Log-Affine Expectation Property¶

The defining characteristic of an affine process is that its log-moment generating function is linear in the current state. This single property---seemingly a modest algebraic constraint---is the engine that reduces option pricing and term structure computation from solving partial differential equations to solving ordinary differential equations. This section states the log-affine expectation property precisely, derives it from the backward Kolmogorov equation, and explores its consequences.

Learning Objectives

By the end of this section, you will be able to:

State the log-affine expectation formula \(\mathbb{E}[e^{\langle u, X_T \rangle} \mid X_t = x] = \exp(\phi(\tau, u) + \langle \psi(\tau, u), x \rangle)\) with correct domains and initial conditions
Derive the exponential-affine form from the backward Kolmogorov equation using an affine ansatz
Distinguish the moment generating function (real \(u\)) from the characteristic function (imaginary \(u\))
Explain why linearity of the log-transform in \(x\) is the key computational advantage

Intuition¶

Consider a Markov process \(X_t\) and the conditional expectation \(\mathbb{E}[e^{\langle u, X_T \rangle} \mid X_t = x]\). For a general process, this expectation is some complicated, possibly intractable function of \(x\). For an affine process, something remarkable happens: the logarithm of this expectation is a linear function of \(x\). In other words, the state \(x\) enters only through a dot product \(\langle \psi, x \rangle\), with the coefficient \(\psi\) depending on the time horizon and the transform variable \(u\), but not on \(x\) itself.

This linearity means that once we compute the two functions \(\phi\) and \(\psi\) (which depend on time and \(u\) but not on \(x\)), we can evaluate the conditional expectation at any starting point \(x\) instantly. In financial terms, this translates to: once we solve the Riccati ODEs for a given maturity, we can price bonds, options, and other derivatives across all current states of the economy without re-solving anything.

The Log-Affine Expectation Formula¶

Statement¶

Definition (Log-Affine Property). A Markov process \((X_t)_{t \geq 0}\) on \(D = \mathbb{R}^m_+ \times \mathbb{R}^{d-m}\) is affine if there exist functions \(\phi : \mathbb{R}_+ \times \mathcal{U} \to \mathbb{C}\) and \(\psi : \mathbb{R}_+ \times \mathcal{U} \to \mathbb{C}^d\) such that for all \(x \in D\), all \(\tau = T - t \geq 0\), and all \(u\) in the admissible set \(\mathcal{U} \subseteq \mathbb{C}^d\):

\[ \mathbb{E}\!\left[e^{\langle u, X_T \rangle} \mid X_t = x\right] = \exp\!\left(\phi(\tau, u) + \langle \psi(\tau, u), x \rangle\right) \]

with the initial conditions

\[ \phi(0, u) = 0, \qquad \psi(0, u) = u \]

The function \(\phi\) captures the contribution that is independent of the current state, while \(\psi\) captures the state-dependent part. The initial conditions ensure consistency at \(\tau = 0\): setting \(T = t\) gives \(\mathbb{E}[e^{\langle u, X_t \rangle} \mid X_t = x] = e^{\langle u, x \rangle}\), which is trivially true.

The Admissible Set¶

The admissible set \(\mathcal{U}\) consists of those \(u \in \mathbb{C}^d\) for which the expectation \(\mathbb{E}[e^{\langle u, X_T \rangle} \mid X_t = x]\) is finite. For the moment generating function, we take \(u \in \mathbb{R}^d\) (or a subset thereof); for the characteristic function, we take \(u = iv\) with \(v \in \mathbb{R}^d\), giving:

\[ \mathbb{E}\!\left[e^{i\langle v, X_T \rangle} \mid X_t = x\right] = \exp\!\left(\phi(\tau, iv) + \langle \psi(\tau, iv), x \rangle\right) \]

The characteristic function always exists (no integrability issues), so \(iv \in \mathcal{U}\) for all \(v \in \mathbb{R}^d\). The moment generating function may only exist for \(u\) in a proper subset of \(\mathbb{R}^d\), determined by the tail behavior of \(X_T\).

Derivation from the Backward Kolmogorov Equation¶

Setup¶

To see why affine coefficients produce the log-affine form, consider a scalar affine diffusion (the argument extends to the multidimensional case):

\[ dX_t = (\kappa_0 + \kappa_1 X_t)\,dt + \sqrt{\sigma_0 + \sigma_1 X_t}\,dW_t \]

Define the function

\[ g(\tau, x) = \mathbb{E}\!\left[e^{u X_{t+\tau}} \mid X_t = x\right] \]

By the backward Kolmogorov equation, \(g\) satisfies the PDE:

\[ \frac{\partial g}{\partial \tau} = (\kappa_0 + \kappa_1 x)\frac{\partial g}{\partial x} + \frac{1}{2}(\sigma_0 + \sigma_1 x)\frac{\partial^2 g}{\partial x^2} \]

with initial condition \(g(0, x) = e^{ux}\).

The Affine Ansatz¶

Motivated by the initial condition, we try the ansatz

\[ g(\tau, x) = \exp\!\left(\phi(\tau) + \psi(\tau) x\right) \]

with \(\phi(0) = 0\) and \(\psi(0) = u\). Computing the required derivatives:

\[ \frac{\partial g}{\partial \tau} = (\phi'(\tau) + \psi'(\tau) x)\,g \]

\[ \frac{\partial g}{\partial x} = \psi(\tau)\,g, \qquad \frac{\partial^2 g}{\partial x^2} = \psi(\tau)^2\,g \]

Substitution and Separation¶

Substituting into the PDE and dividing both sides by \(g\) (which is strictly positive):

\[ \phi'(\tau) + \psi'(\tau) x = (\kappa_0 + \kappa_1 x)\psi(\tau) + \frac{1}{2}(\sigma_0 + \sigma_1 x)\psi(\tau)^2 \]

Collecting terms by powers of \(x\):

\[ \phi'(\tau) + \psi'(\tau) x = \left[\kappa_0 \psi(\tau) + \frac{1}{2}\sigma_0 \psi(\tau)^2\right] + \left[\kappa_1 \psi(\tau) + \frac{1}{2}\sigma_1 \psi(\tau)^2\right] x \]

Since this must hold for all \(x \in D\), we can match the constant and linear terms separately.

Constant terms (the \(\phi\)-equation):

\[ \phi'(\tau) = \kappa_0 \psi(\tau) + \frac{1}{2}\sigma_0 \psi(\tau)^2 \]

Linear terms (the \(\psi\)-equation):

\[ \psi'(\tau) = \kappa_1 \psi(\tau) + \frac{1}{2}\sigma_1 \psi(\tau)^2 \]

This separation works precisely because the drift and diffusion coefficients are affine in \(x\). If the coefficients had any nonlinear dependence on \(x\), the ansatz would fail and the log-affine property would not hold.

Identification with \(F\) and \(R\)¶

Writing \(F(u) = \kappa_0 u + \frac{1}{2}\sigma_0 u^2\) and \(R(u) = \kappa_1 u + \frac{1}{2}\sigma_1 u^2\), the system becomes

\[ \phi'(\tau) = F(\psi(\tau)), \qquad \psi'(\tau) = R(\psi(\tau)) \]

These are the generalized Riccati ODEs. The function \(R\) is a Riccati equation when \(\sigma_1 \neq 0\) (quadratic in \(\psi\)) and a simple linear ODE when \(\sigma_1 = 0\) (Gaussian case). The detailed study of this ODE system is the subject of the next section.

Why Linearity in x Matters¶

Computational Advantage¶

The log-affine property means that to evaluate \(\mathbb{E}[e^{\langle u, X_T \rangle} \mid X_t = x]\) at a new starting point \(x'\), we do not need to re-solve any differential equation. We simply replace \(x\) by \(x'\) in the formula \(\exp(\phi(\tau, u) + \langle \psi(\tau, u), x' \rangle)\). This is a single dot product plus exponentiation---an \(O(d)\) operation.

For bond pricing, this means that \(P(t, T) = e^{A(\tau) + B(\tau)^\top x}\) can be evaluated for any current state \(x\) once \(A\) and \(B\) have been computed. For an entire yield curve (multiple maturities \(T\)), we solve the Riccati ODEs once per maturity and then evaluate the formula across all states. This is vastly more efficient than solving a PDE on a grid or running Monte Carlo simulations.

Contrast with Non-Affine Models¶

In a non-affine model---for instance, the SABR model with \(\sigma(X_t) = X_t^\beta\) for \(\beta \notin \{0, 1/2, 1\}\)---the ansatz \(g = e^{\phi + \psi x}\) fails. The PDE does not separate into terms proportional to \(1\) and \(x\), and we cannot reduce the problem to ODEs. Instead, we must solve the full PDE numerically or use asymptotic approximations.

Log-Affine Property for the Ornstein-Uhlenbeck Process

The Ornstein-Uhlenbeck process \(dX_t = -\kappa X_t\,dt + \sigma\,dW_t\) has \(\kappa_0 = 0\), \(\kappa_1 = -\kappa\), \(\sigma_0 = \sigma^2\), \(\sigma_1 = 0\). The Riccati system is:

\[ \psi'(\tau) = -\kappa\psi(\tau), \qquad \psi(0) = u \]

\[ \phi'(\tau) = \frac{1}{2}\sigma^2\psi(\tau)^2, \qquad \phi(0) = 0 \]

Solving: \(\psi(\tau) = u e^{-\kappa\tau}\) and

\[ \phi(\tau) = \frac{\sigma^2 u^2}{2} \int_0^\tau e^{-2\kappa s}\,ds = \frac{\sigma^2 u^2}{4\kappa}(1 - e^{-2\kappa\tau}) \]

Therefore:

\[ \mathbb{E}\!\left[e^{u X_{t+\tau}} \mid X_t = x\right] = \exp\!\left(\frac{\sigma^2 u^2}{4\kappa}(1 - e^{-2\kappa\tau}) + u e^{-\kappa\tau} x\right) \]

The log of this expression is indeed linear in \(x\): \(\log \mathbb{E}[\cdots] = \frac{\sigma^2 u^2}{4\kappa}(1 - e^{-2\kappa\tau}) + u e^{-\kappa\tau} x\). This confirms that \(X_T \mid X_t = x\) is Gaussian with mean \(x e^{-\kappa\tau}\) and variance \(\frac{\sigma^2}{2\kappa}(1 - e^{-2\kappa\tau})\). \(\square\)

Moment Generating Function vs Characteristic Function¶

The log-affine formula applies to both the moment generating function (MGF) and the characteristic function (CF), with different choices of \(u\).

Moment generating function (\(u \in \mathbb{R}^d\)):

\[ M_X(\tau, u, x) = \mathbb{E}\!\left[e^{\langle u, X_T \rangle} \mid X_t = x\right] = e^{\phi(\tau, u) + \langle \psi(\tau, u), x \rangle} \]

This exists only for \(u\) in the domain where the expectation is finite. For CIR-type components, the MGF may explode in finite time for large \(u\), a phenomenon called moment explosion.

Characteristic function (\(u = iv\), \(v \in \mathbb{R}^d\)):

\[ \Phi_X(\tau, v, x) = \mathbb{E}\!\left[e^{i\langle v, X_T \rangle} \mid X_t = x\right] = e^{\phi(\tau, iv) + \langle \psi(\tau, iv), x \rangle} \]

The characteristic function always exists and is bounded by 1 in absolute value. It is the primary tool for Fourier-based pricing methods.

Moment Explosion

For the CIR process with parameters \(\kappa\), \(\theta\), \(\xi\), the moment generating function \(\mathbb{E}[e^{u r_T}]\) is finite only for \(u\) below a critical threshold that depends on \(\tau\). As \(\tau \to \infty\), this threshold converges to \(u^* = 2\kappa/\xi^2\). For \(u \geq u^*\), the expectation is infinite. This has practical consequences: heavy-tailed distributions and extreme scenarios cannot be captured by the MGF alone.

The Multidimensional Case¶

For a \(d\)-dimensional affine process with drift \(b(x) = b_0 + \sum_{i=1}^d b_i x_i\), diffusion \(a(x) = a_0 + \sum_{i=1}^d a_i x_i\), and jump compensator \(m(x, dz) = m_0(dz) + \sum_{i=1}^d x_i m_i(dz)\), the log-affine property takes the form:

\[ \mathbb{E}\!\left[e^{\langle u, X_T \rangle} \mid X_t = x\right] = \exp\!\left(\phi(\tau, u) + \langle \psi(\tau, u), x \rangle\right) \]

where now \(\phi : \mathbb{R}_+ \times \mathbb{C}^d \to \mathbb{C}\) and \(\psi : \mathbb{R}_+ \times \mathbb{C}^d \to \mathbb{C}^d\) satisfy the system

\[ \frac{\partial \phi}{\partial \tau} = F(\psi), \qquad \frac{\partial \psi}{\partial \tau} = R(\psi) \]

with

\[ F(u) = \langle b_0, u \rangle + \frac{1}{2}\langle u, a_0 u \rangle + \int_{D \setminus \{0\}} (e^{\langle u, z \rangle} - 1)\,m_0(dz) \]

\[ R_i(u) = \langle b_i, u \rangle + \frac{1}{2}\langle u, a_i u \rangle + \int_{D \setminus \{0\}} (e^{\langle u, z \rangle} - 1)\,m_i(dz) \]

The structure is identical to the scalar case: \(\phi\) collects the state-independent terms and \(\psi\) collects the state-dependent terms, with \(R\) now a vector-valued function governing a coupled system of Riccati equations.

Summary¶

The log-affine expectation property states that \(\log \mathbb{E}[e^{\langle u, X_T \rangle} \mid X_t = x]\) is a linear function of \(x\), decomposing into a state-independent part \(\phi(\tau, u)\) and a state-dependent part \(\langle \psi(\tau, u), x \rangle\). This property follows from the affine structure of the drift and diffusion coefficients: the exponential ansatz in the backward Kolmogorov equation separates into constant and linear terms, yielding the generalized Riccati ODEs \(\phi' = F(\psi)\) and \(\psi' = R(\psi)\). The log-affine form is the computational engine of the affine framework, enabling bond pricing, characteristic function evaluation, and Fourier inversion to be performed efficiently across all states \(x\) once the Riccati solutions are known.

Exercises¶

Exercise 1. For the CIR process \(dX_t = \kappa(\theta - X_t)\,dt + \xi\sqrt{X_t}\,dW_t\), identify the functions \(F(u)\) and \(R(u)\) in the generalized Riccati framework. Write down the ODE for \(\psi(\tau)\) with initial condition \(\psi(0) = u\) and verify that it is a scalar Riccati equation. Under what condition on \(u\) does the moment generating function \(\mathbb{E}[e^{uX_T} \mid X_0 = x]\) remain finite for all \(T > 0\)?

Solution to Exercise 1

For the CIR process \(dX_t = \kappa(\theta - X_t)\,dt + \xi\sqrt{X_t}\,dW_t\), we identify the drift and diffusion:

Drift: \(b(x) = \kappa\theta - \kappa x\), so \(\kappa_0 = \kappa\theta\) and \(\kappa_1 = -\kappa\)
Diffusion: \(a(x) = \xi^2 x\), so \(\sigma_0 = 0\) and \(\sigma_1 = \xi^2\)

The functions \(F\) and \(R\) are:

\[ F(u) = \kappa_0 u + \frac{1}{2}\sigma_0 u^2 = \kappa\theta\, u \]

\[ R(u) = \kappa_1 u + \frac{1}{2}\sigma_1 u^2 = -\kappa u + \frac{1}{2}\xi^2 u^2 \]

The ODE for \(\psi\) is:

\[ \psi'(\tau) = R(\psi(\tau)) = -\kappa\psi(\tau) + \frac{1}{2}\xi^2\psi(\tau)^2, \qquad \psi(0) = u \]

This is a scalar Riccati equation because the right-hand side is quadratic in \(\psi\) (the \(\frac{1}{2}\xi^2\psi^2\) term arises from the state-dependent diffusion \(\sigma_1 = \xi^2 \neq 0\)).

Finiteness condition: The moment generating function \(\mathbb{E}[e^{uX_T} \mid X_0 = x]\) remains finite for all \(T > 0\) if and only if the solution \(\psi(\tau, u)\) does not blow up in finite time. The Riccati equation has the equilibrium \(\psi^* = 2\kappa/\xi^2\) (from \(-\kappa\psi + \frac{1}{2}\xi^2\psi^2 = 0\)). If \(u < 2\kappa/\xi^2\), then \(\psi(\tau)\) remains bounded for all \(\tau > 0\), and the MGF is finite. If \(u \geq 2\kappa/\xi^2\), then \(\psi(\tau)\) blows up in finite time, and the MGF is infinite for sufficiently large \(T\). Therefore, the condition is \(u < 2\kappa/\xi^2\).

Exercise 2. Starting from the backward Kolmogorov equation for the two-dimensional process \((X_t^{(1)}, X_t^{(2)})\) with independent Ornstein-Uhlenbeck components \(dX_t^{(i)} = -\kappa_i X_t^{(i)}\,dt + \sigma_i\,dW_t^{(i)}\), verify the affine ansatz

\[ g(\tau, x_1, x_2) = \exp(\phi(\tau) + \psi_1(\tau)x_1 + \psi_2(\tau)x_2) \]

by substituting into the PDE and showing that the system separates into three ODEs: one for \(\phi\), one for \(\psi_1\), and one for \(\psi_2\).

Solution to Exercise 2

The two-dimensional process with independent OU components satisfies:

\[ dX_t^{(i)} = -\kappa_i X_t^{(i)}\,dt + \sigma_i\,dW_t^{(i)}, \qquad i = 1, 2 \]

The backward Kolmogorov equation for \(g(\tau, x_1, x_2) = \mathbb{E}[e^{u_1 X_{t+\tau}^{(1)} + u_2 X_{t+\tau}^{(2)}} \mid X_t^{(1)} = x_1, X_t^{(2)} = x_2]\) is:

\[ \frac{\partial g}{\partial \tau} = -\kappa_1 x_1 \frac{\partial g}{\partial x_1} - \kappa_2 x_2 \frac{\partial g}{\partial x_2} + \frac{1}{2}\sigma_1^2 \frac{\partial^2 g}{\partial x_1^2} + \frac{1}{2}\sigma_2^2 \frac{\partial^2 g}{\partial x_2^2} \]

Substituting the ansatz \(g = \exp(\phi(\tau) + \psi_1(\tau)x_1 + \psi_2(\tau)x_2)\):

\[ \frac{\partial g}{\partial \tau} = (\phi' + \psi_1' x_1 + \psi_2' x_2)\,g \]

\[ \frac{\partial g}{\partial x_i} = \psi_i\,g, \qquad \frac{\partial^2 g}{\partial x_i^2} = \psi_i^2\,g \]

Substituting and dividing by \(g\):

\[ \phi' + \psi_1' x_1 + \psi_2' x_2 = -\kappa_1 x_1 \psi_1 - \kappa_2 x_2 \psi_2 + \frac{1}{2}\sigma_1^2 \psi_1^2 + \frac{1}{2}\sigma_2^2 \psi_2^2 \]

Matching coefficients of \(x_1\), \(x_2\), and the constant term:

\[ \psi_1'(\tau) = -\kappa_1\psi_1(\tau), \qquad \psi_1(0) = u_1 \]

\[ \psi_2'(\tau) = -\kappa_2\psi_2(\tau), \qquad \psi_2(0) = u_2 \]

\[ \phi'(\tau) = \frac{1}{2}\sigma_1^2\psi_1(\tau)^2 + \frac{1}{2}\sigma_2^2\psi_2(\tau)^2, \qquad \phi(0) = 0 \]

The system separates into three ODEs as claimed. The \(\psi_i\) equations are independent of each other and of \(\phi\), while \(\phi\) depends on \(\psi_1\) and \(\psi_2\) but can be obtained by quadrature after solving for them.

Exercise 3. Prove that the characteristic function \(\Phi_X(\tau, v, x) = \mathbb{E}[e^{iv X_T} \mid X_t = x]\) of any affine process satisfies \(|\Phi_X(\tau, v, x)| \leq 1\) for all real \(v\), and explain why the moment generating function \(M_X(\tau, u, x) = \mathbb{E}[e^{uX_T} \mid X_t = x]\) for real \(u > 0\) does not share this boundedness property.

Solution to Exercise 3

Boundedness of the characteristic function: For any real \(v\), we have \(|e^{ivX_T}| = 1\) for all outcomes of \(X_T\). Therefore:

\[ |\Phi_X(\tau, v, x)| = |\mathbb{E}[e^{ivX_T} \mid X_t = x]| \leq \mathbb{E}[|e^{ivX_T}| \mid X_t = x] = \mathbb{E}[1 \mid X_t = x] = 1 \]

The first inequality is the modulus inequality for conditional expectations (or Jensen's inequality applied to the convex function \(|\cdot|\)).

The MGF is not bounded: For real \(u > 0\), we have \(e^{uX_T} \geq 1\) whenever \(X_T \geq 0\), and \(e^{uX_T} \to \infty\) as \(X_T \to \infty\). For any process with unbounded support (e.g., a Gaussian or CIR process), \(\mathbb{E}[e^{uX_T}]\) grows without bound as \(u \to \infty\) or as \(\tau \to \infty\) (for fixed \(u\)). Concretely, for the OU process with \(X_T \mid X_t = x \sim \mathcal{N}(xe^{-\kappa\tau}, \frac{\sigma^2}{2\kappa}(1 - e^{-2\kappa\tau}))\):

\[ M_X(\tau, u, x) = \exp\!\left(uxe^{-\kappa\tau} + \frac{\sigma^2 u^2}{4\kappa}(1 - e^{-2\kappa\tau})\right) \]

This grows as \(e^{cu^2}\) for large \(u\), so \(M_X\) is unbounded as a function of \(u\).

Exercise 4. Using the semiflow property \(\psi(t+s, u) = \psi(t, \psi(s, u))\), show that if \(\psi(\tau, u) = ue^{-\kappa\tau}\) (as in the Ornstein-Uhlenbeck case), then the semiflow equation is automatically satisfied. Then verify the companion equation \(\phi(t+s, u) = \phi(t, \psi(s, u)) + \phi(s, u)\) using the explicit formula

\[ \phi(\tau, u) = \frac{\sigma^2 u^2}{4\kappa}(1 - e^{-2\kappa\tau}) \]

Solution to Exercise 4

Semiflow for \(\psi\): We need to verify \(\psi(t+s, u) = \psi(t, \psi(s, u))\) with \(\psi(\tau, u) = ue^{-\kappa\tau}\).

Left-hand side:

\[ \psi(t+s, u) = ue^{-\kappa(t+s)} \]

Right-hand side:

\[ \psi(t, \psi(s, u)) = \psi(t, ue^{-\kappa s}) = (ue^{-\kappa s})e^{-\kappa t} = ue^{-\kappa(t+s)} \]

These are equal, so the semiflow equation holds.

Semiflow for \(\phi\): We need to verify \(\phi(t+s, u) = \phi(t, \psi(s, u)) + \phi(s, u)\) with \(\phi(\tau, u) = \frac{\sigma^2 u^2}{4\kappa}(1 - e^{-2\kappa\tau})\).

Left-hand side:

\[ \phi(t+s, u) = \frac{\sigma^2 u^2}{4\kappa}(1 - e^{-2\kappa(t+s)}) \]

Right-hand side, first term:

\[ \phi(t, \psi(s, u)) = \frac{\sigma^2 (ue^{-\kappa s})^2}{4\kappa}(1 - e^{-2\kappa t}) = \frac{\sigma^2 u^2 e^{-2\kappa s}}{4\kappa}(1 - e^{-2\kappa t}) \]

Right-hand side, second term:

\[ \phi(s, u) = \frac{\sigma^2 u^2}{4\kappa}(1 - e^{-2\kappa s}) \]

Summing:

\[ \frac{\sigma^2 u^2}{4\kappa}\left[e^{-2\kappa s}(1 - e^{-2\kappa t}) + (1 - e^{-2\kappa s})\right] = \frac{\sigma^2 u^2}{4\kappa}\left[e^{-2\kappa s} - e^{-2\kappa(t+s)} + 1 - e^{-2\kappa s}\right] \]

\[ = \frac{\sigma^2 u^2}{4\kappa}(1 - e^{-2\kappa(t+s)}) \]

This equals the left-hand side, confirming the companion semiflow equation.

Exercise 5. Consider a process with drift \(b(x) = \alpha x\) and diffusion \(a(x) = \beta x^2\) for constants \(\alpha, \beta > 0\). Attempt the affine ansatz \(g(\tau, x) = \exp(\phi(\tau) + \psi(\tau)x)\) in the backward Kolmogorov equation and show explicitly where the separation into constant and linear terms in \(x\) fails. What does this tell you about whether geometric Brownian motion (in the price variable \(S_t\), not the log-price) is an affine process?

Solution to Exercise 5

With \(b(x) = \alpha x\) and \(a(x) = \beta x^2\), the backward Kolmogorov equation for \(g(\tau, x) = \mathbb{E}[e^{uX_{t+\tau}} \mid X_t = x]\) is:

\[ \frac{\partial g}{\partial \tau} = \alpha x\frac{\partial g}{\partial x} + \frac{1}{2}\beta x^2 \frac{\partial^2 g}{\partial x^2} \]

Trying the affine ansatz \(g = \exp(\phi(\tau) + \psi(\tau)x)\):

\[ \frac{\partial g}{\partial x} = \psi\,g, \qquad \frac{\partial^2 g}{\partial x^2} = \psi^2\,g \]

Substituting and dividing by \(g\):

\[ \phi' + \psi' x = \alpha x\psi + \frac{1}{2}\beta x^2 \psi^2 \]

The right-hand side contains a term \(\frac{1}{2}\beta\psi^2 x^2\) that is quadratic in \(x\). The left-hand side is at most linear in \(x\) (the term \(\psi' x\)). Since \(\frac{1}{2}\beta\psi^2 x^2\) cannot be matched by any constant or linear term in \(x\), the separation fails.

This demonstrates that geometric Brownian motion in the price variable \(S_t\) (where \(dS_t = \mu S_t\,dt + \sigma S_t\,dW_t\) gives \(b(S) = \mu S\) and \(a(S) = \sigma^2 S^2\)) is not an affine process. The quadratic dependence of the diffusion coefficient on the state variable violates the affine requirement that \(a(x)\) be linear in \(x\).

Exercise 6. For a \(d\)-dimensional affine process with no jumps (\(m_i = 0\) for all \(i\)), the functions \(F\) and \(R\) reduce to

\[ F(u) = \langle b_0, u \rangle + \tfrac{1}{2}\langle u, a_0 u \rangle, \qquad R_i(u) = \langle b_i, u \rangle + \tfrac{1}{2}\langle u, a_i u \rangle \]

Show that if all diffusion matrices \(a_i = 0\) (pure Gaussian case with no state-dependent volatility), then the Riccati system for \(\psi\) becomes a linear ODE system \(\psi'(\tau) = B\psi(\tau)\) for an appropriate matrix \(B\). Express \(B\) in terms of the drift parameters \(b_1, \ldots, b_d\).

Solution to Exercise 6

With \(a_i = 0\) for all \(i = 1, \ldots, d\) and no jumps, \(R_i(u)\) simplifies to:

\[ R_i(u) = \langle b_i, u \rangle + \frac{1}{2}\langle u, a_i u \rangle = \langle b_i, u \rangle \]

The \(i\)-th component of the Riccati system \(\psi' = R(\psi)\) is:

\[ \psi_i'(\tau) = R_i(\psi(\tau)) = \langle b_i, \psi(\tau) \rangle = \sum_{j=1}^{d} (b_i)_j \psi_j(\tau) \]

In vector form, this is:

\[ \psi'(\tau) = B\,\psi(\tau), \qquad \psi(0) = u \]

where \(B\) is the \(d \times d\) matrix whose \(i\)-th row is \(b_i^T\):

\[ B = \begin{pmatrix} b_1^T \\ b_2^T \\ \vdots \\ b_d^T \end{pmatrix} \]

Equivalently, \(B_{ij} = (b_i)_j\), where \((b_i)_j\) is the \(j\)-th component of the vector \(b_i\). The solution is \(\psi(\tau) = e^{B\tau}u\), a matrix exponential applied to the initial condition.

The quadratic terms \(\frac{1}{2}\langle u, a_i u \rangle\) in \(R_i\) are what make the Riccati system nonlinear. When \(a_i = 0\), these terms vanish, and the ODE for \(\psi\) becomes linear. This is the hallmark of the pure Gaussian (Ornstein-Uhlenbeck) case: the characteristic exponent is quadratic in \(u\) and the Riccati equations are exactly solvable via matrix exponentials.