A Markov Chain conditional probability question from first principle I am stuck proving following result using properties of conditional expectations.
Let random variables $X$, $Y$, and $Z$ be such that for all $A\in\sigma(X)$ and $C\in\sigma(Z)$ almost surely,
$$P(A\cap C|\sigma(Y))=P(A|\sigma(Y))P(C|\sigma(Y))$$
Then show that for any $B\in\sigma(Y)$ 
$$P(A\cap C|B)=P(A|B)P(C|B).$$
Here conditional probabilities are interpreted as conditional expectations of indicator random variables and $P(A|C)=\dfrac{P(A\cap C)}{P(C)}$.
I started by writing out following
$$
P(A\cap B\cap C)=\int_{B}  1\{A\cap C\}\,dP \\
=\int_{B}\mathbb{E}[1\{A\cap C\}|\sigma(Y)]\,dP \\
=\int_{B}\mathbb{E}[1\{A\}|\sigma(Y)]\mathbb{E}[1\{C\}|\sigma(Y)]\,dP
$$
But I am unable to manipulate this further
 A: Consider $\sigma(1_{\{ B\} })\subset \sigma(Y).$ Then
$$P(A\cap C|B)=\frac{P(A\cap B\cap C)}{P(B)}=\frac{1}{P(B)}\mathbb E[\mathbb{E}[1_{\{A\}}1_{\{C\}}1_{\{B\}}|\sigma(1_{ \{B \} })]]$$
$$=\frac{1}{P(B)}\mathbb E[1_{\{B\}}\mathbb{E}[1_{\{A\}}1_{\{C\}}|\sigma(1_{ \{B \} })]]=\frac{1}{P(B)}\mathbb E[1_{\{B\}}\mathbb{E}[1_{\{A\}}|\sigma(1_{ \{B \} })]\mathbb E[1_{\{C\}}|\sigma(1_{ \{B \} })]]=$$
$$=\frac{1}{P(B)}\int_BP(A|\sigma(1_{ \{B \} }))P(C|\sigma(1_{ \{B \} }))dP= \cdots$$
where
$$[P(A|\sigma(1_{ \{B \} }))P(C|\sigma(1_{ \{B \} }))](\omega)=\begin{cases}P(A|B)P(C|B),& \text{ if }&\omega\in B\\
0,& \text{ if }&\omega \not\in B.\end{cases}$$
So,
$$\cdots=\frac{1}{P(B)}\int_BP(A|\sigma(1_{ \{B \} }))P(C|\sigma(1_{ \{B \} }))dP=$$
$$=\frac{1}{P(B)}\int_BP(A|B)P(C|B)dP=P(A|B)P(C|B).$$
A: We'll assume Y is discrete for simplicity at first. (the general case will be discussed later)
Consider the following three conditions:


*

*For all $A\in\sigma(X)$ and $C\in\sigma(Z)$, $P(A\cap C|\sigma(Y))=P(A|\sigma(Y))P(C|\sigma(Y))$.

*For all $A\in\sigma(X)$ and $C\in\sigma(Z)$ and each atom $B$ of $\sigma(Y)$ (i.e. for each event $B$ of the form $Y=y$), $P(A\cap C|B)=P(A|B)P(C|B)$.

*For all $A\in\sigma(X)$ and $C\in\sigma(Z)$ and all $B\in\sigma(Y)$, $P(A\cap C|B)=P(A|B)P(C|B)$.
It seems you are asking about how to prove $1\implies3$ which has a counter-example (which will be discussed soon).
The random variable $P(A\cap C|\sigma(Y))$ is just the random variable that takes value $P(A\cap C|Y=y)$ on the event $Y=y$. So, condition 1 and 2 are equivalent and is called conditional independence.
On the other hand, condition 3 is too strong. Note that condition 3 implies independence between X and Z. For a counter-example to $1\implies 3$, imagine a simple random walk on $\mathbb Z$, then the walker's position at time $n=2015$ is not independent to that at time $n=2015+2$. Indeed, knowing the position at time 2015 would restrict the possible positions at time 2015+2 to just three values. Nonetheless, the positions at 2015, 2016, 2017 together satisfy condition 1 (where Y is the position at 2016).
For general (not necessarily discrete or continuous) real-valued random variables, the analogue of the condition 2 would be a bit more involved and would require the use of disintegration of measures, which is usually considered to be an advanced topic. On the other hand, the condition 1 easily transfers without modification to the general case and that is part of why condition 1 is usually taken to be the definition of conditional independence in the general case. 
