Given an SDE in a Hilbert Space $H$ and an underlying probability space $(\Omega, \mathcal{F}, P)$ with solution $(X_t)_{t\ge 0}$ and writing $X_t^x$ as the solution with initial condition $x$, define the following semigroup $(P_t)$ on the space of bounded Borel-measurable functions on $H$ via $$ (P_t\varphi)(x) = \mathbb{E}[ \varphi(X_t^x)]. $$ Next, define a semigroup $(P_t^*)$ on the space of all probability measures on $H$ via $$ (P_t^*\mu)(\varphi) = \mu(P_t\varphi), $$ where we use the notation $\mu(f) = \int_H f(x)\,d\mu(x)$.

Now for my question, if $X_t$ has law $\nu_t$, then the claim is that $P^*_{t}\nu_s = \nu_{t+s}$. I'm having trouble unwinding all of the definitions to see why this is true. But moreover, this seems like a standard procedure in defining semigroups. I'm not too familiar with subject, but is this somehow related to transition probabilities of a Markov process?

There are a lot of little things going on here that I'm somewhat familiar with, but I'm having a bit of trouble synthesizing to see the big picture.

  • $\begingroup$ What exactly do you mean by "$X_t$ has law $\nu_t$"? What is the initial distribution of $X_t$? (I take it that you are assuming that $(P_t)_t$ is a semigroup... because, in general, the solution to an SDE does not need to be Markovian...) $\endgroup$ – saz Mar 2 '18 at 6:46
  • $\begingroup$ @saz The author of the text I'm reading uses the term "law" for "probability distribution." So here I have a family of distributions. And indeed, there is a theorem that proves the Markovian property for my specific SDE so for the purposes of this post, assume $P_t$ is a semigroup. The initial condition to my SDE, is some measurable function taking values in $H$, so whatever it's probability distribution is would be the initial one (I suppose simply $P(X_0\in A)$). $\endgroup$ – Curious Mar 4 '18 at 17:24

I ended up figuring it out. We show it for $s=0$ and then use the semigroup property to conclude. First, by changing variables, we note if $X$ has distribution $\mu$, then with my notation above, $\mu(f) = E[f(X)]$. With this, for $f\in C_b(H)$ (i.e. continuous bounded functions on $H$), we see $$ (P^*_tf)(\nu_0) = \nu_0(P_tf)=E[(P_tf)(X_0)] = (P_0(P_tf))=P_tf=E[f(X_t)] = \nu_t(f), $$ which gives $P_t^* \nu_0 = \nu_t$.

We then conclude $P^*_t \nu_s = P^*_tP^*_s \nu_0 = P^*_{t+s}\nu_0 = \nu_{t+s}$.

  • 1
    $\begingroup$ I doubt that the third "=" in your first calculation holds. Note that $P_0$ is a mapping from $H$ to $H$; in particular $P_0(P_t f)$ is a (non-trivial) mapping $H \to H$. On the other hand, $E[(P_t f)(X_0)]$ is just a fixed number, and this indicates that there is something off. The problem is the initial distribution. The function $P_0 g(x)= \mathbb{E}g(X_0^x)=g(x)$ gives the expected value for the initial distribution $\delta_x$; you are, however, interested in the expectation $\mathbb{E}g(X_0^{\nu_0})$ (...for $g=P_t f$). Clearly, $\mathbb{E}g(X_0^{\nu_0})$ does not equal $P_0 g$ $\endgroup$ – saz Mar 7 '18 at 14:36
  • $\begingroup$ ... but that's what you are claiming in the third equality. $\endgroup$ – saz Mar 7 '18 at 14:37
  • $\begingroup$ ah yes, you're right. Then I'm back to square one, not quite sure how to see it. I mean, it seems set up in such a way that it works. Do you have any hints? $\endgroup$ – Curious Mar 7 '18 at 21:36
  • $\begingroup$ Do you know whether the solution to the SDE (which you are considering) is unique (for any initial distribution)? If so, then $$\mathbb{E}g(X_t^{\nu}) = \int \mathbb{E}(g(X_t^x)) \, \nu(dx);$$ using this identity and the fact that $(P_t)_t$ is a semigroup, it is then not difficult to prove the assertion. $\endgroup$ – saz Mar 7 '18 at 21:43
  • $\begingroup$ @saz thank you, I'll think about it some more. Just to make sure I understand your notation, when you write $X_t^{\nu}$, you mean that the initial condition (in the superscript) has distribution $\nu$, correct? $\endgroup$ – Curious Mar 7 '18 at 21:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.