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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.

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  • $\begingroup$ 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
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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 $P^*_t \nu_s = P^*_tP^*_s \nu_0 = P^*_{t+s}\nu_0 = \nu_{t+s}$.

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    $\begingroup$ 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

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