# Markov process and transition semigroup

I read in my lecture notes an introduction chapter about Markov process. Here is the setting:

Let $(\Omega,\mathcal{F},P)$ a probability space, $(S,\mathcal{S})$ a measurable space and $X=(X_t)$ a S-valued Markov process on $(\Omega,\mathcal{F},P)$ with transition semigroup $K=(K_t)$. Let $\nu$ be the law on $\mathcal{S}$ of $X_0$ under $P$ (the initial distribution), i.e.

$$\nu(A):=P(X_0^{-1}(A))$$

for all $A\in \mathcal{S}$. And by $\mathbb{P}_\nu$ we denote the distribution on $S^{[0,\infty)}$ of $X$ under $P$. Moreover on the space $S^{[0,\infty)}$ we define the coordinate process $Y=(Y_t)$ as, $Y_t:S^{[0,\infty)}\to S$, $\omega\mapsto Y_t(\omega):=\omega(t)$, and the $\sigma$-algebra $\mathcal{S}^{[0,\infty)}:=\sigma(Y_s;s\ge0)$.

There is a sentence in my lectures notes which seems to me not that obvious. They say: The law of $X$ under $P$ is the same as the law of $Y$ on $S^{[0,\infty)}$ under $\mathbb{P}_\nu$, i.e.

$$P(X_{t_1}\in A_1,\dots,X_{t_n}\in A_n)=\mathbb{P}_\nu(Y_{t_1}\in A_1,\dots,Y_{t_n}\in A_n)$$

for $A_i\in \mathcal{S}$.

Unfortunately for me it is not that obvious. Therefore a short clarification would be appreciated.

cheers

math

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The displayed equation at the end of your post is not equivalent to the fact that the processes $X$ and $Y$ are indentically distributed. Likewise, the displayed equation defining $\nu$ does not correspond to the law of $X_0$. All in all, I fear that some confusions between the distributions of random variables and the distributions of processes might be lurking in the background here. (And, as already mentioned elsewhere, to accept an answer 18 minutes after the question appeared on the site might not be the best practice available.) – Did Mar 24 '12 at 13:08
@ Didier Piau: Thanks for your comment. In the first equation I forget the $0$, sorry for that. I also fixed the second equation. Hopefully it's correct now. I would appreciate if you could explain in an answer how to prove this correctly. About accepting an answer: Since I was not aware of the mistake in the second equation, I thought the proof given by martini is correct. – math Mar 24 '12 at 13:39
The critiques about the immediate acceptation of answers are not related to the correctness of said answers, see here on meta for a discussion. – Did Mar 24 '12 at 14:55

we have, following your last line \begin{align*} P(X_t \in A) &= P(Y_t \circ X \in A) & \text{as $Y_t: S^{[0,\infty)} \to S$ is the evaluation}\\ &= P\bigl((Y_t \circ X)^{-1}(A)\bigr) & \text{just rewritten}\\ &= P\Bigl(X^{-1}\bigl(Y_t^{-1}(A)\bigr)\Bigr)\\ &= (P \circ X^{-1})\bigl(Y_t^{-1}(A)\bigr)\\ &= \mathbb P_\nu\bigl(Y_t^{-1}(A)\bigr) & \text{definition of $\mathbb P_\nu$}\\ &= \mathbb P_\nu(Y_t \in A) \end{align*}

HTH, AB,

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As everybody knows, the entirety of mathematics is nothing but a concatenation of rigorous trivialities... and the solution below is no exception.

Call $T=[0,\infty)$, $E=S^T$ and $\mathcal E=\mathcal S^{\otimes T}$, and let $X:\Omega\to E$ denote any function measurable with respect to $\mathcal F$ and $\mathcal E$. Equivalently, $X^{-1}(B)$ is in $\mathcal F$ for every $B$ in $\mathcal E$. Still equivalently, since $\mathcal E$ is a product sigma-algebra, $X_t^{-1}(B_t)$ is in $\mathcal F$ for every $t$ in $T$ and every $B_t$ in $\mathcal S$, where by definition $X=(X_t)_{t\in T}$ hence $X_t:\Omega\to S$ for each $t$ in $T$.

(There is a subtlety here, which is that the former condition (on $X$) is a priori stronger than the latter (on every $X_t$) since $T$ is uncountable. It happens that the two conditions become equivalent (as asserted above) as soon as one adds some regularity hypothesis on the process $X=(X_t)_t$, for example the hypothesis that the paths $t\mapsto X_t(\omega)$ are almost surely càdlàg, or the Markovianity hypothesis mentioned in the post. We now forget these subtleties and proceed.)

For every probability measure $\nu$ on $(S,\mathcal S)$, if $\nu$ is the image of $P$ by $X_0$, one calls $\mathbb P_\nu$ the probability measure on $(E,\mathcal E)$ image of $P$ by $X$, that is, $\mathbb P_\nu=P\circ X^{-1}$.

For each $t$ in $T$, $Y_t:E\to S$ is defined by $Y_t(x)=x(t)$ for every $x$ in $E$ defined as $x:T\to S$, $t\mapsto x(t)$. (Note that the choice of $\omega$ as the generic element of $E=S^T$, as in the post above, might be seen as unfortunate in such a context since another set $\Omega$, a priori different from $E$, is in the picture.)

In other words, introducing $Y=(Y_t)_{t\in T}$, one sees that $Y:\color{red}{E}\to \color{blue}{E}$ is simply the identity function $Y:\color{red}{x}\mapsto \color{blue}{x}$. In particular, $Y$ is indeed measurable as a function $Y:(\color{red}{E},\color{red}{\mathcal E})\to(\color{blue}{E},\color{blue}{\mathcal E})$ and, as soon as $(\color{red}{E},\color{red}{\mathcal E})$ is endowed with any probability measure $\color{red}{Q}$, then the distribution of $Y$ becomes $\color{blue}{Q}=\color{red}{Q}\circ Y^{-1}$ on $(\color{blue}{E},\color{blue}{\mathcal E})$.

In particular, under $\mathbb P_\nu$, the distribution of $Y=(Y_t)_{t\in T}$ is $\mathbb P_\nu$. Since $\mathbb P_\nu$ was defined from the start as the distribution of $X$, you are done.

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@ Didier Piau: Thank you a lot for taking the time and explain it in such a rigorous way. It really helps me to get a better understanding of these things. – math Mar 24 '12 at 17:10