From a Diffusion to an Equation for Its Density
A diffusion describes the motion of a random state, while the Fokker-Planck equation describes how the distribution of that state changes over time. Its drift term records the systematic local motion; its diffusion term records the spreading caused by random fluctuations. When volatility depends on the state, that term differentiates the product of the local variance and the density—not the density alone.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Consider a nonexplosive, continuous, adapted solution of the Ito stochastic differential equation \(dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t\), where \(\mu\) is the drift coefficient and \(\sigma\) is the diffusion coefficient. The coefficients here are time-independent. Write \(a(x)=\sigma(x)^2\) for the local variance. The equation is interpreted in the Ito sense; a Stratonovich equation \(dX_t=\mu(X_t)\,dt+\sigma(X_t)\circ dB_t\) has Ito drift \(\mu+\tfrac12\sigma\sigma'\), when the coefficients are differentiable, and so generally gives a different forward equation.
For a twice continuously differentiable test function \(f\), the diffusion generator is \(\mathcal{L}f(x)=\mu(x)f'(x)+\tfrac12a(x)f''(x)\). Suppose \(f\) and the coefficients satisfy the integrability conditions needed below: on each finite time interval \([0,T]\), \(\mathbb{E}\int_0^T|\mathcal{L}f(X_s)|\,ds<\infty\) and \(\mathbb{E}\int_0^T\sigma(X_s)^2 f'(X_s)^2\,ds<\infty\). In particular, the Ito integrand \(H_s=\sigma(X_s)f'(X_s)\) is progressively measurable and belongs to \(\mathcal{L}^2\), the class for which \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). These conditions make \(\int_0^t H_s\,dB_s\) a square-integrable martingale, so its expectation is zero.
The Weak Equation and Its Derivation
Proof. Apply Ito’s formula to \(f(X_t)\). Its first derivative contributes \(f'(X_s)\mu(X_s)\,ds\), and its second derivative contributes the explicit quadratic-variation correction \(\tfrac12 f''(X_s)\sigma(X_s)^2\,ds\), because \(dB_s\,dB_s=ds\), \(dB_s\,ds=0\), and \(ds\,ds=0\). Thus \(f(X_t)=f(X_0)+\int_0^t\mathcal{L}f(X_s)\,ds+\int_0^t\sigma(X_s)f'(X_s)\,dB_s\). The last integral is a square-integrable martingale: its integrand is progressively measurable and in \(\mathcal{L}^2\), and the Ito integral has conditional mean-zero increments. Taking expectations therefore removes that integral. The assumed drift integrability permits taking expectation inside the time integral. Using the law of \(X_s\) to write \(\mathbb{E}[\mathcal{L}f(X_s)]=\int\mathcal{L}f(x)p(s,x)\,dx\) gives the stated identity.
To identify the differential equation, integrate by parts in the spatial variable against a compactly supported \(f\). There are no boundary terms because \(f\) has compact support, and \(\int \mathcal{L}f(x)p(s,x)\,dx=\int f(x)\{-\partial_x(\mu p)(s,x)+\tfrac12\partial_{xx}(ap)(s,x)\}\,dx\), where derivatives may be distributional. The integrated identity says that the time change of the pairing with \(f\) is this pairing with the displayed right-hand side. Equivalently, against every smooth compactly supported test function of both time and space, \(\partial_t p=-\partial_x(\mu p)+\tfrac12\partial_{xx}(ap)\). This is precisely the distributional Fokker-Planck equation. \(\square\)
This argument gives an integrated identity without claiming that \(\partial_t p(t,x)\) exists pointwise at every time. Under just the stated integrability, the pairing with each fixed test function is absolutely continuous in time, and its derivative identity holds almost everywhere. Additional regularity is needed to assert a classical, pointwise PDE.
How the Two Terms Act
The weak equation makes the roles of the coefficients precise. The generator applies one derivative to \(f\) for drift and two derivatives for diffusion. Moving those derivatives onto the density reverses the sign of the first derivative and leaves the second derivative’s sign unchanged. The result is the forward, or adjoint, operator.
The drift term is not in general \(-\mu\,\partial_xp\): it differentiates the whole product \(\mu p\). Similarly, for state-dependent volatility the diffusion term differentiates \(a p=\sigma^2p\). This product structure accounts for spatial variation in the coefficients as well as variation in the density.
Worked Example: Constant Drift and Volatility
Let \(X_t=x_0+2t+3B_t\), with deterministic \(x_0\). Its density for \(t>0\) is \(p(t,x)=\dfrac{1}{3\sqrt{2\pi t}}\exp\!\left(-\dfrac{y^2}{18t}\right)\), where \(y=x-x_0-2t\). Verify the forward equation.
Here \(\mu=2\) and \(a=9\), so the equation is \(p_t=-2p_x+\tfrac92p_{xx}\). Differentiating in \(x\) gives \(p_x=-\dfrac{y}{9t}p\) and \(p_{xx}=(\dfrac{y^2}{81t^2}-\dfrac{1}{9t})p\). For the time derivative, \(y_t=-2\); differentiating both the prefactor and exponent gives \(p_t=(-\dfrac{1}{2t}+\dfrac{2y}{9t}+\dfrac{y^2}{18t^2})p\). The right-hand side is \(-2p_x+\tfrac92p_{xx}=(\dfrac{2y}{9t}+\dfrac{y^2}{18t^2}-\dfrac{1}{2t})p\), exactly \(p_t\).
Check: The density is normal with mean \(x_0+2t\) and variance \(9t\), matching \(\mathbb{E}[X_t]=x_0+2t\) and \(\operatorname{Var}(X_t)=9t\) computed directly from \(X_t=x_0+2t+3B_t\).
Worked Example: Expanding a State-Dependent Diffusion Term
Consider \(dX_t=(1-X_t)\,dt+\sqrt{1+X_t^2}\,dB_t\). The drift is \(\mu(x)=1-x\), and the local variance is \(a(x)=1+x^2\). Write the Fokker-Planck equation in expanded form.
Start with the divergence form: \(\partial_t p=-\partial_x((1-x)p)+\tfrac12\partial_{xx}((1+x^2)p)\). The drift contribution is \(p-(1-x)p_x\), since \(\partial_x((1-x)p)=-p+(1-x)p_x\). For the diffusion contribution, applying the product rule twice gives \(\tfrac12\partial_{xx}((1+x^2)p)=\tfrac12(1+x^2)p_{xx}+2xp_x+p\). Adding the terms yields \(\partial_t p=\tfrac12(1+x^2)p_{xx}+(-1+3x)p_x+2p\).
Check: At \(x=0\), the expanded operator has coefficients \(\tfrac12\) on \(p_{xx}\), \(-1\) on \(p_x\), and \(2\) on \(p\). The divergence form gives the same values there: its drift contribution is \(p-p_x\), and its diffusion contribution is \(\tfrac12p_{xx}+p\). This check also shows why replacing the diffusion term with \(\tfrac12(1+x^2)p_{xx}\) alone would omit the product-rule contributions.
Worked Example: Verifying the Geometric Brownian Motion Density
Let \(dS_t=\alpha S_t\,dt+\beta S_t\,dB_t\), with \(S_0=s_0>0\) and \(\beta>0\). Its density is supported on \(x>0\). Set \(m=\alpha-\tfrac12\beta^2\), \(y=\log(x/s_0)\), and \(q(t,y)=\dfrac{1}{\beta\sqrt{2\pi t}}\exp(-\dfrac{(y-mt)^2}{2\beta^2t})\). The density of \(S_t\) is \(p(t,x)=q(t,\log(x/s_0))/x\). Verify the equation on \(x>0\).
For this diffusion, the forward equation is \(p_t=-\partial_x(\alpha xp)+\tfrac12\beta^2\partial_{xx}(x^2p)\). Since \(q=xp\), the left side multiplied by \(x\) is \(q_t\). On the right, use \(xp=q\) and \(x^2p=xq\): \(x[-\alpha q_x+\tfrac12\beta^2(xq)_{xx}]\). The identities \(xq_x=q_y\) and \(x^2q_{xx}=q_{yy}-q_y\) turn this expression into \((\tfrac12\beta^2-\alpha)q_y+\tfrac12\beta^2q_{yy}=-m q_y+\tfrac12\beta^2q_{yy}\).
Writing \(z=y-mt\), direct differentiation of the displayed Gaussian gives \(q_y=-\dfrac{z}{\beta^2t}q\), \(q_{yy}=(\dfrac{z^2}{\beta^4t^2}-\dfrac{1}{\beta^2t})q\), and \(q_t=(-\dfrac{1}{2t}+\dfrac{mz}{\beta^2t}+\dfrac{z^2}{2\beta^2t^2})q\). Substitution yields \(-m q_y+\tfrac12\beta^2q_{yy}=(\dfrac{mz}{\beta^2t}+\dfrac{z^2}{2\beta^2t^2}-\dfrac{1}{2t})q=q_t\), which verifies the original forward equation.
Check: Since \(y\) is normal with mean \(mt\) and variance \(\beta^2t\), the density gives \(\mathbb{E}[S_t]=s_0\mathbb{E}[e^y]=s_0e^{mt+\beta^2t/2}=s_0e^{\alpha t}\). Ito’s formula applied to \(S_t\) gives drift \(\alpha S_t\,dt\), consistent with this mean.
Weak Versus Classical Solutions
The weak formulation is useful even when a density is not twice differentiable or the derivatives in the PDE are not ordinary functions. It says exactly how the law evolves when tested against smooth functions. If \(p\), \(\mu p\), and \(ap\) have sufficient regularity, the distributional identity can be read pointwise; otherwise, the weak identity is the appropriate meaning of the equation.
Nor does the existence of an SDE solution by itself guarantee that its law has a density at every time. Degenerate diffusion, in particular, can leave mass concentrated at points. The equation for a law can still be expressed weakly using probability measures, but writing it as a density equation requires a density. At \(t=0\), the initial law may itself be a point mass even when the law has a density for every positive time.
Common Pitfall
A frequent error is to write the state-dependent diffusion term as \(\tfrac12\sigma^2(x)p_{xx}\). That expression drops derivatives of \(\sigma^2\). For example, with \(\sigma^2(x)=1+x^2\), \(\tfrac12\partial_{xx}((1+x^2)p)=\tfrac12(1+x^2)p_{xx}+2xp_x+p\), not merely \(\tfrac12(1+x^2)p_{xx}\). The correction follows from applying the product rule twice. The diffusion coefficient in the SDE is \(\sigma\), whereas the coefficient multiplying the second derivative inside the forward operator is the local variance \(\sigma^2\).
Check Your Understanding
Use the weak equation and the product rule to check your answers.
- For drift \(\mu(x)=c\) and constant diffusion coefficient \(\sigma(x)=v\), write the forward equation and identify its local variance.
- Expand \(\tfrac12\partial_{xx}((2+x)p)\) and state which terms would be lost by differentiating \(p\) alone.
- Explain why the theorem gives an integrated identity for every time even when its derivative form is known only almost everywhere.
- For \(dX_t=-X_t\,dt+2\,dB_t\), write the Fokker-Planck equation in divergence form.
- Why must the second derivative in the state-dependent diffusion term act on \(\sigma^2p\), rather than on \(p\) alone?