Assume a process with Itô dynamics of the generic form $$dX_t=\mu(t,X_t)dt+\sigma(t,X_t)dW_t$$

and let $f:\mathbb{R}\to\mathbb{R}$ be borel-measurable. Is the following function smooth ? $$g(t,x)=\mathbb{E}[f(X_T)|\mathcal{F}_t]$$

I remember comming upon the proof of above once but I cannot find it any longer.


This depends on what you assume of $\mu$, $\sigma$, and $f$. I'll provide a few examples. Let's assume $f$ is bounded throughout, since what I'll say below is still true in the unbounded case under suitable growth assumptions.

First, as long the SDE is well-posed (in the sense of weak existence and uniqueness in law), then the solution is a Feller process (this is somewhere in Stroock and Varadhan's book), and your function $g$ is continuous in $x$ whenever $f$ is continuous. If $f \in C^2$, then in fact $g$ is in $C^{1,2}$ (see Theorem 24.1 of Kallenberg's "Foundations of modern probability").

If you're interested in higher order derivatives, consult the vast literature on stochastic flows of diffeomorphisms. For example, if $\mu = \mu(x)$ and $\sigma = \sigma(x)$ are time-homogeneous and $C^\infty$ with bounded first derivative, then $g(t,\cdot)$inherits derivatives from $f$; i.e. whenever $f \in C^k$ then $g(t,\cdot) \in C^k$ for each $t$. (See Theorem V.13.8 of Rogers & Williams, "Diffusions, Markov processes, and martingales" vol 2.)

Finally, if you are really only willing to assume $f$ is measurable, the story is a bit more complicated. You'll want uniformly nondegenerate volatility, i.e. $\sigma\sigma^\top \ge \delta I$ for some $\delta > 0$, so that the noise helps you recover some smoothness. In this case, at least if $\mu = \mu(x)$ and $\sigma = \sigma(x)$ are time-homogeneous, then the map from the initial condition $x$ to $\text{Law}(X^x_t)$ (the distribution of the SDE solution at some time $t > 0$) is typically continuous in total variation. See sections 11.3 and 11.4 of Stroock and Varadhan's book.

  • $\begingroup$ May I know by Feller process, do you mean the Feller property that if $f$ is continuous and vanishes at infinity, then $g(x)=E^x[f(X_t)]$ is continuous and vanishes at infinity? I am particularly interested why $g(x)$ vanishes at infinity. May you suggest some references about it? $\endgroup$ – John May 10 '16 at 17:47

I assume here the $x$ variable is the initial condition of you process? Here is a partial answer: if $(\mathcal F_t)$ is the brownian filtration, then by Itô's martingale representation theorem for any $x$ there is a predictable process $(H_t^x)$ such that $g(t,x)=\mathbb E(f(X_T))+ \int_0^t H^x_s dW_s$. So in that case there is at least continuity in $t$.

  • $\begingroup$ okey - this helps somewhat for we have reduced the problem to showing smoothness of $\int^t_0 H^x_s dW_s$ $\endgroup$ – Boldwing Mar 28 '14 at 8:53

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