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From: Richard Hayden on 1 Jun 2010 11:25
Hi,

I was wondering about a simple generalisation of the well-known
Dynkin's formula for Markov processes.

I'm interested in a discrete-state continuous-time Markov process (not
sure the discrete-state bit matters though), say X(t). Denote the
state-space by X and the infinitesimal generator (matrix) by A. Let
also T <= S be stopping times. Then does it hold that (for suitably
nice f : X -> R):

E[ f(X(S)) | F_T ] - f(X(T)) = E[ \int_T^S Af (X(s)) ds | F_T ]

where F_T is the sigma-algebra where A \in F_T iff A \cap {T <= t} \in
F_t, and F_t is the natural filtration of X(t).

If this is true, I'd be very grateful if someone could show me a
careful proof.

Best regards and many thanks,

Richard.
From: ArtflDodgr on 3 Jun 2010 13:56
In article
<845a9b69-8cf9-4294-b2d5-79dcd9e6bd87(a)d12g2000vbr.googlegroups.com>,
Richard Hayden <r.hayden(a)gmail.com> wrote:

> Hi,
>
> I was wondering about a simple generalisation of the well-known
> Dynkin's formula for Markov processes.
>
> I'm interested in a discrete-state continuous-time Markov process (not
> sure the discrete-state bit matters though), say X(t). Denote the
> state-space by X and the infinitesimal generator (matrix) by A. Let
> also T <= S be stopping times. Then does it hold that (for suitably
> nice f : X -> R):
>
> E[ f(X(S)) | F_T ] - f(X(T)) = E[ \int_T^S Af (X(s)) ds | F_T ]
>
> where F_T is the sigma-algebra where A \in F_T iff A \cap {T <= t} \in
> F_t, and F_t is the natural filtration of X(t).
>
> If this is true, I'd be very grateful if someone could show me a
> careful proof.

The process M(t) = f(X(t)) - int_0^t Af(X(s)) ds is a martingale.
Under conditions permitting the application of the Optional Stopping
Theorem, we have

E[ M(S) | F_T] = M(T),

which is your identity slightly rearranged.

--
A.
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