condition probability of a condition probability

Question

Given a probability space ($\Omega,\mathscr{F},\mathbb{P}$).Suppose A $\in\mathscr{F}$ satisfies $\mathbb{P}(A)>0$.Define $\mathbb{Q}(B):=\mathbb{P}(B|A)$ for any B$\in\mathscr{F}$.Prove for any $B\in\mathscr{F}$ and sub $\sigma$ algebra $\mathscr{C}\subset\mathscr{F}$, $\mathbb{Q}(B|\mathscr{C})=\frac{\mathbb{P}(A,B|\mathscr{C})}{\mathbb{P}(A|\mathscr{C})}$.

I think, we must proof $\forall C\in\mathscr{C}$,$\mathbb{Q}(B,C)=\int_C\frac{\mathbb{P}(A,B|\mathscr{C})}{\mathbb{P}(A|\mathscr{C})}$,but I don't know how to calculate the last integral

Answer 1

Since OP asked Speltzu for an explanation I add how I would prove this:

By definition: $$\tag1 \mathbb Q(B)=\mathbb P(B|A)=\frac{\mathbb P(B\cap A)}{\mathbb P(A)}\,. $$ Claim. For a $\sigma$-algebra $\mathscr C$ we have $$\tag2 \boxed{\phantom{\Bigg|}\quad\mathbb Q\big[B\,\big|\,\mathscr C\big]=\frac{\mathbb P[B\cap A\,|\,\mathscr C]}{\mathbb P[A\,|\,\mathscr C]}\,.\quad} $$ Proof. By defintion of the conditional probability we have for any $C\in\mathscr C$ $$\tag3 \int_C\mathbb P\big[B\cap A\,\big|\,\mathscr C\big]\,d\mathbb P=\mathbb P\big[B\cap A\cap C\big]\,. $$ The measure $\mathbb Q$ has Radon-Nikodym density $$\tag4 \frac{d\mathbb Q}{d\mathbb P}=\frac{1_A}{\mathbb P(A)} $$ with respect to $\mathbb P\,.$ Then, \begin{align}\tag5\require{cancel} &\int_C \frac{\mathbb P[B\cap A\,|\,\mathscr C]}{\mathbb P[A\,|\,\mathscr C]}\,d\mathbb Q\stackrel{(4)}{=} \int_C\,\frac{\mathbb P[B\cap A\,|\,\mathscr C]}{\mathbb P[A\,|\,\mathscr C]}\,\frac{1_A}{\mathbb P(A)}\,d\mathbb P\\[2mm]\tag6 &\stackrel{\textstyle\color{red}{(*)}}= \int_C\,\frac{\mathbb P[B\cap A\,|\,\mathscr C]}{\cancel{\mathbb P[A\,|\,\mathscr C]}}\frac{\cancel{\mathbb P[A\,|\,\mathscr C]}}{\mathbb P(A)}\,d\mathbb P\\[2mm]\tag7 &=\int_C\,\frac{\mathbb P[B\cap A\,|\,\mathscr C]}{\mathbb P(A)}\,d\mathbb P \stackrel{(3)}{=} \frac{\mathbb P[B\cap A\cap C]}{\mathbb P(A)}\stackrel{(1)}=\mathbb Q(B\cap C)\,. \end{align} $$\tag*{ $\Box$} \phantom{\quad} $$

To answer the question by Speltzu why in the equals sign marked by $\color{red}{(*)}$ we are allowed to switch from $1_A$ to $\mathbb P[A\,|\,{\mathscr C}]\,:$

All other ingredients of that integral are ${\mathscr C}$-measurable. The very definition of $\mathbb P[A|{\mathscr C}]$ says $$\tag8 \int_\Omega 1_C\mathbb P[A\,|\,{\mathscr C}]\,d\mathbb P=\int_\Omega 1_C1_A\,d\mathbb P\,,\quad \forall C\in{\mathscr C}\,.$$ This carries over from the indicator function $1_C$ to all simple ${\mathscr C}$-measurable functions and then to all ${\mathscr C}$-measurable functions.

Answer 2

Difficult problem where there are. I rectify my previous answer. The proposition is true.The reasoning would be as follows: \begin{gather*} \mathbb{E}_{\mathbb{Q}}\left[\frac{\mathbb{P}[A\cap B|\mathscr{C}]}{\mathbb{P}[A|\mathscr{C}]}. 1_C\right] =\mathbb{E}_{\mathbb{Q}}\left[\frac{\mathbb{P}[A\cap B\cap C|\mathscr{C}]}{\mathbb{P}[A|\mathscr{C}]}\right] =\int \left[\frac{\mathbb{P}[A\cap B\cap C|\mathscr{C}]}{\mathbb{P}[A|\mathscr{C}]}\right]d\mathbb{P}_{\Omega/A}=\\ =\frac{1}{\mathbb{P}(A)}\int 1_A.\left[\frac{\mathbb{P}[A\cap B\cap C|\mathscr{C}]}{\mathbb{P}[A|\mathscr{C}]}\right]d\mathbb{P}_{\Omega} =\frac{1}{\mathbb{P}(A)}\int \mathbb{P}[A\cap B\cap C|\mathscr{C}]d\mathbb{P}_{\Omega}=\\ =\frac{\mathbb{P}(A\cap B\cap C)}{\mathbb{P}(A)}=\mathbb{Q}(B\cap C) =\mathbb{E}_{\mathbb{Q}}[1_B.1_C] \end{gather*} By the way, this means that: $\mathbb{Q}[B|\mathscr{C}]=\mathbb{P}[B|\mathscr{C},A]$.