Probability theory: given a calculation process $ X $ and a measure μ in the space of Skorohod, how can we prove that μ is the law of $ X $ up to the dependence of the initial value?

Leave

  • $ ( Omega, mathcal A, operatorname P) $ be a probability space
  • $ (E, d) $ be a compact locally separable complete metric space
  • $ D ([0infty)E):=left{x:[0infty)aEmidxtext{iscàdlàg}right}$[0infty)E):=left{x:[0infty)toEmidxtext{iscàdlàg}right}$[0infty)E):=left{x:[0infty)aEmidxtext{iscàdlàg}right}$[0infty)E):=left{x:[0infty)toEmidxtext{iscàdlàg}right}$ Be equipped with the Skorohod topology.
  • $ X: Omega a D ([0infty)E)$[0infty)E)$[0infty)E)$[0infty)E)$ be $ ( mathcal A, mathcal B (D ([0infty)E)))$[0infty)E)))$[0infty)E)))$[0infty)E)))$-measurable
  • $ mu $ be a measure of probability in $ mathcal B (D ([0infty)E))$[0infty)E))$[0infty)E))$[0infty)E))$

How can we formalize the condition of what? $ mu $ is "essentially the distribution of $ X $ Until the dependence of the initial value "?

As you probably do not know exactly what I mean, let me explain it in more detail: from $ E $ It is Polish, $ D ([0infty)E)$[0infty)E)$[0infty)E)$[0infty)E)$ It is also Polish and, therefore, there is a Markov core. $ kappa $ with fountain $ (E, mathcal B (E)) $ and objective $ (D ([0infty)E)mathcalB(D([0infty)E)))$[0infty)E)mathcalB(D([0infty)E)))$[0infty)E)mathcalB(D([0infty)E)))$[0infty)E)mathcalB(D([0infty)E)))$ with $$ operatorname P left[Xin Bmid X_0right]= kappa (X_0, B) ; ; ; text {almost safe for all} B in mathcal B (D ([0infty)E))Tag1$$[0infty)E))tag1$$[0infty)E))Tag1$$[0infty)E))tag1$$ Note that $$ operatorname P left[Xin Bright]= int operatorname P left[X_0in{rm d}x_0right] kappa (x_0, B) ; ; ; text {for all} B in mathcal B (D ([0infty)E))Tag2$$[0infty)E))tag2$$[0infty)E))Tag2$$[0infty)E))tag2$$ Now leave $ mu_0 $ be a measure of probability in $ mathcal B (E) $ Y $ mu: = mu_0 kappa $ be the composition of $ mu_0 $ Y $ kappa $. So $$ mu left ( left {x in D ([0infty)E):x(0)inB_0right}right)=mu_0(B_0);;;text{forall}B_0inmathcalB(E)tag3$$[0infty)E):x(0)inB_0right}right)=mu_0(B_0);;;text{forall}B_0inmathcalB(E)tag3$$[0infty)E):x(0)inB_0right}right)=mu_0(B_0);;;text{paratodos}B_0inmathcalB(E)tag3$$[0infty)E):x(0)inB_0right}right)=mu_0(B_0);;;text{forall}B_0inmathcalB(E)tag3$$ Y $$ mu (B) = int mu_0 ({ rm d} x_0) kappa (x_0, B) ; ; ; text {for all} B in mathcal B (D ([0infty)E))Tag4$$[0infty)E))tag4$$[0infty)E))Tag4$$[0infty)E))tag4$$

Yes $ mu $ it's the way $ (4) $, then we see from $ (2) $ that $ mu $ is "essentially the distribution of $ X $ Until the dependence on the initial value. "How can we formalize this?

Intuitively, $ mu $ Y $ operatorname P left[Xin;cdot;right]$ It must coincide in some kind of "trace" of $ mathcal B (D ([0infty)E)$[0infty)E)$[0infty)E)$[0infty)E)$. In particular, in $$ left {B in mathcal B ([0, infty), E): mu_0 ( pi_0 (B)) = operatorname P left[0, infty), E): mu_0 ( pi_0 (B)) = operatorname P left[0, infty), E): mu_0 ( pi_0 (B)) = operatorname P left[0,infty),E):mu_0(pi_0(B))=operatorname Pleft[X_0inpi_0(B)right] right }, tag5 $$ where $$ pi_t: D ([0infty)E)aE;;;xmapstox(t)Tag6$$[0infty)E)toE;;;xmapstox(t)tag6$$[0infty)E)aE;;;xmapstox(t)Tag6$$[0infty)E)toE;;;xmapstox(t)tag6$$ But it is $ pi_0 (B) en mathcal B (E) $ for all $ B en mathcal B ([0infty)E)$[0infty)E)$[0infty)E)$[0infty)E)$. We clearly know that $$ mathcal B ([0infty)E)supseteqsigma(pi_t:tge0)tag7$$[0infty)E)supseteqsigma(pi_t:tge0)tag7$$[0infty)E)supseteqsigma(pi_t:tge0)tag7$$[0infty)E)supseteqsigma(pi_t:tge0)tag7$$ (and by separability of $ E $ we even get equality).