Variance Process Moments

The moments of the CIR variance process -- both conditional (given \(v_t\)) and unconditional (stationary) -- are essential building blocks in the Heston framework. The conditional mean \(\mathbb{E}[v_T | v_t]\) reveals the mean-reversion dynamics, the conditional variance \(\text{Var}(v_T | v_t)\) quantifies uncertainty about future variance levels, and the integrated variance \(\mathbb{E}[\int_t^T v_s\,ds | v_t]\) determines variance swap fair values. All these moments have closed-form expressions, derivable either from the non-central chi-squared transition density or directly from the SDE.

This section derives the first two conditional moments by two methods (SDE-based and distribution-based), extends to higher moments via a recursive formula, and computes the moments of the integrated variance. We assume familiarity with the CIR transition density.

Learning Objectives

After completing this section, you should be able to:

• Derive \(\mathbb{E}[v_T | v_t]\) and \(\text{Var}(v_T | v_t)\) from the CIR SDE
• Verify these moments using the non-central chi-squared distribution
• Compute higher conditional moments via the recursive ODE method
• Derive \(\mathbb{E}[\int_t^T v_s\,ds | v_t]\) and \(\text{Var}(\int_t^T v_s\,ds | v_t)\)
• State the stationary (unconditional) mean, variance, skewness, and kurtosis

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Conditional Mean

Derivation from the SDE

Proposition: Conditional Mean of v_T

For the CIR process \(dv_t = \kappa(\theta - v_t)\,dt + \sigma_v\sqrt{v_t}\,dW_t\):

\[ \mathbb{E}[v_T \mid v_t] = \theta + (v_t - \theta)\,e^{-\kappa\tau} \]

where \(\tau = T - t\).

Proof. Define \(m(\tau) = \mathbb{E}[v_{t+\tau} | v_t]\). Taking expectations of the SDE:

\[ \frac{dm}{d\tau} = \kappa(\theta - m(\tau)) \]

The stochastic integral \(\mathbb{E}[\sigma_v\sqrt{v_s}\,dW_s] = 0\) vanishes because it is a martingale. This is a first-order linear ODE with initial condition \(m(0) = v_t\). The solution is:

\[ m(\tau) = \theta + (v_t - \theta)\,e^{-\kappa\tau} \]

\(\square\)

Interpretation

The conditional mean is a weighted average of the current variance \(v_t\) and the long-run level \(\theta\), with the weight on \(v_t\) decaying exponentially at rate \(\kappa\):

• At \(\tau = 0\): \(\mathbb{E}[v_T | v_t] = v_t\) (trivially)
• As \(\tau \to \infty\): \(\mathbb{E}[v_T | v_t] \to \theta\) (mean reversion)
• At \(\tau = \ln 2/\kappa\) (the half-life): \(\mathbb{E}[v_T | v_t] = \frac{1}{2}(v_t + \theta)\)

Verification from the Non-Central Chi-Squared Distribution

From the transition density, \(v_T | v_t \sim c\,\chi^2_\delta(\lambda)\) where \(c = \sigma_v^2(1-e^{-\kappa\tau})/(4\kappa)\), \(\delta = 4\kappa\theta/\sigma_v^2\), and \(\lambda = 4\kappa v_t e^{-\kappa\tau}/(\sigma_v^2(1-e^{-\kappa\tau}))\). The mean of a non-central chi-squared is \(\delta + \lambda\), so:

\[ \mathbb{E}[v_T | v_t] = c(\delta + \lambda) = \frac{\sigma_v^2(1-e^{-\kappa\tau})}{4\kappa}\left(\frac{4\kappa\theta}{\sigma_v^2} + \frac{4\kappa v_t e^{-\kappa\tau}}{\sigma_v^2(1-e^{-\kappa\tau})}\right) = \theta(1-e^{-\kappa\tau}) + v_t e^{-\kappa\tau} \]

which matches the SDE-derived formula. \(\square\)

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Conditional Variance

Proposition: Conditional Variance of v_T

For the CIR process:

\[ \text{Var}(v_T \mid v_t) = v_t\,\frac{\sigma_v^2 e^{-\kappa\tau}}{\kappa}(1 - e^{-\kappa\tau}) + \theta\,\frac{\sigma_v^2}{2\kappa}(1 - e^{-\kappa\tau})^2 \]

Proof. Define \(s(\tau) = \mathbb{E}[v_{t+\tau}^2 | v_t]\). Applying Ito's lemma to \(v^2\):

\[ d(v_t^2) = 2v_t\,dv_t + (dv_t)^2 = 2v_t[\kappa(\theta - v_t)\,dt + \sigma_v\sqrt{v_t}\,dW_t] + \sigma_v^2 v_t\,dt \]

\[ = [2\kappa\theta v_t - 2\kappa v_t^2 + \sigma_v^2 v_t]\,dt + 2\sigma_v v_t^{3/2}\,dW_t \]

Taking expectations:

\[ \frac{ds}{d\tau} = (2\kappa\theta + \sigma_v^2)\,m(\tau) - 2\kappa\,s(\tau) \]

where \(m(\tau) = \mathbb{E}[v_{t+\tau}|v_t] = \theta + (v_t-\theta)e^{-\kappa\tau}\). This is a first-order linear ODE in \(s\) with initial condition \(s(0) = v_t^2\).

Solving via the integrating factor \(e^{2\kappa\tau}\):

\[ s(\tau) = e^{-2\kappa\tau}v_t^2 + (2\kappa\theta + \sigma_v^2)\int_0^\tau e^{-2\kappa(\tau-u)}m(u)\,du \]

After evaluating the integral and simplifying:

\[ s(\tau) = \left[\theta + (v_t - \theta)e^{-\kappa\tau}\right]^2 + v_t\,\frac{\sigma_v^2 e^{-\kappa\tau}}{\kappa}(1-e^{-\kappa\tau}) + \theta\,\frac{\sigma_v^2}{2\kappa}(1-e^{-\kappa\tau})^2 \]

Since \(\text{Var}(v_T|v_t) = s(\tau) - m(\tau)^2\) and the first term equals \(m(\tau)^2\), the result follows. \(\square\)

Interpretation

The conditional variance has two components:

1. Initial-condition-dependent term: \(v_t \cdot \frac{\sigma_v^2 e^{-\kappa\tau}}{\kappa}(1 - e^{-\kappa\tau})\), which is proportional to \(v_t\) and decays with \(\tau\)
2. Stationary term: \(\theta \cdot \frac{\sigma_v^2}{2\kappa}(1 - e^{-\kappa\tau})^2\), which grows from zero and saturates at \(\theta\sigma_v^2/(2\kappa)\)

For short \(\tau\): \(\text{Var}(v_T|v_t) \approx v_t\sigma_v^2\tau\) (the variance grows linearly, proportional to the current level).

For large \(\tau\): \(\text{Var}(v_T|v_t) \to \theta\sigma_v^2/(2\kappa)\) (the stationary variance, independent of \(v_t\)).

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Higher Conditional Moments

Higher conditional moments \(\mathbb{E}[v_T^n | v_t]\) can be computed recursively.

Proposition: Recursive Moment Formula

Define \(m_n(\tau) = \mathbb{E}[v_{t+\tau}^n | v_t]\). Then for \(n \geq 1\):

\[ \frac{dm_n}{d\tau} = n\kappa\theta\,m_{n-1}(\tau) + \left(\tfrac{1}{2}n(n-1)\sigma_v^2 - n\kappa\right)m_n(\tau) \]

with initial condition \(m_n(0) = v_t^n\).

Proof. Apply Ito's lemma to \(v^n\):

\[ d(v_t^n) = nv_t^{n-1}\,dv_t + \tfrac{1}{2}n(n-1)v_t^{n-2}(dv_t)^2 \]

\[ = nv_t^{n-1}[\kappa(\theta-v_t)\,dt + \sigma_v\sqrt{v_t}\,dW_t] + \tfrac{1}{2}n(n-1)\sigma_v^2 v_t^{n-1}\,dt \]

Taking expectations and noting \(\mathbb{E}[dW_t] = 0\):

\[ \frac{dm_n}{d\tau} = n\kappa\theta\,m_{n-1} - n\kappa\,m_n + \tfrac{1}{2}n(n-1)\sigma_v^2\,m_n \]

\(\square\)

The recursion shows that \(m_n\) depends on \(m_{n-1}\), so moments can be computed sequentially: \(m_0 = 1 \to m_1 \to m_2 \to m_3 \to \cdots\)

Third Conditional Moment

The third moment satisfies:

\[ \frac{dm_3}{d\tau} = 3\kappa\theta\,m_2(\tau) + (3\sigma_v^2 - 3\kappa)\,m_3(\tau) \]

This ODE can be solved given \(m_2(\tau)\) (already computed), but the closed-form expression is lengthy. In practice, the first two moments suffice for most applications (moment matching in simulation schemes, variance swap pricing).

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Integrated Variance Moments

The integrated variance \(\int_t^T v_s\,ds\) appears in variance swap pricing and in the Broadie-Kaya exact simulation scheme. Its moments have closed-form expressions.

Proposition: Integrated Variance Moments

Define \(V_\tau = \int_t^{t+\tau} v_s\,ds\). Then:

\[ \mathbb{E}[V_\tau \mid v_t] = \theta\tau + (v_t - \theta)\,\frac{1 - e^{-\kappa\tau}}{\kappa} \]

\[ \text{Var}(V_\tau \mid v_t) = \frac{\sigma_v^2}{\kappa^2}\left[\theta\tau - 2(v_t - \theta)\frac{e^{-\kappa\tau} - 1}{\kappa} - \frac{v_t(e^{-\kappa\tau} - 1)^2}{\kappa} - \frac{\theta(1-e^{-\kappa\tau})^2}{2\kappa}\right] \]

Proof (mean). Integrate the conditional mean formula:

\[ \mathbb{E}[V_\tau | v_t] = \int_0^\tau \mathbb{E}[v_{t+s} | v_t]\,ds = \int_0^\tau \left[\theta + (v_t - \theta)e^{-\kappa s}\right]ds = \theta\tau + (v_t - \theta)\frac{1 - e^{-\kappa\tau}}{\kappa} \]

\(\square\)

Variance Swap Fair Value

The fair strike of a variance swap with maturity \(T\) is:

\[ K_{\text{var}} = \frac{1}{T}\mathbb{E}^{\mathbb{Q}}\!\left[\int_0^T v_s\,ds\right] = \theta + (v_0 - \theta)\frac{1 - e^{-\kappa T}}{\kappa T} \]

For short maturities (\(\kappa T \ll 1\)), \(K_{\text{var}} \approx v_0\). For long maturities (\(\kappa T \gg 1\)), \(K_{\text{var}} \approx \theta\). This formula is one of the rare closed-form results for exotic pricing under Heston.

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Stationary (Unconditional) Moments

As \(\tau \to \infty\), the CIR process reaches its stationary Gamma distribution (see CIR transition density). The unconditional moments are:

Moment	Formula	Value (typical parameters)
Mean	\(\theta\)	\(0.04\)
Variance	\(\frac{\theta\sigma_v^2}{2\kappa}\)	\(\frac{0.04 \times 0.25}{4} = 0.0025\)
Skewness	\(\frac{\sigma_v}{\sqrt{\kappa\theta}} \cdot \sqrt{2}\)	\(\frac{0.5}{\sqrt{0.08}} \cdot 1.414 = 2.5\)
Excess kurtosis	\(\frac{3\sigma_v^2}{\kappa\theta}\)	\(\frac{0.75}{0.08} = 9.375\)

The typical parameters used: \(\kappa = 2\), \(\theta = 0.04\), \(\sigma_v = 0.5\).

Heavy Tails of the Stationary Distribution

The stationary distribution of the CIR process is a Gamma distribution, which has positive skewness and excess kurtosis. With typical calibrated parameters, the skewness exceeds 2 and the kurtosis exceeds 9, indicating a strongly right-skewed, heavy-tailed distribution for the variance. This means variance spikes (e.g., during market crises) are much more likely than a Gaussian model would suggest.

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Worked Example: Moment Calculations

Full Moment Calculation

Parameters: \(\kappa = 2.0\), \(\theta = 0.04\), \(\sigma_v = 0.5\), \(v_t = 0.06\), \(\tau = 0.25\) years.

Conditional mean:

\[ \mathbb{E}[v_T|v_t] = 0.04 + (0.06 - 0.04)e^{-0.5} = 0.04 + 0.02 \times 0.6065 = 0.04 + 0.01213 = 0.05213 \]

Conditional variance:

\[ \text{Var}(v_T|v_t) = 0.06 \cdot \frac{0.25 \times 0.6065}{2}(1-0.6065) + 0.04 \cdot \frac{0.25}{4}(1-0.6065)^2 \]

\[ = 0.06 \times 0.0758 \times 0.3935 + 0.04 \times 0.0625 \times 0.1548 = 0.001789 + 0.000387 = 0.002176 \]

Standard deviation: \(\sqrt{0.002176} \approx 0.0467\).

Coefficient of variation: \(0.0467 / 0.05213 \approx 0.90\). The standard deviation is 90\% of the mean, confirming the high uncertainty in future variance levels.

Integrated variance mean:

\[ \mathbb{E}[V_{0.25}|v_t] = 0.04 \times 0.25 + (0.06 - 0.04)\frac{1-e^{-0.5}}{2} = 0.01 + 0.02 \times 0.1967 = 0.01 + 0.003935 = 0.01394 \]

The annualized variance swap strike is \(K_{\text{var}} = 0.01394/0.25 = 0.05574\), corresponding to a volatility of \(\sqrt{0.05574} \approx 23.6\%\).

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Summary

The CIR variance process has closed-form conditional moments at all orders, derivable from a recursive ODE system or from the non-central chi-squared transition density. The conditional mean \(\mathbb{E}[v_T|v_t] = \theta + (v_t - \theta)e^{-\kappa\tau}\) exhibits exponential mean-reversion, while the conditional variance captures both the current-state-dependent and stationary components. The integrated variance \(\int_t^T v_s\,ds\) has a closed-form mean that directly gives the variance swap fair value. The stationary distribution is Gamma with mean \(\theta\), variance \(\theta\sigma_v^2/(2\kappa)\), and substantial positive skewness.

The next section examines the mean-reversion mechanism in detail, including the half-life, the term structure of variance, and the distinction between physical and risk-neutral mean-reversion parameters.

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Exercises

Exercise 1. Derive \(\mathbb{E}[V_T \mid V_t] = \theta + (V_t - \theta)e^{-\kappa\tau}\) by solving the ODE \(\frac{d}{d\tau}\mathbb{E}[V_{t+\tau}] = \kappa(\theta - \mathbb{E}[V_{t+\tau}])\) with initial condition \(\mathbb{E}[V_t] = V_t\).

Solution to Exercise 1

We solve the ODE \(\frac{d}{d\tau}m(\tau) = \kappa(\theta - m(\tau))\) with initial condition \(m(0) = V_t\), where \(m(\tau) = \mathbb{E}[V_{t+\tau} \mid V_t]\).

Step 1: Rewrite the ODE.

\[ \frac{dm}{d\tau} + \kappa\,m = \kappa\theta \]

This is a first-order linear ODE with constant coefficients.

Step 2: Find the integrating factor. The integrating factor is \(\mu(\tau) = e^{\kappa\tau}\). Multiplying both sides:

\[ \frac{d}{d\tau}\!\left(e^{\kappa\tau}m\right) = \kappa\theta\,e^{\kappa\tau} \]

Step 3: Integrate both sides from \(0\) to \(\tau\).

\[ e^{\kappa\tau}m(\tau) - m(0) = \kappa\theta\int_0^\tau e^{\kappa s}\,ds = \kappa\theta \cdot \frac{e^{\kappa\tau} - 1}{\kappa} = \theta(e^{\kappa\tau} - 1) \]

Step 4: Solve for \(m(\tau)\).

\[ m(\tau) = e^{-\kappa\tau}m(0) + \theta(1 - e^{-\kappa\tau}) = e^{-\kappa\tau}V_t + \theta(1 - e^{-\kappa\tau}) \]

Rearranging:

\[ \mathbb{E}[V_T \mid V_t] = \theta + (V_t - \theta)e^{-\kappa\tau} \]

This confirms the conditional mean formula. The derivation uses only the linearity of expectation and the fact that \(\mathbb{E}[\sigma_v\sqrt{V_s}\,dW_s] = 0\) (the stochastic integral is a martingale with zero expectation).

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Exercise 2. Compute the conditional variance \(\operatorname{Var}(V_T \mid V_t) = V_t\frac{\sigma_v^2}{\kappa}(e^{-\kappa\tau} - e^{-2\kappa\tau}) + \frac{\theta\sigma_v^2}{2\kappa}(1 - e^{-\kappa\tau})^2\) for \(\kappa = 2\), \(\theta = 0.04\), \(\sigma_v = 0.3\), \(V_t = 0.04\), \(\tau = 1\). Is the conditional standard deviation of \(V_T\) large relative to its mean?

Solution to Exercise 2

With \(\kappa = 2\), \(\theta = 0.04\), \(\sigma_v = 0.3\), \(V_t = 0.04\), \(\tau = 1\):

First compute the needed quantities:

\[ e^{-\kappa\tau} = e^{-2} = 0.1353, \qquad 1 - e^{-\kappa\tau} = 0.8647 \]

Conditional mean:

\[ \mathbb{E}[V_T \mid V_t] = 0.04 + (0.04 - 0.04)e^{-2} = 0.04 \]

Since \(V_t = \theta\), the expected value remains at \(\theta\) for all \(\tau\).

Conditional variance:

\[ \operatorname{Var}(V_T \mid V_t) = V_t\frac{\sigma_v^2}{\kappa}(e^{-\kappa\tau} - e^{-2\kappa\tau}) + \frac{\theta\sigma_v^2}{2\kappa}(1 - e^{-\kappa\tau})^2 \]

Computing each term:

First term:

\[ 0.04 \times \frac{0.09}{2}(e^{-2} - e^{-4}) = 0.04 \times 0.045 \times (0.1353 - 0.01832) = 0.04 \times 0.045 \times 0.1170 = 0.0002106 \]

Second term:

\[ \frac{0.04 \times 0.09}{4}(0.8647)^2 = 0.0009 \times 0.7477 = 0.0006729 \]

Total:

\[ \operatorname{Var}(V_T \mid V_t) = 0.0002106 + 0.0006729 = 0.0008835 \]

Conditional standard deviation:

\[ \sigma_{V_T} = \sqrt{0.0008835} = 0.02972 \]

Relative to the mean: The coefficient of variation is:

\[ \text{CV} = \frac{0.02972}{0.04} = 0.743 \]

The conditional standard deviation is about 74% of the mean. This is large, indicating substantial uncertainty about future variance levels even when starting at the long-run mean. This high variability is characteristic of the CIR process with its square-root diffusion and is consistent with the empirically observed leptokurtic (heavy-tailed) distribution of realized variance.

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Exercise 3. The integrated variance \(\bar{V} = \frac{1}{\tau}\int_t^{t+\tau} V_s\,ds\) has conditional mean \(\mathbb{E}[\bar{V} \mid V_t] = \theta + (V_t - \theta)\frac{1 - e^{-\kappa\tau}}{\kappa\tau}\). Compute this for \(\tau = 0.25\) (3-month variance swap) with \(V_t = 0.05\), \(\theta = 0.04\), \(\kappa = 2\).

Solution to Exercise 3

The conditional mean of the average variance (variance swap fair strike) is:

\[ \mathbb{E}[\bar{V} \mid V_t] = \theta + (V_t - \theta)\frac{1 - e^{-\kappa\tau}}{\kappa\tau} \]

With \(\tau = 0.25\), \(V_t = 0.05\), \(\theta = 0.04\), \(\kappa = 2\):

\[ \kappa\tau = 2 \times 0.25 = 0.5 \]

\[ \frac{1 - e^{-\kappa\tau}}{\kappa\tau} = \frac{1 - e^{-0.5}}{0.5} = \frac{1 - 0.6065}{0.5} = \frac{0.3935}{0.5} = 0.7870 \]

\[ \mathbb{E}[\bar{V} \mid V_t] = 0.04 + (0.05 - 0.04) \times 0.7870 = 0.04 + 0.01 \times 0.7870 = 0.04 + 0.007870 = 0.04787 \]

The 3-month variance swap fair strike is \(K_{\text{var}} = 0.04787\), corresponding to a fair volatility strike of \(\sqrt{0.04787} = 21.88\%\).

Note that this is between \(V_t = 0.05\) (current annualized variance, or 22.36% volatility) and \(\theta = 0.04\) (long-run variance, or 20% volatility). The factor \(0.787\) indicates that over 3 months, the average variance is still dominated by the current variance level, with only partial mean reversion toward \(\theta\).

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Exercise 4. The stationary variance of \(V_t\) is \(\operatorname{Var}_\infty(V) = \theta\sigma_v^2/(2\kappa)\). Compute the coefficient of variation \(\text{CV} = \sqrt{\operatorname{Var}_\infty(V)}/\theta\) for the parameters in Exercise 2. A CV greater than 1 indicates very heavy-tailed variance dynamics.

Solution to Exercise 4

The stationary variance of \(V_t\) is:

\[ \operatorname{Var}_\infty(V) = \frac{\theta\sigma_v^2}{2\kappa} \]

With \(\kappa = 2\), \(\theta = 0.04\), \(\sigma_v = 0.3\):

\[ \operatorname{Var}_\infty(V) = \frac{0.04 \times 0.09}{4} = \frac{0.0036}{4} = 0.0009 \]

The coefficient of variation is:

\[ \text{CV} = \frac{\sqrt{\operatorname{Var}_\infty(V)}}{\theta} = \frac{\sqrt{0.0009}}{0.04} = \frac{0.03}{0.04} = 0.75 \]

A coefficient of variation of 0.75 means the standard deviation is 75% of the mean. This is less than 1, so the variance dynamics are not extremely heavy-tailed, but they are still substantially more variable than a Gaussian model would suggest.

For comparison, if we increased \(\sigma_v\) to \(0.5\) (keeping other parameters fixed):

\[ \text{CV} = \frac{\sqrt{0.04 \times 0.25 / 4}}{0.04} = \frac{\sqrt{0.0025}}{0.04} = \frac{0.05}{0.04} = 1.25 \]

With \(\sigma_v = 0.5\), the CV exceeds 1, indicating very heavy-tailed variance dynamics. The threshold CV \(= 1\) corresponds to \(\sigma_v = \sqrt{2\kappa\theta} = \sqrt{0.16} = 0.4\), which is also the Feller condition boundary. When \(\sigma_v > \sqrt{2\kappa\theta}\) (Feller condition violated), the CV exceeds 1 and the variance process can touch zero.

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Exercise 5. Show that \(\mathbb{E}[\int_t^T V_s\,ds \mid V_t] = \theta\tau + (V_t - \theta)\frac{1 - e^{-\kappa\tau}}{\kappa}\) by integrating \(\mathbb{E}[V_s \mid V_t]\) from \(t\) to \(T\). This formula gives the fair value of a variance swap starting at \(t\) with maturity \(T\).

Solution to Exercise 5

Starting from the conditional mean \(\mathbb{E}[V_s \mid V_t] = \theta + (V_t - \theta)e^{-\kappa(s-t)}\) for \(s \geq t\):

\[ \mathbb{E}\!\left[\int_t^T V_s\,ds \,\middle|\, V_t\right] = \int_t^T \mathbb{E}[V_s \mid V_t]\,ds \]

The interchange of expectation and integration is justified by Fubini's theorem (the integrand is non-negative and integrable).

Substituting the conditional mean:

\[ = \int_t^T \left[\theta + (V_t - \theta)e^{-\kappa(s-t)}\right]ds \]

Using the substitution \(u = s - t\) with \(du = ds\), and the limits changing to \(u \in [0, \tau]\) where \(\tau = T - t\):

\[ = \int_0^\tau \left[\theta + (V_t - \theta)e^{-\kappa u}\right]du \]

\[ = \theta\int_0^\tau du + (V_t - \theta)\int_0^\tau e^{-\kappa u}\,du \]

\[ = \theta\tau + (V_t - \theta)\left[-\frac{1}{\kappa}e^{-\kappa u}\right]_0^\tau \]

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

This is the variance swap fair value formula. For a variance swap with strike \(K_{\text{var}}\) and maturity \(T\), the annualized fair strike is:

\[ K_{\text{var}} = \frac{1}{\tau}\mathbb{E}^{\mathbb{Q}}\!\left[\int_t^T V_s\,ds \,\middle|\, V_t\right] = \theta + (V_t - \theta)\frac{1 - e^{-\kappa\tau}}{\kappa\tau} \]

For short maturities (\(\kappa\tau \ll 1\)), the factor \(\frac{1 - e^{-\kappa\tau}}{\kappa\tau} \approx 1\), so \(K_{\text{var}} \approx V_t\). For long maturities (\(\kappa\tau \gg 1\)), the factor vanishes and \(K_{\text{var}} \approx \theta\).

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Exercise 6. For the higher moments, the conditional third central moment of \(V_T\) is needed for skewness. Without deriving it fully, explain why the CIR process has positive skewness (heavier right tail than left) and relate this to the square-root diffusion.

Solution to Exercise 6

The CIR process \(dV_t = \kappa(\theta - V_t)\,dt + \sigma_v\sqrt{V_t}\,dW_t\) has positive skewness for two related reasons:

1. The square-root diffusion creates asymmetric volatility. The diffusion coefficient \(\sigma_v\sqrt{V_t}\) is an increasing function of \(V_t\). When \(V_t\) is above its mean \(\theta\), the diffusion is larger, producing larger random fluctuations that can push \(V_t\) even further above the mean. When \(V_t\) is below the mean, the diffusion is smaller, limiting how far below the mean the process can go. This creates a natural asymmetry: upward excursions are amplified while downward excursions are dampened.

2. The non-negativity constraint truncates the left tail. The process satisfies \(V_t \geq 0\) a.s. (and \(V_t > 0\) a.s. under the Feller condition). This provides a hard lower bound but no upper bound, automatically creating a right-skewed distribution.

3. The stationary distribution is Gamma. The CIR process converges to a \(\text{Gamma}(\alpha, \beta)\) stationary distribution with \(\alpha = 2\kappa\theta/\sigma_v^2\) and \(\beta = 2\kappa/\sigma_v^2\). The skewness of the Gamma distribution is \(2/\sqrt{\alpha}\), which is always positive. For typical parameters (\(\alpha\) in the range 1--4), the skewness ranges from 1 to 2, confirming a strongly right-skewed distribution.

Connection to the conditional third central moment: The positive skewness at the distributional level implies that the conditional third central moment \(\mathbb{E}[(V_T - \mathbb{E}[V_T|V_t])^3 | V_t]\) is positive. Heuristically, from the recursive moment ODE, the third moment depends on the second moment and inherits the asymmetry of the square-root diffusion. The detailed derivation (through the \(n=3\) case of the recursive formula) produces a lengthy but positive expression for all positive parameter values.

Financial implication: The positive skewness of the variance process means that variance spikes (large upward moves in \(V_t\)) are more common than variance troughs (large downward moves) relative to a symmetric model. This is consistent with the empirical observation that realized volatility has a long right tail -- extreme volatility spikes during market crises are much more dramatic than periods of unusually low volatility.