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In the paper Existence de processus Markoviens pour des systèmes infinis de particules by Cocozza, C. and Kipnis, C. (Ann. lnst. H. Poincaré, Sect. B, 13, 239-257, 1977), one reads

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A transition probability from $H_1$ to $H_2$ is a function $P: H_1 \times A_2 \to [0,1] $ such that

  1. The map $h_1 \mapsto P(h_1,C)$ is $A_1$ measurable for every $C \in A_2$

  2. the map $C \mapsto P(h_1,C)$ is a probability measure for every $h_1 \in H_1$

Consider the function $X_{h_1}:h_2 \mapsto X(h_1, h_2)$. It is assumed that $$\Bbb{E} [X_{h_1} \mid B_2] = Y_{h_1} $$ $\mu$ - almost surely, where $\mu$ is a probability measure on $H_1$

It seems that if we define $\Bbb{P}(dh_1,dh_2) = \mu(h_1) P(h_1, dh_2)$ we obtain

$$ \Bbb{E}_{\Bbb{P}}[X] = \int_{H_1} \int_{H_2}X_{h_1}(h_2)P(h_1,dh_2) d\mu(h_1)= \int_{H_1} \int_{H_2}Y_{h_1}(h_2)P(h_1,dh_2) d\mu(h_1)= \Bbb{E}_{\Bbb{P}}[Y] $$

But how do we obtain

$$\Bbb{E}_{\Bbb{P}}[X] = Y ?$$

Edit: Perhaps the question is a notation question: what is the meaning of $$\Bbb{E}^{B_2\otimes A_1}_{\Bbb{P}}[X]?$$

Translation: Let $(H_1,A_1)$ and $(H_2,A_2)$ be two measurable spaces and let $B_2$ be a subsigma-algebra of $A_2$. Let $P(h_1,dh2)$ be a transition probability from $H_1$ to $H_2$, and let $\mu$ be a probability on $H_1$. If for $\mu$ - almost every $h_1$ the conditional expectation of a random variable $X(h_1,\cdot)$ with respect to the subsigma algebra $B_2$ and the probability $P(h_1,\cdot)$ is $Y(h_1,\cdot)$ then $$\Bbb{E}^{B_2\otimes A_1}_{\Bbb{P}}[X]$$

where $\Bbb{P}(dh_1,dh_2) = \mu(dh_1) P(h_1, dh_2)$

  • You might provide a more complete and literal translation and put it in a quoted box so that it is clear to English readers exactly what is being stated. I can get the gist from the passage you provide as well as the rest of your answer, but I fear I'm missing something critical in the passage. Doesn't affect this post too much I think, just a good idea in general on an English mathematics forum :) good question though otherwise! – Brevan Ellefsen Oct 17 '16 at 13:24

0 Answers0