2
$\begingroup$

Let $X_t$ be a semi-martingale and $H_t$ be a predictable process and $g$ be a measurable bijective function with measurable inverse. Does there exist a function $f(h,x)$ satisfying $$ \int_0^Tf(H_t,X_t) dg(X_t) = \int_0^TH_t dX_t? $$

If not what conditions do we need for that to hold?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ When $X_t$ is a martingale you would expect $g(X_t)$ must also be. That must constrain g a lot. $\endgroup$
    – user83457
    Commented Oct 18, 2016 at 15:46
  • $\begingroup$ I don't want $X_t$ to be a strict martingale in general and yes g will be very particular, in general I've seen this. $\endgroup$
    – ABIM
    Commented Oct 18, 2016 at 15:55

1 Answer 1

Reset to default
2
$\begingroup$

The LHS might not make sense if g is very irregular.

Consider, pth-Holder function $g(x)=x^{p}$ for $0<p<1$ and $\frac{1}{p}$ odd integer and Brownian motion $X_{t}=B_{t}$, then the process $g(X_{t})= (B_{t})^{p}$ for $p<\frac{1}{3}$ is not even in the usual regime for rough integrals (Uniqueness of solutions of Young differential equations). So we would also need to formulate the LHS in some way.

Generally, we also need $C^2$/Convex in order to apply change of variables see here https://almostsuremath.com/2020/10/12/the-ito-tanaka-meyer-formula/

In even weaker settings of rough paths, there is still an Ito formula (In Proposition 6.9 "Rough Path Theory" by Andrew L. Allan).

enter image description here

Even for just Brownian motion for the Ito-formulation, in most cases of $g$ there will be a drift-term (if there was a formula with no-drift term, by taking expectation or using Ito-isometry we would likely get contradictions to the Ito formula).

But if you interpet the RHS as the Stratonovich-formulation there is the regular chain rule

$$ g'(W_t) \circ \mathrm{d} W_t = dg(W_t),$$

and so one can use the above Ito-formulas to get something close to the OP statement (see here for semimartingales and Stratonovich).

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks for posting the comment on this old question. Nevertheless, even with added regularity and by enriching the process (as in Andy's coursenotes) this isn't really an answer to the question. $\endgroup$
    – ABIM
    Commented Dec 12, 2022 at 4:01
  • 1
    $\begingroup$ For such general g and/or Ito formulation, it will generally be false. But if you interpret the RHS in the Stratonovich-formulation, then there is chain rule. $\endgroup$
    – Thomas Kojar
    Commented Dec 12, 2022 at 5:08

You must log in to answer this question.