Model Definition¶

The Heston model is one of the most widely used stochastic volatility models in equity, FX, and commodity markets. It extends Black–Scholes by modeling variance as a mean-reverting stochastic process while retaining analytical tractability through its affine structure. This section provides a complete specification of the model and its properties.

Dynamics Under the Risk-Neutral Measure¶

Recall (see § Heston SDE and Parameters) the risk-neutral SDE system

\[ \begin{aligned} dS_t &= (r - q)S_t\,dt + \sqrt{V_t}\,S_t\,dW_t^S \\ dV_t &= \kappa(\theta - V_t)\,dt + \xi\sqrt{V_t}\,dW_t^V \\ d\langle W^S, W^V \rangle_t &= \rho\,dt \end{aligned} \]

with parameters $(\kappa, \theta, \xi, \rho, V_0)$ and standing assumptions $\kappa, \theta, \xi > 0$, $\rho \in (-1, 1)$. The log-price form

\[ dX_t = \left(r - q - \frac{1}{2}V_t\right)dt + \sqrt{V_t}\,dW_t^S, \qquad X_t = \log S_t, \]

makes $(X_t, V_t)$ a two-dimensional affine diffusion (see § Affine Processes).

Interpretation of Parameters¶

Each parameter controls a distinct smile feature (full discussion: § Heston SDE and Parameters, calibration mapping: § Calibration).

Parameter	Primary effect on implied vol	Quick rule
$V_0$	ATM level (short maturity)	$\sigma_{\text{impl}}(0,T)\approx\sqrt{V_0}$ for small $T$
$\theta$	Long-maturity ATM level	$\sigma_{\text{impl}}^2(T\to\infty)\approx\theta$
$\kappa$	Term-structure speed	Half-life $t_{1/2}=\ln 2/\kappa$
$\rho$	Skew (slope)	$\partial\sigma_{\text{impl}}/\partial k\big
$\xi$	Smile curvature (wings)	$\partial^2\sigma_{\text{impl}}/\partial k^2\propto\xi^2$

Stationary Distribution¶

Recall (see § Variance Dynamics) that the CIR variance has stationary law

\[ V_{\infty} \sim \text{Gamma}\!\left(\tfrac{2\kappa\theta}{\xi^2}, \tfrac{2\kappa}{\xi^2}\right), \qquad \mathbb{E}[V_\infty]=\theta,\quad \text{Var}[V_\infty]=\tfrac{\xi^2\theta}{2\kappa}. \]

The distribution is right-skewed and supported on $[0,\infty)$.

Relation to Black–Scholes¶

Deterministic Limit¶

If volatility is deterministic ($\xi = 0$), the variance satisfies:

\[ V_t = V_0 e^{-\kappa t} + \theta(1 - e^{-\kappa t}) \]

The model reduces to time-dependent Black–Scholes:

\[ C = C^{\text{BS}}\left(S_0, K, T, \bar{\sigma}(T)\right) \]

where $\bar{\sigma}^2(T) = \frac{1}{T}\int_0^T V_s\,ds$.

Small Vol-of-Vol Expansion¶

For small $\xi$, Heston prices can be expanded:

\[ C^{\text{Heston}} = C^{\text{BS}} + \xi \cdot C_1 + \xi^2 \cdot C_2 + O(\xi^3) \]

The correction terms introduce smile and skew.

Implied Volatility Approximation¶

For short maturities and near-ATM options:

\[ \sigma_{\text{impl}}(k, T) \approx \sqrt{V_0} + \frac{\rho\xi}{4\sqrt{V_0}}k + \frac{\xi^2(1-\rho^2)}{8V_0^{3/2}}k^2 + O(T, k^3) \]

This shows:

ATM level: $\sqrt{V_0}$
Skew: $\propto \rho\xi$
Convexity: $\propto \xi^2$

Parameter Constraints and Simulation¶

Recall (see § Feller Condition and Boundary Behavior) that the Feller condition $2\kappa\theta \geq \xi^2$ ensures strict positivity of $V_t$.

Moment existence: $\mathbb{E}^{\mathbb{Q}}[S_T^n] < \infty$ iff $n < n^*(\kappa, \theta, \xi, \rho)$ (Andersen–Piterbarg moment-explosion bound).

Recall (see § Variance Dynamics and § Euler/QE Simulation) the standard simulation toolkit: Euler with full truncation, Andersen's QE scheme, and exact non-central $\chi^2$ sampling.

Model Summary¶

Property	Heston	Black–Scholes
Volatility	Stochastic	Constant
# Parameters	5 ($\kappa, \theta, \xi, \rho, V_0$)	1 ($\sigma$)
Smile	Yes (endogenous)	No
Skew	Yes (from $\rho$)	No
Term structure	Yes (from $\kappa, \theta$)	Flat
Closed-form European	Yes (semi-analytic)	Yes (analytic)
Market completeness	No	Yes
Computational cost	Moderate	Low

Key Takeaways¶

Heston introduces stochastic variance with mean reversion
Five risk-neutral parameters: $\kappa, \theta, \xi, \rho, V_0$
Captures skew ($\rho$), smile curvature ($\xi$), and term structure ($\kappa, \theta$)
Analytical tractability via affine structure and characteristic functions
The Feller condition $2\kappa\theta \geq \xi^2$ ensures strict positivity
Heston is the minimal extension of Black–Scholes that generates realistic smiles

Exercises¶

Exercise 1. Given the Heston parameters $\kappa = 2.5$, $\theta = 0.04$, $\xi = 0.6$, $\rho = -0.7$, and $V_0 = 0.06$, compute: (a) the half-life of variance shocks $t_{1/2} = \ln 2 / \kappa$; (b) the Feller ratio $\nu = 2\kappa\theta/\xi^2$ and whether the Feller condition is satisfied; (c) the stationary variance $\text{Var}[V_\infty] = \xi^2\theta/(2\kappa)$; and (d) the expected path of $V_t$ in the deterministic limit ($\xi = 0$) at $t = 0.5$ and $t = 2.0$.

Solution to Exercise 1

We are given $\kappa = 2.5$, $\theta = 0.04$, $\xi = 0.6$, $\rho = -0.7$, $V_0 = 0.06$.

(a) Half-life of variance shocks:

\[ t_{1/2} = \frac{\ln 2}{\kappa} = \frac{0.6931}{2.5} = 0.2773 \text{ years} \approx 3.3 \text{ months} \]

(b) Feller ratio:

\[ \nu = \frac{2\kappa\theta}{\xi^2} = \frac{2 \times 2.5 \times 0.04}{0.36} = \frac{0.20}{0.36} = 0.5556 \]

Since $\nu < 1$, the Feller condition is not satisfied. The variance process can reach zero.

(c) Stationary variance:

\[ \text{Var}[V_\infty] = \frac{\xi^2\theta}{2\kappa} = \frac{0.36 \times 0.04}{2 \times 2.5} = \frac{0.0144}{5} = 0.00288 \]

The stationary standard deviation is $\sqrt{0.00288} \approx 0.0537$.

(d) In the deterministic limit ($\xi = 0$), the variance follows $V_t = V_0 e^{-\kappa t} + \theta(1 - e^{-\kappa t})$.

At $t = 0.5$:

\[ V_{0.5} = 0.06 \, e^{-2.5 \times 0.5} + 0.04(1 - e^{-1.25}) = 0.06 \times 0.2865 + 0.04 \times 0.7135 = 0.01719 + 0.02854 = 0.04573 \]

At $t = 2.0$:

\[ V_{2.0} = 0.06 \, e^{-5.0} + 0.04(1 - e^{-5.0}) = 0.06 \times 0.00674 + 0.04 \times 0.99326 = 0.00040 + 0.03973 = 0.04013 \]

As $t$ grows, $V_t \to \theta = 0.04$, consistent with the mean-reverting dynamics.

Exercise 2. Using the short-maturity implied volatility approximation

\[ \sigma_{\text{impl}}(k, T) \approx \sqrt{V_0} + \frac{\rho\xi}{4\sqrt{V_0}}k + \frac{\xi^2(1-\rho^2)}{8V_0^{3/2}}k^2 \]

with $V_0 = 0.04$, $\rho = -0.65$, and $\xi = 0.50$, compute $\sigma_{\text{impl}}$ at log-moneyness $k = -0.10$ (OTM put), $k = 0$ (ATM), and $k = 0.10$ (OTM call). Plot or sketch the resulting smile and identify the skew and curvature.

Solution to Exercise 2

We have $V_0 = 0.04$, $\rho = -0.65$, $\xi = 0.50$, so $\sqrt{V_0} = 0.20$.

The approximation formula is:

\[ \sigma_{\text{impl}}(k) \approx \sqrt{V_0} + \frac{\rho\xi}{4\sqrt{V_0}}k + \frac{\xi^2(1-\rho^2)}{8V_0^{3/2}}k^2 \]

First, compute the coefficients:

Slope (skew): $\frac{\rho\xi}{4\sqrt{V_0}} = \frac{(-0.65)(0.50)}{4 \times 0.20} = \frac{-0.325}{0.80} = -0.40625$
Curvature: $\frac{\xi^2(1-\rho^2)}{8V_0^{3/2}} = \frac{0.25 \times (1 - 0.4225)}{8 \times 0.008} = \frac{0.25 \times 0.5775}{0.064} = \frac{0.144375}{0.064} = 2.2559$

Now evaluate at each moneyness:

ATM ($k = 0$):

\[ \sigma_{\text{impl}}(0) = 0.20 + 0 + 0 = 20.0\% \]

OTM put ($k = -0.10$):

\[ \sigma_{\text{impl}}(-0.10) = 0.20 + (-0.40625)(-0.10) + 2.2559 \times 0.01 = 0.20 + 0.04063 + 0.02256 = 0.26319 \approx 26.3\% \]

OTM call ($k = 0.10$):

\[ \sigma_{\text{impl}}(0.10) = 0.20 + (-0.40625)(0.10) + 2.2559 \times 0.01 = 0.20 - 0.04063 + 0.02256 = 0.18194 \approx 18.2\% \]

The smile exhibits a pronounced negative skew: OTM puts have higher implied volatility than OTM calls. The skew is driven by $\rho < 0$ (the slope coefficient is negative). The curvature term $\propto \xi^2$ adds convexity, lifting both wings relative to the ATM level.

Exercise 3. The Heston model reduces to time-dependent Black–Scholes when $\xi = 0$. In this limit, the effective variance is

\[ \bar{\sigma}^2(T) = \frac{1}{T}\int_0^T V_s\,ds = \theta + \frac{(V_0 - \theta)(1 - e^{-\kappa T})}{\kappa T} \]

Compute $\bar{\sigma}(T)$ for $T = 0.25, 1.0, 5.0$ with $V_0 = 0.09$, $\theta = 0.04$, $\kappa = 1.5$. Verify that $\bar{\sigma}(T) \to \sqrt{\theta}$ as $T \to \infty$.

Solution to Exercise 3

With $V_0 = 0.09$, $\theta = 0.04$, $\kappa = 1.5$, the effective variance is:

\[ \bar{\sigma}^2(T) = \theta + \frac{(V_0 - \theta)(1 - e^{-\kappa T})}{\kappa T} = 0.04 + \frac{0.05(1 - e^{-1.5T})}{1.5T} \]

At $T = 0.25$:

\[ \bar{\sigma}^2(0.25) = 0.04 + \frac{0.05(1 - e^{-0.375})}{0.375} = 0.04 + \frac{0.05 \times 0.3127}{0.375} = 0.04 + 0.04169 = 0.08169 \]

\[ \bar{\sigma}(0.25) = \sqrt{0.08169} \approx 28.58\% \]

At $T = 1.0$:

\[ \bar{\sigma}^2(1.0) = 0.04 + \frac{0.05(1 - e^{-1.5})}{1.5} = 0.04 + \frac{0.05 \times 0.7769}{1.5} = 0.04 + 0.02590 = 0.06590 \]

\[ \bar{\sigma}(1.0) = \sqrt{0.06590} \approx 25.67\% \]

At $T = 5.0$:

\[ \bar{\sigma}^2(5.0) = 0.04 + \frac{0.05(1 - e^{-7.5})}{7.5} = 0.04 + \frac{0.05 \times 0.99945}{7.5} = 0.04 + 0.00666 = 0.04666 \]

\[ \bar{\sigma}(5.0) = \sqrt{0.04666} \approx 21.60\% \]

Verification: As $T \to \infty$, the exponential term vanishes and $\bar{\sigma}^2(T) \to \theta = 0.04$, so $\bar{\sigma}(\infty) = \sqrt{0.04} = 20\%$. Since $V_0 = 0.09 > \theta = 0.04$, the effective volatility starts high and decays monotonically toward $\sqrt{\theta} = 20\%$.

Exercise 4. The stationary distribution of the CIR variance process is Gamma with shape $\alpha = 2\kappa\theta/\xi^2$ and rate $\beta = 2\kappa/\xi^2$. For $\kappa = 3$, $\theta = 0.05$, $\xi = 0.4$: (a) compute the shape and rate parameters; (b) find the mode of the Gamma distribution; (c) compute $\mathbb{P}(V_\infty > 0.10)$ (variance exceeding 10%) using the Gamma CDF. What does this tell you about the frequency of high-volatility regimes?

Solution to Exercise 4

With $\kappa = 3$, $\theta = 0.05$, $\xi = 0.4$:

(a) Shape and rate parameters:

\[ \alpha = \frac{2\kappa\theta}{\xi^2} = \frac{2 \times 3 \times 0.05}{0.16} = \frac{0.30}{0.16} = 1.875 \]

\[ \beta = \frac{2\kappa}{\xi^2} = \frac{6}{0.16} = 37.5 \]

(b) The mode of the Gamma$(\alpha, \beta)$ distribution (for $\alpha > 1$) is:

\[ \text{mode} = \frac{\alpha - 1}{\beta} = \frac{0.875}{37.5} = 0.02333 \]

The mode is less than the mean $\theta = 0.05$ because the Gamma distribution is right-skewed.

(c) We need $\mathbb{P}(V_\infty > 0.10) = 1 - F_{\text{Gamma}}(0.10; 1.875, 37.5)$, where $F_{\text{Gamma}}$ is the Gamma CDF. Using the substitution $x = \beta \cdot v = 37.5 \times 0.10 = 3.75$, this equals $1 - \frac{\gamma(1.875, 3.75)}{\Gamma(1.875)}$, where $\gamma$ is the lower incomplete gamma function. Numerically, $\mathbb{P}(V_\infty > 0.10) \approx 0.057$, or about 5.7%.

This means that in the stationary regime, variance exceeds 10% (volatility exceeds $\sqrt{0.10} \approx 31.6\%$) only about 5.7% of the time. High-volatility regimes are relatively rare under these parameters, though not negligible.

Exercise 5. In the Euler simulation scheme, one step of the variance process is

\[ V_{t+\Delta} = V_t + \kappa(\theta - V_t^+)\Delta + \xi\sqrt{V_t^+}\sqrt{\Delta}\,Z \]

with $Z \sim \mathcal{N}(0,1)$. Starting from $V_0 = 0.01$ with $\kappa = 2$, $\theta = 0.04$, $\xi = 0.5$, $\Delta = 1/252$, compute the threshold value $Z^*$ such that $V_\Delta < 0$ when $Z < Z^*$. What is $\mathbb{P}(Z < Z^*)$? How does this probability change if $V_0 = 0.04$?

Solution to Exercise 5

Starting from $V_0 = 0.01$, with $\kappa = 2$, $\theta = 0.04$, $\xi = 0.5$, $\Delta = 1/252$. The Euler step is:

\[ V_\Delta = V_0 + \kappa(\theta - V_0)\Delta + \xi\sqrt{V_0}\sqrt{\Delta}\,Z \]

Compute the deterministic part:

\[ V_0 + \kappa(\theta - V_0)\Delta = 0.01 + 2(0.04 - 0.01)/252 = 0.01 + 0.06/252 = 0.01 + 0.000238 = 0.010238 \]

Compute the stochastic coefficient:

\[ \xi\sqrt{V_0}\sqrt{\Delta} = 0.5 \times \sqrt{0.01} \times \sqrt{1/252} = 0.5 \times 0.1 \times 0.06299 = 0.003150 \]

Setting $V_\Delta = 0$:

\[ 0.010238 + 0.003150 \, Z^* = 0 \implies Z^* = -\frac{0.010238}{0.003150} = -3.250 \]

The probability is $\mathbb{P}(Z < -3.250) \approx 0.058\%$.

For $V_0 = 0.04$:

\[ V_0 + \kappa(\theta - V_0)\Delta = 0.04 + 0 = 0.04 \]

\[ \xi\sqrt{V_0}\sqrt{\Delta} = 0.5 \times 0.2 \times 0.06299 = 0.006299 \]

\[ Z^* = -\frac{0.04}{0.006299} = -6.350 \]

The probability $\mathbb{P}(Z < -6.350)$ is essentially zero ($< 10^{-10}$). Starting from the long-run mean $V_0 = \theta$ makes negative variance in a single Euler step virtually impossible, whereas starting from $V_0 = 0.01$ (well below $\theta$) gives a small but nonzero probability.

Exercise 6. Compare the five Heston parameters $(\kappa, \theta, \xi, \rho, V_0)$ in terms of which implied volatility surface features each primarily controls. For each parameter, state whether it mainly affects (a) the ATM level, (b) the skew (slope), (c) the smile curvature (wings), or (d) the term structure. Justify each answer with the relevant approximation formula or economic intuition.

Solution to Exercise 6

$V_0$ (initial variance): Mainly affects (a) ATM level, especially for short maturities. The short-maturity approximation gives $\sigma_{\text{impl}}(0, T) \approx \sqrt{V_0}$. It has negligible impact on skew, curvature, or long-dated term structure.

$\theta$ (long-run variance): Mainly affects (d) term structure and long-maturity ATM level. As $T \to \infty$, $\sigma_{\text{impl}}^2 \approx \theta$. Together with $\kappa$, it determines how the ATM volatility interpolates between $\sqrt{V_0}$ and $\sqrt{\theta}$.

$\kappa$ (mean-reversion speed): Mainly affects (d) term structure. Higher $\kappa$ causes the term structure to flatten more quickly toward $\sqrt{\theta}$. The half-life $t_{1/2} = \ln 2/\kappa$ sets the characteristic time scale.

$\rho$ (correlation): Mainly affects (b) skew. From the approximation, the skew slope is $\partial\sigma_{\text{impl}}/\partial k|_{k=0} \propto \rho\xi/(4\sqrt{V_0})$. Negative $\rho$ produces the equity-like downside skew.

$\xi$ (vol-of-vol): Mainly affects (c) smile curvature. The curvature coefficient is $\propto \xi^2(1-\rho^2)/(8V_0^{3/2})$. Higher $\xi$ produces more pronounced wings. It also contributes to skew through its product with $\rho$.

Exercise 7. A trader calibrates the Heston model to S&P 500 options and obtains $V_0 = 0.025$, $\theta = 0.04$, $\kappa = 1.8$, $\xi = 0.55$, $\rho = -0.72$. She then uses $\sigma_{\text{BS}} = \sqrt{V_0} = 15.8\%$ in the Black–Scholes formula for an ATM 1-year call. Using the Heston effective variance formula, compute $\bar{\sigma}(1)$ and the percentage error in using $\sqrt{V_0}$ instead. Explain why this error systematically under- or over-prices the option depending on whether $V_0 < \theta$ or $V_0 > \theta$.

Solution to Exercise 7

With $V_0 = 0.025$, $\theta = 0.04$, $\kappa = 1.8$, $T = 1$:

\[ \bar{\sigma}^2(1) = \theta + \frac{(V_0 - \theta)(1 - e^{-\kappa T})}{\kappa T} = 0.04 + \frac{(0.025 - 0.04)(1 - e^{-1.8})}{1.8} \]

\[ = 0.04 + \frac{(-0.015)(1 - 0.16530)}{1.8} = 0.04 + \frac{(-0.015)(0.83470)}{1.8} = 0.04 - 0.006956 = 0.033044 \]

\[ \bar{\sigma}(1) = \sqrt{0.033044} \approx 18.18\% \]

The trader uses $\sqrt{V_0} = \sqrt{0.025} = 15.81\%$.

Percentage error:

\[ \frac{\bar{\sigma}(1) - \sqrt{V_0}}{\bar{\sigma}(1)} = \frac{18.18\% - 15.81\%}{18.18\%} \approx 13.0\% \]

The trader underestimates the effective volatility by about 13%, which leads to underpricing the call.

Systematic direction: When $V_0 < \theta$, the variance is expected to increase over the option's life, so the effective average variance $\bar{\sigma}^2(T) > V_0$, and using $\sqrt{V_0}$ underprices. Conversely, when $V_0 > \theta$, variance is expected to decrease, $\bar{\sigma}^2(T) < V_0$, and using $\sqrt{V_0}$ overprices. The Heston effective variance accounts for this mean-reverting drift, while the naive Black–Scholes approach ignores it.

Parameter	Primary effect on implied vol	Quick rule
\(V_0\)	ATM level (short maturity)	\(\sigma_{\text{impl}}(0,T)\approx\sqrt{V_0}\) for small \(T\)
\(\theta\)	Long-maturity ATM level	\(\sigma_{\text{impl}}^2(T\to\infty)\approx\theta\)
\(\kappa\)	Term-structure speed	Half-life \(t_{1/2}=\ln 2/\kappa\)
\(\rho\)	Skew (slope)	$\partial\sigma_{\text{impl}}/\partial k\big
\(\xi\)	Smile curvature (wings)	\(\partial^2\sigma_{\text{impl}}/\partial k^2\propto\xi^2\)