# 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?

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• $$( Omega, mathcal A, operatorname P)$$ 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).