Here is a problem I couldn't solve.

Given a probability space $ (\Omega, \mathcal{A},\mathbb{P}) $, and $ \mathcal{F}, \mathcal{G}, \mathcal{B} $ sub-$\sigma$-algebras of $\mathcal{A}$.

Is it true that coniditional independence of $\mathcal{B}$ and $\mathcal{F}$ given $\mathcal{G}$ which we can note $\mathcal{F}\;\amalg_{\mathcal{G}} \mathcal{B}$ implies the following property :

$$\mathbb{P}(F\cap B \mid G) = \mathbb{P}(F \mid G)\mathbb{P}(B \mid G),$$ almost surely $ \;\; \forall \;(F,G,B)\in \mathcal{F} \times \mathcal{G}\times \mathcal{B}$

Where I note :
$$\mathbb{P}(A \mid G) =\mathbb{E}[1_A|\sigma(G)], \forall A\in \mathcal{A}$$ or more generally given any sub sigma algebra $\mathcal{C}$ of $\mathcal{A}$ :
$$\mathbb{P}(A \mid \mathcal{C}) =\mathbb{E}[1_A|\mathcal{C}], \forall A\in \mathcal{A}$$

And where conditional independence $\mathcal{F}\;\amalg_{\mathcal{G}} \mathcal{B}$ is defined by :

$$\mathbb{P}(F\cap B \mid \mathcal{G}) = \mathbb{P}(F \mid \mathcal{G})\mathbb{P}(B \mid \mathcal{G}),$$ almost surely $ \;\; \forall \;(F,B)\in \mathcal{F} \times \mathcal{B}$.

I think this implication is not true but I can't find a counter-example.

Moreover I think the reverse implication holds true.(check edit)

Best Regards

Edit : I thought I had a proof for the reverse implication but as gnometorule suggested it wasn't true, I rechecked it and found a mistake, so I take that last comment back, as I don't know if it is true or not.

  • $\begingroup$ I haven't done this in a long while, but isn't this essentially the definition of conditional expectation - you have equality on each set of the sub- sigma algebra? Both s $\endgroup$ – gnometorule Jan 6 '12 at 15:07
  • $\begingroup$ @gnometorule: Hi the property is not far from the definition but $\mathbb{P}(A|G)$ can be quiet different from $\mathbb{P}(A|\mathcal{G})$. If the sigma algebra $\mathcal{G}$ is discrete then you can write over its atoms that :$\mathbb{P}(A|\mathcal{G})=\sum_{G \in atoms \; of\; \mathcal{G}}1_G.\mathbb{P}(A|G)$ but I think the matter gets more complex in the general case. Regards $\endgroup$ – TheBridge Jan 6 '12 at 15:40
  • $\begingroup$ There are different, of course, but the sigma algebra definition of conditional expectation is that you have equality on the sets of the sigma algebra, so it seems to restate the definition. Your last equality follows from the definition. $\endgroup$ – gnometorule Jan 6 '12 at 15:44
  • $\begingroup$ The other direction would't be true necessarily.Then again, this is distant past for me unfortunately, and I might be wrong. $\endgroup$ – gnometorule Jan 6 '12 at 15:54
  • $\begingroup$ @gnometorule: Unless there is something obvious I miss (which I hope), it only seems to be a restatement, because rewriting all this under the expectation operator I couldn"t derive the result. Best Regards $\endgroup$ – TheBridge Jan 6 '12 at 16:10

First implication is not true, since for $G = \Omega$ it would mean that $\mathcal{F}$ and $\mathcal{B}$ are unconditionally independent as soon as they are conditionally independent given some $\sigma $-algebra $\mathcal{G}$, which is absurd.

Edit: I feel the need to write this down in more detail:

$\mathbb{E}\left[ {{1_F}|\sigma \left( \Omega \right)} \right] = \frac{{\mathbb{E}\left[ {{1_\Omega }{1_F}} \right]}} {{\mathbb{P}\left( \Omega \right)}}{1_\Omega } = \mathbb{E}\left[ {{1_F}} \right] = \mathbb{P}\left( F \right)$, similiarly $\mathbb{E}\left[ {{1_B}|\sigma \left( \Omega \right)} \right] = \mathbb{P}\left( B \right)$ and $\mathbb{E}\left[ {{1_{B \cap F}}|\sigma \left( \Omega \right)} \right] = \mathbb{P}\left( {B \cap F} \right)$, which would give us $\mathbb{P}\left( {B \cap F} \right) = \mathbb{P}\left( F \right)\mathbb{P}\left( B \right),\forall \left( {F,B} \right) \in \mathcal{F} \times \mathcal{B}$, which is to say that $\mathcal{F}$ and $\mathcal{B}$ are independent.

Regarding the other answer posted here, if $\mathcal{G} = \mathcal{A}$, there is no contradiction there. It is then simply true that $\mathcal{F}$ and $\mathcal{B}$ are independent, as you have neatly shown. However, $\mathcal{G}$ here is given and fixed, as are $\mathcal{F}$ and $\mathcal{B}$, what is variable is $G \in \mathcal{G}$.

On the other hand, reverse implication holds true, because

$\mathbb{E}\left[ {\mathbb{E}\left[ {{1_{F \cap B}}|\mathcal{G}} \right]{1_G}} \right] = \mathbb{E}\left[ {\mathbb{E}\left[ {{1_{F \cap B}}|G} \right]{1_G}} \right] = \mathbb{E}\left[ {\mathbb{E}\left[ {{1_F}|G} \right]\mathbb{E}\left[ {{1_B}|G} \right]{1_G}} \right] = \mathbb{E}\left[ {\mathbb{E}\left[ {\mathbb{E}\left[ {{1_B}|G} \right]{1_F}|G} \right]{1_G}} \right] = $ $ = \mathbb{E}\left[ {\mathbb{E}\left[ {\mathbb{E}\left[ {{1_B}|G} \right]{1_F}|\mathcal{G}} \right]{1_G}} \right] = \mathbb{E}\left[ {\mathbb{E}\left[ {\mathbb{E}\left[ {{1_F}|G} \right]{1_B}|\mathcal{G}} \right]{1_G}} \right] = \mathbb{E}\left[ {\mathbb{E}\left[ {{1_F}|G} \right]\underbrace {{1_G}\mathbb{E}\left[ {{1_B}|\mathcal{G}} \right]}_{ = {1_{G'}},G' \in \mathcal{G}}} \right] = $

$\mathbb{E}\left[ {\mathbb{E}\left[ {{1_F}|\mathcal{G}} \right]\mathbb{E}\left[ {{1_B}|\mathcal{G}} \right]{1_G}} \right]$

Intuitively, if you reduce the information given, you can't guarantee that two events will remain independent. On the other hand, if two events are independent, adding more information atop of existing one won't change that.

  • $\begingroup$ @ Alen : thank's very good answer. $\endgroup$ – TheBridge Jan 6 '12 at 23:54
  • $\begingroup$ By the way, I think you can simplify a little your proof of the second assertion, as it is not necessary to prove the existence of a set $G'$, because $1_G.\mathbb{E}[ 1_B|\mathcal{G}]$ is a positive integrable $\mathcal{G}$-measurable random variable which is enough to go to the next step (the conclusion). $\endgroup$ – TheBridge Jan 9 '12 at 17:20
  • $\begingroup$ That is true, I was thinking about that same part after I wrote the answer $\endgroup$ – Alen Jan 9 '12 at 18:01

Suppose we take $\mathcal{G}$ as large as possible, i.e. $\mathcal{G}=\mathcal{A}$.

Then, for each random variable $Z$ we have $\mathbb{E}[Z|\mathcal{G}]=Z$, since $Z$ is already $\mathcal{G}$-measurable. Applying this to your definition of conditional independence gives $\mathcal{F}\;\amalg_{\mathcal{G}} \mathcal{B}$. Now $\Omega\in \mathcal{G}$, so if the result were to hold we'd have $\mathbb{P}[B\cap F]=\mathbb{P}[B]\mathbb{P}[F]$, for all $B\in\mathcal{B}$, $F\in\mathcal{F}$. This is only true if $B$ and $F$ are independent.


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