqrt{\bar{v}} + \mathcal{O}(T^{-1}) \]

The rate of convergence depends on the speed of mean reversion \(\kappa\).

Large deviations and tail behavior¶

For non-ergodic components (like \(\log S\) itself), large-time behavior is governed by large deviations theory. The probability that the log-return deviates from its drift satisfies

\[ \mathbb{P}\!\left(\frac{\log(S_T/S_0)}{T} \notin [a,b]\right) \sim e^{-T \cdot I} \]

where \(I\) is a rate function determined by the model. In Black–Scholes:

\[ I(x) = \frac{(x - \mu)^2}{2\sigma^2} \]

with \(\mu = r - \sigma^2/2\) under the risk-neutral measure. This connects to the exponential decay of deep OTM option prices at long maturities.

Practical relevance¶

Long-time asymptotics matter for:

LEAPS and long-dated warrants: pricing and hedging options with maturities of several years.
Insurance-linked products: equity-indexed annuities and variable annuity guarantees often have 10–30 year horizons.
Model calibration: the long end of the implied volatility term structure constrains the ergodic properties of the variance process.
Risk management: long-horizon VaR and expected shortfall calculations depend on whether variance factors are mean-reverting.

What to remember¶

Black–Scholes is not ergodic in \(S\); long-horizon option prices are dominated by drift and the option degenerates toward the underlying.
Mean-reverting factors (e.g., stochastic variance) can be ergodic, and their long-run averages determine the effective volatility for long-dated options.
Long-time asymptotics are model-dependent and linked to large deviations or ergodicity of latent factors.
The rate of convergence to ergodic limits is governed by the mean-reversion speed.

Exercises¶

Exercise 1. For a European call in Black--Scholes, show that \(C(S_0, K, T) \to S_0\) as \(T \to \infty\) by verifying that \(N(d_1) \to 1\) and \(Ke^{-rT}N(d_2) \to 0\). What is the economic interpretation of this limit?

Solution to Exercise 1

In Black--Scholes, \(C = S_0 N(d_1) - Ke^{-rT}N(d_2)\) where

\[ d_1 = \frac{\ln(S_0/K) + (r + \frac{1}{2}\sigma^2)T}{\sigma\sqrt{T}}, \quad d_2 = d_1 - \sigma\sqrt{T} \]

Step 1: Show \(d_1 \to +\infty\). For fixed \(S_0, K, r, \sigma > 0\):

\[ d_1 = \frac{\ln(S_0/K)}{\sigma\sqrt{T}} + \frac{(r + \frac{1}{2}\sigma^2)\sqrt{T}}{\sigma} \]

The first term is \(\mathcal{O}(T^{-1/2}) \to 0\) and the second grows as \(\sqrt{T} \to +\infty\). Therefore \(d_1 \to +\infty\), so \(N(d_1) \to 1\).

Step 2: Show \(Ke^{-rT}N(d_2) \to 0\). We have \(d_2 = d_1 - \sigma\sqrt{T}\). Similarly:

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

If \(r > \sigma^2/2\), then \(d_2 \to +\infty\) and \(N(d_2) \to 1\), but \(Ke^{-rT} \to 0\), so the product vanishes.

If \(r < \sigma^2/2\), then \(d_2 \to -\infty\), so \(N(d_2) \to 0\), and \(Ke^{-rT}N(d_2) \to 0\).

If \(r = \sigma^2/2\), then \(d_2 \to 0\), \(N(d_2) \to 1/2\), but \(Ke^{-rT} \to 0\) still dominates.

In all cases, \(Ke^{-rT}N(d_2) \to 0\).

Conclusion: \(C \to S_0 \times 1 - 0 = S_0\).

Economic interpretation. With infinite time, the forward price \(S_0 e^{rT} \to \infty\), so any finite strike \(K\) becomes deeply in the money. The discounted strike \(Ke^{-rT}\) vanishes, so paying \(K\) at expiry costs nothing in present-value terms. The call effectively becomes the stock itself --- you are certain to exercise, and the present value of the strike payment is zero.

Exercise 2. In the CIR variance process \(dv_t = \kappa(\bar{v} - v_t)\,dt + \xi\sqrt{v_t}\,dW_t\), the stationary distribution is Gamma with shape \(\alpha = 2\kappa\bar{v}/\xi^2\) and rate \(\beta = 2\kappa/\xi^2\). For \(\kappa = 2\), \(\bar{v} = 0.04\), \(\xi = 0.3\), compute \(\alpha\) and \(\beta\) and verify that the Feller condition \(2\kappa\bar{v} > \xi^2\) is satisfied.

Solution to Exercise 2

With \(\kappa = 2\), \(\bar{v} = 0.04\), \(\xi = 0.3\):

\[ \alpha = \frac{2\kappa\bar{v}}{\xi^2} = \frac{2 \times 2 \times 0.04}{0.09} = \frac{0.16}{0.09} = 1.778 \]

\[ \beta = \frac{2\kappa}{\xi^2} = \frac{2 \times 2}{0.09} = \frac{4}{0.09} = 44.44 \]

Feller condition: \(2\kappa\bar{v} = 0.16\) and \(\xi^2 = 0.09\), so \(2\kappa\bar{v} = 0.16 > 0.09 = \xi^2\). The Feller condition is satisfied.

Verification. The stationary Gamma distribution has mean \(\alpha/\beta = 1.778/44.44 = 0.04 = \bar{v}\) and variance \(\alpha/\beta^2 = 1.778/1975 = 0.0009\), giving a standard deviation of \(0.03\). Since the mean is \(0.04\) and the standard deviation is \(0.03\), the coefficient of variation is \(0.75\), indicating significant variance fluctuation around the long-run mean.

The Feller condition ensures \(v_t > 0\) almost surely, which is necessary for the process to be well-defined and ergodic.

Exercise 3. The implied volatility for long-dated ATM options converges to \(\sqrt{\bar{v}}\) with corrections of order \(T^{-1}\). For the parameters in Exercise 2, compute \(\sqrt{\bar{v}}\) and explain why the speed of convergence depends on \(\kappa\). What happens to the implied vol term structure when \(\kappa\) is very small?

Solution to Exercise 3

From Exercise 2, \(\bar{v} = 0.04\), so

\[ \sqrt{\bar{v}} = \sqrt{0.04} = 0.20 \]

This is the long-run implied volatility: as \(T \to \infty\), ATM implied vol converges to \(20\%\).

Dependence on \(\kappa\). The implied volatility term structure satisfies

\[ \sigma_{\text{implied}}(T) \approx \sqrt{\bar{v}} + \frac{c(v_0 - \bar{v})}{\kappa T} + \mathcal{O}(T^{-2}) \]

where \(c\) is a constant and \(v_0\) is the initial variance. The correction decays as \(1/(\kappa T)\), so:

Large \(\kappa\): Fast mean reversion. The variance quickly reaches its stationary distribution, and implied vol converges to \(\sqrt{\bar{v}}\) at shorter maturities. The term structure flattens quickly.
Small \(\kappa\): Slow mean reversion. The current variance \(v_0\) influences pricing for a long time. If \(v_0 \neq \bar{v}\), the term structure takes a long time to flatten.

When \(\kappa\) is very small: The variance process is nearly a martingale (very slow reversion). The time-averaged variance \(\frac{1}{T}\int_0^T v_s\,ds\) converges extremely slowly to \(\bar{v}\). The implied vol term structure shows a strong slope: if \(v_0 < \bar{v}\), the term structure is upward-sloping (higher vol for longer maturities), and vice versa. In the extreme case \(\kappa \to 0\), ergodicity is lost and the term structure never flattens --- it depends on \(v_0\) at all horizons.

Exercise 4. For a European put in Black--Scholes, show that \(P(S_0, K, T) \to 0\) as \(T \to \infty\). Explain why the put price vanishes but the call price does not, and relate this to the asymmetry of the payoff functions.

Solution to Exercise 4

The put price is \(P = Ke^{-rT}N(-d_2) - S_0 N(-d_1)\).

From Exercise 1, \(d_1 \to +\infty\) as \(T \to \infty\), so \(N(-d_1) \to 0\), giving \(S_0 N(-d_1) \to 0\).

For the first term, regardless of the behavior of \(N(-d_2)\) (which is bounded between 0 and 1):

\[ 0 \leq Ke^{-rT}N(-d_2) \leq Ke^{-rT} \to 0 \quad \text{as } T \to \infty \]

Therefore \(P \to 0 - 0 = 0\).

Why call and put behave differently. The asymmetry comes from the payoff structure and discounting:

Call payoff \((S_T - K)^+\): The upside is unbounded. As \(T \to \infty\), \(\mathbb{E}[S_T] = S_0 e^{rT} \to \infty\), so the expected payoff grows without bound. Discounting at rate \(r\) exactly offsets the forward growth, leaving \(C \to S_0\).
Put payoff \((K - S_T)^+\): The upside is bounded by \(K\). As \(T \to \infty\), the probability \(\mathbb{P}(S_T < K)\) does not vanish (in fact, \(\mathbb{P}(S_T < K) \to 1\) under the real-world measure if \(\mu < \sigma^2/2\)), but the present value of receiving at most \(K\) in the far future is \(Ke^{-rT} \to 0\).

Alternatively, put-call parity gives \(P = C - S_0 + Ke^{-rT} \to S_0 - S_0 + 0 = 0\), consistent with the direct calculation.

Exercise 5. The large deviations rate function in Black--Scholes is \(I(x) = (x - \mu)^2/(2\sigma^2)\) with \(\mu = r - \sigma^2/2\). For \(r = 0.05\), \(\sigma = 0.20\), compute \(I(x)\) at \(x = 0\) (no growth), \(x = \mu\) (drift), and \(x = 2\mu\) (double the drift). Which deviation costs the most probability?

Solution to Exercise 5

The rate function is \(I(x) = \frac{(x - \mu)^2}{2\sigma^2}\) with \(\mu = r - \sigma^2/2 = 0.05 - 0.02 = 0.03\) and \(\sigma = 0.20\).

At \(x = 0\) (no growth):

\[ I(0) = \frac{(0 - 0.03)^2}{2 \times 0.04} = \frac{0.0009}{0.08} = 0.01125 \]

At \(x = \mu = 0.03\) (drift):

\[ I(0.03) = \frac{(0.03 - 0.03)^2}{0.08} = 0 \]

At \(x = 2\mu = 0.06\) (double the drift):

\[ I(0.06) = \frac{(0.06 - 0.03)^2}{0.08} = \frac{0.0009}{0.08} = 0.01125 \]

The deviation \(x = \mu\) costs zero --- this is the most likely growth rate (mode of the large deviations principle). Both \(x = 0\) and \(x = 2\mu\) deviate equally from \(\mu\) and incur the same cost \(I = 0.01125\).

The probability of observing annualized log-return \(x\) over horizon \(T\) decays as \(\exp(-TI(x))\). For \(T = 20\) years and \(x = 0\):

\[ \mathbb{P}\!\left(\frac{\log(S_T/S_0)}{T} \approx 0\right) \sim e^{-20 \times 0.01125} = e^{-0.225} \approx 0.80 \]

This shows that even over long horizons, the Gaussian rate function is not very penalizing for moderate deviations. The cost \(I(x) = 0.01125\) is small because the drift is small relative to \(\sigma^2\). The deviation that "costs the most" among the three is a tie between \(x = 0\) and \(x = 2\mu\), both with \(I = 0.01125\).

Exercise 6. A pension fund holds equity-linked guarantees with a 20-year horizon. Using the large-time asymptotics framework, explain why the choice between a stochastic volatility model (with ergodic variance) and a constant-volatility Black--Scholes model matters significantly for pricing these guarantees. Which model features are most important to calibrate for long-dated products?

Solution to Exercise 6

Why model choice matters for long-dated guarantees:

In a constant-volatility Black--Scholes model, the total variance over \(T = 20\) years is \(\sigma^2 T = 0.04 \times 20 = 0.80\) (using \(\sigma = 0.20\)). This is deterministic, so the distribution of terminal wealth is exactly log-normal.

In a stochastic volatility model with ergodic variance (e.g., Heston), the total integrated variance \(\int_0^T v_s\,ds\) is random. By ergodicity, \(\frac{1}{T}\int_0^T v_s\,ds \to \bar{v}\) a.s., so the average variance is approximately \(\bar{v}\). However:

Variance of integrated variance. Even though the average converges, the fluctuation around \(\bar{v}\) over 20 years is not negligible. The variance of \(\frac{1}{T}\int_0^T v_s\,ds\) decays as \(\mathcal{O}(1/(\kappa T))\), so slow mean reversion (\(\kappa\) small) keeps uncertainty high.
Fat tails in returns. Random variance creates a mixture of Gaussians, producing heavier tails than log-normal. For guarantees that pay in tail scenarios (e.g., minimum return guarantees), this dramatically affects pricing.
Volatility-of-volatility risk. The parameter \(\xi\) (vol-of-vol) determines how much the integrated variance can deviate from \(\bar{v}T\), directly affecting guarantee costs.
Correlation effects. In stochastic vol models, correlation between spot and variance (\(\rho < 0\) typically) creates skewness in the return distribution, making downside guarantees more expensive than a symmetric model would suggest.

Most important features to calibrate:

Long-run variance \(\bar{v}\): Determines the baseline volatility level for the guarantee horizon.
Mean reversion speed \(\kappa\): Controls how quickly the term structure flattens; slow reversion means current conditions persist and the model departs significantly from Black--Scholes.
Vol-of-vol \(\xi\): Drives the fatness of tails and the cost of tail guarantees; underfitting \(\xi\) can severely underprice guarantees.
Spot-vol correlation \(\rho\): Affects skewness of long-horizon returns; critical for directional guarantees (e.g., minimum return floors).
Feller condition (\(2\kappa\bar{v} > \xi^2\)): Must be verified to ensure the model is well-behaved and truly ergodic over the guarantee horizon.