Time-Dependent Coefficients in the Local Volatility Fokker-Planck Equation

# 1D Fokker-Planck derivation for Local Volatility model


# 1D Fokker-Planck derivation for Local Volatility model












In the calibration of a local volatility model with diffusion $dX_t = b(X_t) dt + \sigma(t, X_t)dW_t$, one relies on the Dupire equation which itself derives from the Fokker-Planck equation $$\partial_t p(x, t) = - \partial_x[b(x)p(x, t)] + \frac{1}{2}\partial_{xx}^2[\sigma^2(t, x)p(x, t)]$$ My question concerns this Fokker-Planck equation. In all resources I could find, the hypothesis from which this equation derives is that both $b$ and $\sigma$ are functions of $x$ only. But obviously, we need to allow $\sigma$ to also depend on $t$.

So my question is the following: is the proof of Fokker-Planck in the case where $b$ and $sigma$ only depend on $x$ easily extended to the case where $\sigma$ can also depend on $t$?

======================================================================

Additionnaly, I wrote the following proof, how can it be generalized?

Consider the infinitesimal generator $\mathcal{L} f(x)$, defined as \begin{equation} \mathcal{L} f(x) := b(x) \partial_x f(x) + \frac{1}{2}\sigma^2(x)\partial_{xx}^2f(x) \end{equation} From Itô's lemma, one notices that \begin{align*} &f(X_t) = f(x) + \int_0^t \partial_x f(X_s) dX_s + \frac{1}{2}\int_0^t \partial_{xx}^2 f(X_s) d\langle X, X\rangle_s\\ &= f(x) + \int_0^t b(X_s) \partial_x f(X_s) + \frac{1}{2} \sigma^2(X_s)\partial_{xx}^2f(X_s)ds + \int_0^t \sigma(X_s) \partial_x f(X_s) dW_s\\ &= f(x) + \int_0^t \mathcal{L} f(X_s)ds + M_t \end{align*} with $M_t$ a local martingale. Using the above formule for times $t$ and $t+s$, taking the expectation and substracting leads to \begin{equation} \mathbb{E}[f(X_{t+s}] - \mathbb{E}[f(X_t)] = \mathbb{E}\left[\int_{t}^{t+s}\mathcal{L} f(X_r)dr\right] \end{equation} Taking $s\to 0$, we conclude that $\partial_t \mathbb{E}[f(X_t)] = \mathbb{E}[\mathcal{L} f(X_t)]$.\ Letting $(p_t(x, \cdot))_{t\geq 0, x\in \mathbb{R}}$ denote the transition kernel of $(X_t)$, we can introduce a semi-group operator: \begin{equation} P_tf(y) := \int_x f(x) p_t(x, dy) \end{equation} Because a transition kernel verifies by definition the Chapman-Kolmogorov equation \begin{equation} p_t(x, A) = \int_y p_s(x, dy) p_{t-s}(y, A)\quad \forall x\in \mathbb{R},\quad \forall 0\leq s \leq t \end{equation} and from the fact that $(X_t)$ is Markovian, we conclude that $\mathbb{E}[h(X_t)\mid \mathcal{F}_s] = P_{t-s}h(X_s)$ for any almost surely bounded measurable function $h$, so that \begin{equation} \label{eq:penultian_fokker_planck} \partial_t \mathbb{E}[f(X_t)] = \mathbb{E}[\mathcal{L} f(X_t)] \quad \Leftrightarrow \quad \partial_t P_tf(x) = P_t \mathcal{L} f(x) \end{equation} because $X_0 = x$ almost surely.\ Now, we can introduce the adjoint of $\mathcal{L}$, denoted $\mathcal{L}^*$ and defined by \begin{equation} \mathcal{L}^*f(x) := - \partial_x(bf)(x) + \frac{1}{2}\partial_{xx}^2(\sigma^2f)(x) \end{equation} We have \begin{equation} \int_x f(x)\mathcal{L} g(x)dx = \int_x \mathcal{L}^* f(x)g(x)dx \end{equation} for any two infinitely differentiable functions $f$ and $g$ with compact support. In particular, we can can write in terms of density: $\partial_t p_t(x,y) = \mathcal{L}^*p_t(x,y)$. Indeed, with $p_t(x, dy) = p_t(x, y) dy$: \begin{equation} P_t \mathcal{L} f(x) = \int_y \mathcal{L} f(y) p_t(x, y)dy = \int_y f(y) ^*p_t(x, y)dy \end{equation} and \begin{equation} \partial_t P_tf(x) = \partial_t \int_y f(y) p_t(x, y)dy = \int_y f(y) \partial_t p_t(x, y)dy \end{equation} i.e $\partial_t p_t(x,y) = \mathcal{L}^*p_t(x,y)$. Replacing by the definition of $\mathcal{L}^*$ allows us to conclude.

## Answer by Frido (score 1, accepted)

https://quant.stackexchange.com/a/81254

I can't quite follow your proof (maybe because of notation), which doesn't mean that it's wrong. To answer your question: yes it holds if $b,\sigma$ depend on $t$ as well.

I'll give a slightly simpler proof below. If you're proficient in analysis you can fill in the technical conditions yourself (or otherwise look them up or just accept them).

First of all let $F: \mathbb R \times [0,T] \to \mathbb R$ be a test function such that $\forall y \in \mathbb R$ $F(y, T) = F(y,t) = 0$.

Also, let $p(y,u|x,t)$ be the probability density with $(x,t)$ the backward variables and $(y,u)$ the forward variables

By It$\hat{\text{o}}$'s lemma $$ F(X_T,T) = F(X_t,t) + \int_t^T \left( \partial_u + \mathcal{L} \right)F(X_u,u) du + \int_t^T (\cdot) dW_u. $$

Now take expectations and using the property of the test function $$ \int_0^\infty \int_t^T \left( \partial_u + \mathcal{L} \right)F(y,u) p(y,u | x,t) \, du \, dy = 0 $$ Carrying out integration by parts wrt $u$ for $\partial_u$ and wrt to $y$ for $\mathcal L$ gives $$ \int_0^\infty \int_t^T F(y,u) \left( - \partial_u + \mathcal{L^*} \right) p(y,u | x,t) \, du \, dy = 0 $$ Since this is valid for an arbitrary test function $F$ $$ \partial_T p(K,T|x,t) = \mathcal{L^*} p(K,T|x,t) $$ where I have switched to the notation $(K,T)$ for the forward variables, as those are usually used in the context of local volatility.

As you can see, nowhere in the proof is it necessary to require that $b$ and/or $\sigma$ are independent of $t$.