# Several algebras in conditional and join probabilities

There are several cases in combining conditional and joint probabilities, I get confused to show them if there are correct:

Given $A, B, C$ are random variables, $P(A|B)$ denotes conditional probability and $P(B, C)$ denotes joint probability.

1. $P((A|B)|C) = P(A|(B, C))$

We have: $P((A|B)|C) = \frac{P((A|B), C)}{P(C)}$. This requires following

1. $P((A|B), C) = P(A|(B, C))P(C)$

If the equality holds, there would be no ambiguity with $P(A|B, C)$

• The notations $(A|B)$ and $(B, C)$ (even this sometimes can be understood as $B\cap C$, but it's better to be more explicit) are not well defined (not events), so it is meaningless to talk about the expression "P((A|B)|C)". Jul 22, 2015 at 12:42
• Oddly enough, this is not the first time the absurd P((A|B)|C) pops up on the site. I remember some exhausting discussions with a user to make them realize the thing was not correct. (On the other hand, P(A|B,C) to mean P(A|B∩C) is standard.)
– Did
Jul 22, 2015 at 12:50
• @Xingdong Any source for the notation P((A|B)|C)?
– Did
Jul 22, 2015 at 12:51
• @XingdongZuo Note $$P(A|B, C_i)P(C_i|B) = \frac{P(A\cap B \cap C_i)}{P(B\cap C_i)}\times \frac{P(C_i\cap B)}{P(B)} = \frac{P(A \cap B \cap C_i)}{P(B)}.$$ Jul 22, 2015 at 13:44
• Indeed, and sum this over $i$, using that $(C_i)$ is a partition, hence $$\sum_iP(A,B,C_i)=P(A,B).$$
– Did
Jul 22, 2015 at 13:46

We can to understand $\Bbb{P}(A \vert B) = \Bbb{E}[1_A |\sigma(1_B)](\omega)$ for any $\omega \in B$.

In that sense one would interpret $\Bbb{P}((A \vert B) \vert C) = \Bbb{E}\Bbb{E}[1_A |\sigma(1_B)] \vert \sigma(1_C)](\omega)$ for $\omega \in B \cap C$

$$\Bbb{P}((A \vert B) \vert C) = \Bbb{E}\Bbb{E}[1_A |\sigma(1_B)] \vert \sigma(1_C)](\omega) = \Bbb{E}\Bbb{E}[\Bbb{E}[1_A |\sigma(1_B, 1_C)] |\sigma(1_B)] \vert \sigma(1_C)](\omega) = \Bbb{E}\Bbb{E}[1_A |\sigma(1_B, 1_C)] \vert \sigma(1_C)](\omega) = \Bbb{E}[1_A |\sigma(1_B, 1_C)](\omega) = P(A \vert B,C)$$

Note $\Bbb{E}[1_A\vert \sigma(1_B)]$ is a random variable $$\Bbb{E}[1_A\vert \sigma(1_B)] = P(A\vert B) 1_B + P(A\vert B^c)1_{B^c}$$

$$\mathcal{G} \subset \mathcal{F} \Rightarrow\Bbb{E}[\Bbb{E}[1_A |\mathcal{G}] |\mathcal{F}] = \Bbb{E}[\Bbb{E}[1_A |\mathcal{G}]$$

where $\mathcal{F},\mathcal{G}$ are sigma algebras.

Note that we are understanding $P(A\vert B,C) = \Bbb{E}[1_A |\sigma(1_B, 1_C)]$ wich is different than $\Bbb{E}[1_A |\sigma(1_B 1_C)]$ to finally get a number we write

$$\Bbb{E}[1_A |\sigma(1_B, 1_C)] = P(A\vert B \cap C)1_{B \cap C} + P(A\vert B^c \cap C)1_{B^c \cap C} + P(A\vert B \cap C^c)1_{B \cap C^c}+ P(A\vert B^c \cap C^c)1_{B^c \cap C^c}$$

So you define $P(A \vert B,C) = \Bbb{E}[1_A |\sigma(1_B, 1_C)] (\omega)$ for any $\omega \in B \cap C$

As you would have defined $P(A\vert B) = \Bbb{E}[1_A |\sigma(1_B)](\omega)$ for any $\omega \in B$

• Sorry but $P(A\mid B)$ and $E(1_A\mid\sigma(B))$ are two different objects since the former is a real number which is the value of the latter (a random variable) on the event $B$. Thus the first sentence of your post (contradicted by the note) is not true.
– Did
Jul 22, 2015 at 13:35
• The line of computations just before the note is wrong as well (there is no reason why $P(A\mid\sigma(B,C))$ should be always $\sigma(C)$-measurable and this is what you are saying).
– Did
Jul 22, 2015 at 13:39
• OP: In view of the frantic edits above, I am wondering: do you know the meaning of the expression "to take one's losses"?
– Did
Jul 22, 2015 at 13:50