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Is it true that, given a space $X$ and a probability measure $\mu$ on it, given some sets $A, B \subset X$ and a finite number of disjoint sets $C_{\sigma}$ such that $\bigcup_{\sigma} C_{\sigma} =X$, the following inequality holds,

$$\mu (A \cap B ) = \sum_{\sigma}\mu(A \cap C_{\sigma}\cap B) \leq \sum_{\sigma} \frac{ \mu(A \cap C_{\sigma} ) } {\mu(A)} \frac{\mu( C_{\sigma} \cap B)}{\mu(C_{\sigma})}, ?$$

Probabily it is wrong in general ( but I am not sure ), but is it true that if each $C_{sigma}$ is contained into $A$, then the equality holds? I think yes!

Context of the question: approximation of a Markov partition with a partition which is non-Markov (Dynamical Systems). In particular $A$ is the set where the initial condition is contained with a probability given by the measure of this set. I want to estimate the probability that at time $t$ the system will be in a set $D$. The set $B$ of the previous expression corresponds then to the sets of points which satisfy $T^t(B) = D$, where $T$ is the map of the dynamical system. If the inequality holds, then I can write

$$Prob(T^t(x) \in D | x \in A )\leq\sum _{\sigma}Prob (T^{t-1}(x) \in C{\sigma}| x\in A ) Prob (T^t(x) \in D | T^{t-1}(x) \in C\sigma).$$ Estimating the right side of the product of the second term and by iteration, I am then able to treat the dynamical system as a stochastic process, although the partition is not Markov.

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Is it true that, given a space $X$ and a probability measure $\mu$ on it, given some sets $A, B \subset X$ and a finite number of disjoint sets $C_{\sigma}$ such that $\bigcup_{\sigma} C_{\sigma} =X$, the following inequality holds,

$$\mu (A \cap B ) = \sum_{\sigma}\mu(A \cap C_{\sigma}\cap B) \leq \sum_{\sigma} \frac{ \mu(A \cap C_{\sigma} ) } {\mu(A)} \frac{\mu( C_{\sigma} \cap B)}{\mu(C_{\sigma})}, ?$$

Probabily it is wrong in general ( but I am not sure ), but is it true that if $A$ and each $B$ are both equal to the union of a finite number of sets C_{sigma}$is contained into$C_{\sigma}$, than A$, then the equality holds? I think yes!

Context of the question: approximation of a Markov partition with a partition which is non-Markov (Dynamical Systems). In particular $A$ is the set where the initial condition is contained with a probability given by the measure of this set. I want to estimate the probability that at time $t$ the system will be in a set $D$. The set $B$ of the previous expression corresponds then to the sets of points which satisfy $T^t(B) = D$, where $T$ is the map of the dynamical system. If the inequality holds, then I can write

$$Prob(T^t(x) \in D | x \in A )\leq\sum _{\sigma}Prob (T^{t-1}(x) \in C{\sigma}| x\in A ) Prob (T^t(x) \in D | T^{t-1}(x) \in C\sigma).$$ Estimating the right side of the product of the second term and by iteration, I am then able to treat the dynamical system as a stochastic process, although the partition is not Markov.

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Is it true that, given a space $X$ and a probability measure $\mu$ on it, given some sets $A, B \subset X$ and a finite number of disjoint sets $C_{\sigma}$ such that $\bigcup_{\sigma} C_{\sigma} =X$, the following inequality holds,

$$\mu (A \cap B ) = \sum_{\sigma}\mu(A \cap C_{\sigma}\cap B) \leq \sum_{\sigma} \frac{ \mu(A \cap C_{\sigma} ) } {\mu(A)} \frac{\mu( C_{\sigma} \cap B)}{\mu(C_{\sigma})}, ?$$

Probabily it is wrong in general ( but I am not sure ), but is it true that if $A$ and $B$ are both equal to the union of a finite number of sets $C_{\sigma}$, than the equality holds? I think yes!

Contest

Context of the question: approximation of a Markov partition with a partition which is non-Markov (Dynamical Systems). In particular $A$ is the set where the initial condition is contained with a probability given by the measure of this set. I want to estimate the probability that at time $t$ the system will be in a set $D$. The set $B$ of the previous expression corresponds then to the sets of points which satisfy $T^t(B) = D$, where $T$ is the map of the dynamical system. If the inequality holds, then I can write

$$Prob(T^t(x) \in D | x \in A )\leq\sum _{\sigma}Prob (T^{t-1}(x) \in C{\sigma}| x\in A ) Prob (T^t(x) \in D | T^{t-1}(x) \in C\sigma).$$ Estimating the right side of the product of the second term and by iteration, I am then able to treat the dynamical system as a stochastic process, although the partition is not Markov.

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