SDE of conditional expectation

Question

Assume I have an SDE of the form $$ dX_t = a(X_t, t) dt + b(X_t, t) dW_t, X_0 = x_0. $$

Let now with $T>t$ denote $f(X_t)$ the conditional expectation of $X_T$, $$ f(X_t) = E[X_T|X_t] $$ My questions:

1. Is $f(\cdot )$ an Ito-diffusion? If no, why? If yes, always?
2. What form does $df(X_t)$ have? I know this seems to be just some application of Itos lemma but I got a bit confused with the derivative of the conditional expectation.

My steps so far: with the martingale representation theorem I get a representation of $f$ as the sum of an expectation and an Ito integral. I know have an Ito diffusion constant which I can’t fully link to $X_t$. I’ve seen that I can represent that diffusion term with the Clark-Ocone-theorem but before I went down that rabbit hole I wanted to check whether I am really on the right path here. I also know that I can represent the conditional expectation as an integral over the conditional density, here however I was afraid that it will be a lot more difficult to reconstruct a more explicit link to $a$ and $b$.

In the optimal case I want to represent $df(X_t)$ as functions of $a$ and $b$. I have the feeling it’s simpler if those are differentiable for the beginning, maybe if I get the basic mechanics going I can continue by myself. My main issue right now is that I don’t know what to do about $\frac{d}{dx} f(x)$ and how to write down a representation for it.

Possible that this is a simple question - I’d appreciate it if you’d help me a bit with a formal derivation there.

Answer 1

Let $f(t,x) = \mathbb{E}(X_T|X_t=x)$ for any $(t,x) \in [0, T]\times \mathbb{R}$. By Feynman-Kac it follows that the Kolmogorov-Backward PDE is satisfied: $\partial_t f+a(t, x) \partial_x f+ 0.5 b(t,x)^2 \partial_{xx}^2 f = 0$ for $(t,x)\in[0,T)\times \mathbb{R}$ and $f(T,x)=x$ for all $x$. If you apply Ito's formula to $Y_t = f(t, X_t)$ then you get $dY_t = \partial_x u(t, X_t) b(t, X_t) dW_t$ using the PDE. Thus $Y_t = f(t, X_t)$ is an Ito process, driftless and a martingale provided $a$ and $b$ are nice enough. If the solution of the Backward-PDE has a closed form, you can compute $\partial_x u(t,x)$ exactly.

By the way, many authors use the term diffusion to refer to coefficients that are time-independent $a(t,x)=a(x)$ and $b(t,x)=b(x)$ and say Ito processes for the more general case.

Answer 2

For details see Using Markov Property in solving PDE/SDE.

To be clear the deterministic quantity

$$f(x,t):=E[F(X_T)|X_t=x]$$

is different from the random quantity

$$Y_{t}:=E[F(X_T)|\sigma(X_t)]$$

which is equal to

$$f(X_{t},t):=E[F(X_T)|\mathcal{F}_t]$$ by the Markov property of SDEs (see Schilling 21.24 Corollary).

The $f(x,t)$ is a deterministic solution to the backward equation eg. as explained Lesson 5, SDE and PDE and the random $f(X_{t},t)$ is a martingale (as shown in Using Markov Property in solving PDE/SDE)

The random quantity $Z_{t}=f(X_{t},t)$ does not necessarily satisfy a diffusion, it simply satisfies the follow stochastic pde

$$f(X_{t},t)=\int^{t} f_{x}(X_{s},s) b((X_{s},s)) dW_{s},$$

so it is not an Itô diffusion where the only independent variable is time $t$.

For this to return to being an Ito diffusion you would need a relation

$$f_{x}(X_{s},s)=g(f(X_{s},s))$$

for some function $g$, which is generally not true. For example it fails even for Brownian motion where $f(x,t)$ is in terms of Gaussian since the backward-equation is $f_{t}+\frac{1}{2}\Delta f=0$ as done here

$$f(x,t)=E[X_{T}|X_{t}=x]=\int_{\mathbb{R}}y\frac{1}{\sqrt{2\pi (T-t)}}e^{-\frac{(y-x)^{2}}{2(T-t)}}dy.$$