Common Errors in Diffusion Generator and Kernel Calculations

Four Checks for Diffusion Calculations

Generator and transition-kernel calculations are short enough to invite small errors with large consequences. A missing factor of one-half changes the infinitesimal variance; confusing a diffusion matrix with its covariance matrix can lose cross-derivative terms; and a transition kernel started at one point is not automatically the law of a process with a random initial value. Stopping a process introduces another issue: an expectation identity proved at bounded times need not persist at an unbounded stopping time.

We work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a \(d\)-dimensional Brownian motion \(\mathbf{B}_t\). For an \(m\)-dimensional diffusion started at \(x\in\mathbb{R}^m\), write \(dX_t=b(X_t)\,dt+\Sigma(X_t)\,d\mathbf{B}_t\), where \(b\) takes values in \(\mathbb{R}^m\) and \(\Sigma\) is an \(m\times d\) matrix. In every stochastic integral below the integrand is progressively measurable and belongs to \(\mathcal{L}^2\): on each stated finite horizon \(T\), its squared norm has finite expectation when integrated over \([0,T]\). The matrix-integrand condition means that the sum of the expected integrals of the squared entries is finite.

Definition: The diffusion covariance matrix is \(a(x)=\Sigma(x)\Sigma(x)^{\mathsf T}\). For a twice continuously differentiable test function \(f:\mathbb{R}^m\to\mathbb{R}\), the diffusion generator is \(\mathcal{L}f(x)=\sum_i b_i(x)\partial_i f(x)+\tfrac12\sum_{i,j}a_{ij}(x)\partial_{ij}f(x)\), wherever the derivatives and expectations used to define the generator are valid.

The entries of \(a\), not the entries of \(\Sigma\), multiply the second derivatives. In particular, the double sum includes both \((i,j)\) and \((j,i)\). Since \(a\) is symmetric, an off-diagonal entry contributes twice inside the sum, and the overall factor one-half leaves one copy multiplying the mixed derivative.

Check the Factor and the Covariance Matrix

Worked Example: A Correlated Two-Dimensional Diffusion

Take \(b=(1,-1)^{\mathsf T}\) and \(\Sigma=\begin{pmatrix}1&0\\ \tfrac12&\tfrac{\sqrt3}{2}\end{pmatrix}\), and use \(f(x_1,x_2)=x_1x_2\). Find \(\mathcal{L}f\), then check it against the transition law.

First calculate the covariance matrix, rather than reading it off from \(\Sigma\):

a=\Sigma\Sigma^{\mathsf T} =\begin{pmatrix}1&0\\ \tfrac12&\tfrac{\sqrt3}{2}\end{pmatrix} \begin{pmatrix}1&\tfrac12\\ 0&\tfrac{\sqrt3}{2}\end{pmatrix} =\begin{pmatrix}1&\tfrac12\\ \tfrac12&1\end{pmatrix}.

The derivatives are \(f_1=x_2\), \(f_2=x_1\), \(f_{11}=f_{22}=0\), and \(f_{12}=f_{21}=1\). Therefore

\mathcal{L}f =x_2-x_1+\tfrac12\left(a_{12}f_{12}+a_{21}f_{21}\right) =x_2-x_1+\tfrac12.

For an independent check, start at \(x=(x_1,x_2)\). The coordinates at time \(t\) are \(X_t^1=x_1+t+B_t^1\) and \(X_t^2=x_2-t+\tfrac12B_t^1+\tfrac{\sqrt3}{2}B_t^2\), where \(B^1,B^2\) are independent standard Brownian coordinates. Their means are \(x_1+t\) and \(x_2-t\); their covariance is \(t/2\). Thus

\mathbb{E}_x[f(X_t)] =(x_1+t)(x_2-t)+\tfrac{t}{2} =x_1x_2+t(x_2-x_1)+\tfrac{t}{2}-t^2.

Subtract \(f(x)=x_1x_2\), divide by \(t\), and let \(t\downarrow0\): the result is \(x_2-x_1+\tfrac12\), as the generator calculation gives. In particular, replacing \(a\) by \(\Sigma\) would not even give a well-formed coefficient for the second-derivative sum: \(\Sigma\) maps Brownian coordinates to state coordinates, whereas \(a\) records the state-coordinate covariances.

Result: Suppose \(b\) and \(\Sigma\) are bounded and continuous, \(f\in C^2(\mathbb{R}^m)\) has bounded first and second derivatives, and \(X_0=x\). Then \(\displaystyle \lim_{t\downarrow0}\frac{\mathbb{E}_x[f(X_t)]-f(x)}{t}=\mathcal{L}f(x)\).

Proof. Apply Itô’s formula to \(f(X_t)\). Its first-order drift term is \(\sum_i f_i(X_s)b_i(X_s)\,ds\); its second-order correction is \(\tfrac12\sum_{i,j}f_{ij}(X_s)a_{ij}(X_s)\,ds\). Together these are \(\mathcal{L}f(X_s)\,ds\). The stochastic integral has integrand \(\nabla f(X_s)\Sigma(X_s)\), which is progressively measurable and in \(\mathcal{L}^2\) on \([0,T]\), because \(\nabla f\) and \(\Sigma\) are bounded. Its expectation is zero: the square-integrable Itô integral is a martingale, so its conditional expectation at time \(0\) is its value there, zero. Consequently, \(\mathbb{E}_x[f(X_t)]-f(x)=\int_0^t\mathbb{E}_x[\mathcal{L}f(X_s)]\,ds\). Continuity of \(X\), \(b\), \(\Sigma\), and the derivatives of \(f\) gives \(\mathcal{L}f(X_s)\to\mathcal{L}f(x)\) almost surely as \(s\downarrow0\). Boundedness gives dominated convergence, so the time average on the right tends to \(\mathcal{L}f(x)\). Dividing by \(t\) proves the claim. \(\square\)

This computation also explains the one-half factor: it is the second-order correction in Itô’s formula, paired with quadratic covariation \(dB_t^k\,dB_t^\ell=\mathbf{1}_{\{k=\ell\}}dt\). The ordinary first-derivative chain rule does not account for this quadratic variation.

A Kernel Is Conditional on Its Starting State

For a diffusion started deterministically at \(x\), its transition kernel at time \(t\) is \(P_t(x,A)=\mathbb{P}_x(X_t\in A)\), for measurable sets \(A\). The subscript indicates the law with \(X_0=x\); it is not an unconditional law that can be used unchanged when the initial state is random. For a random initial state, the conditional transition law given \(X_0\) is the kernel evaluated at that realized state. An unconditional probability then averages those conditional laws over the distribution of \(X_0\).

Worked Example: Brownian Motion with a Random Initial State

Let \(X_0\) equal \(0\) or \(2\), each with probability \(1/2\), independently of a standard Brownian motion \(B_t\), and let \(X_t=X_0+B_t\). Compare the conditional and unconditional second moments at \(t=1\).

For a fixed starting state \(x\), the transition law is normal with mean \(x\) and variance \(t\). Hence \(\mathbb{E}_x[X_1^2]=x^2+1\): the value is \(1\) when \(x=0\), and \(5\) when \(x=2\). Since both initial states have positive probability, averaging gives

\mathbb{E}[X_1^2] =\tfrac12(0^2+1)+\tfrac12(2^2+1) =\tfrac12+\tfrac52=3.

A mistaken calculation that uses only the kernel started at zero gives \(1\), which is the conditional moment \(\mathbb{E}[X_1^2\mid X_0=0]\), not the unconditional moment. As a second check, \(\mathbb{E}[X_1]=\mathbb{E}[X_0]+\mathbb{E}[B_1]=1\), and \(\operatorname{Var}(X_1)=\operatorname{Var}(X_0)+\operatorname{Var}(B_1)=1+1=2\); therefore \(\mathbb{E}[X_1^2]=1^2+2=3\), agreeing with the kernel mixture.

When \(X_0\) has a continuous distribution, the event \(X_0=x\) may have probability zero. The statement about conditioning is then interpreted using a regular conditional distribution, or equivalently by the family of laws \(\mathbb{P}_x\). One should not treat a zero-probability event as though it were an ordinary event on which conditional probabilities are defined by a ratio.

Stopping-Time Expectations Need Their Hypotheses

A bounded-stopping-time version of Dynkin’s formula follows directly from stopped Itô’s formula. Here is a sufficient form. Let \(X\) solve a diffusion SDE from \(x\), let \(f\in C^2\) have bounded first and second derivatives, and suppose the coefficients are bounded on the time interval \([0,T]\). If \(\tau\) is a stopping time with \(0\le\tau\le T\), then

\mathbb{E}_x[f(X_\tau)] =f(x)+\mathbb{E}_x\left[\int_0^\tau \mathcal{L}f(X_s)\,ds\right].

Proof. Itô’s formula for \(f(X_{t\wedge\tau})\) has drift \(\int_0^{t\wedge\tau}\mathcal{L}f(X_s)\,ds\) and stochastic term \(\int_0^t\mathbf{1}_{\{s\le\tau\}}\nabla f(X_s)\Sigma(X_s)\,d\mathbf{B}_s\). The integrand is progressively measurable and in \(\mathcal{L}^2\) on \([0,T]\), since the derivatives and coefficients are bounded and \(\mathbb{E}\int_0^T\|\mathbf{1}_{\{s\le\tau\}}\nabla f(X_s)\Sigma(X_s)\|^2ds\le C T\) for a finite constant \(C\). This Itô integral is a square-integrable martingale: for \(u\le t\), its increment from \(u\) to \(t\) has conditional expectation zero given \(\mathcal{F}_u\), first for simple adapted integrands by independent Brownian increments and then for \(\mathcal{L}^2\) integrands by the Itô isometry. Its expectation at \(T\) is therefore zero. Since \(T\wedge\tau=\tau\), taking expectations in the stopped formula gives the displayed identity. \(\square\)

Boundedness of \(\tau\) is a convenient sufficient condition, not a decoration. For an unbounded \(\tau\), one can first apply the formula at \(t\wedge\tau\), but passing to the limit requires additional control. For example, it is sufficient that \(f(X_{t\wedge\tau})\to f(X_\tau)\) in \(L^1\) and that \(\int_0^{t\wedge\tau}\mathcal{L}f(X_s)\,ds\) converges in \(L^1\) to \(\int_0^\tau\mathcal{L}f(X_s)\,ds\). Almost-sure convergence alone does not justify passing expectations through the limit.

Worked Example: A Hitting Time Where the Expectation Changes

Let \(B_0=0\) and let \(\tau=\inf\{t\ge0:B_t=1\}\), the first time standard Brownian motion hits \(1\). The function \(f(x)=x\) has \(f'(x)=1\), \(f''(x)=0\), and generator \(\mathcal{L}f=0\) for Brownian motion. A tempting but invalid calculation is \(\mathbb{E}[B_\tau]=B_0+\mathbb{E}\int_0^\tau0\,ds=0\).

For each finite \(t\), stopped Itô’s formula gives \(B_{t\wedge\tau}=\int_0^t\mathbf{1}_{\{s\le\tau\}}\,dB_s\). On \([0,t]\), this integrand is progressively measurable and in \(\mathcal{L}^2\), since \(\mathbb{E}\int_0^t\mathbf{1}_{\{s\le\tau\}}^2ds\le t\). Thus the integral is a square-integrable martingale with expectation zero, and \(\mathbb{E}[B_{t\wedge\tau}]=0\).

On the other hand, the reflection principle gives \(\mathbb{P}(\tau>t)=2\Phi(1/\sqrt{t})-1\), which tends to zero as \(t\to\infty\). Therefore \(\tau<\infty\) almost surely, and path continuity gives \(B_\tau=1\) almost surely. Hence \(\mathbb{E}[B_\tau]=1\), not zero. The stopped variables \(B_{t\wedge\tau}\) converge almost surely to \(1\) but cannot converge in \(L^1\): their expectations are all zero. In particular they are not uniformly integrable. The invalid step was passing from the bounded-time expectation identity to the unbounded hitting time without a convergence condition.

Common Mistakes and a Reliable Audit

Wrong calculation: for the correlated example above, one might write \(\mathcal{L}f=x_2-x_1+\tfrac12 f_{12}\), then report \(x_2-x_1+\tfrac14\), on the theory that the cross covariance is \(1/2\) and the generator has another factor \(1/2\). This counts the factor incorrectly. The double sum contains \(a_{12}f_{12}+a_{21}f_{21}=\tfrac12+\tfrac12=1\); multiplying that sum by one-half gives \(\tfrac12\), not \(\tfrac14\). The transition-moment check independently detects the error.

Before trusting a calculation, check which law it concerns, compute \(a=\Sigma\Sigma^{\mathsf T}\) explicitly, and expand the full second-derivative sum. At a stopping time, first ask whether it is bounded. If it is not, identify the integrability or uniform-integrability argument that permits the limit; the fact that a stopped identity holds for every finite time does not supply that argument.

Check Your Understanding

Use the stated assumptions carefully; distinguish fixed-starting-state expectations from unconditional ones.

For a two-dimensional diffusion with covariance matrix \(a\), write the generator’s mixed-derivative contribution when \(a_{12}=a_{21}=c\).
If \(\Sigma=\begin{pmatrix}2&0\\1&1\end{pmatrix}\), compute \(\Sigma\Sigma^{\mathsf T}\). Which matrix belongs in the generator?
For \(X_t=x+B_t\), compute \(\mathbb{E}_x[X_t^2]\). How would the answer change if \(X_0\) were random?
State the bounded-stopping-time condition in Dynkin’s formula and explain why the stochastic integral there has expectation zero.
In the hitting-time example, which convergence property fails when \(t\wedge\tau\) is replaced by \(\tau\)?