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I'm facing the following issue. Let $X$ be an integrable random variable on the probability space $(\Omega,\mathcal{F},\mathbb{P})$ and $\mathcal{G},\mathcal{H} \subseteq \mathcal{F}$ be two sigma-algebras. We assume that $X$ is independent of $\mathcal{G}$, i.e. $\sigma(X)$ is independent of $\mathcal{G}$. Can I say (and can I prove) that $$ E(X \mid \sigma(\mathcal{G} \cup \mathcal{H})) = E(X\mid \mathcal{H}) ?$$

Thank you very much for your help!

Davide Giraudo
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2 Answers2

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If you're willing to assume that $\sigma(\sigma(X)\cup \mathcal{H})$ is independent of $\sigma(\mathcal{G})$ and that $X$ is integrable, then the assertion is indeed true. We need to show that $$ E[X\mid \sigma(\mathcal{G}\cup\mathcal{H})]=E[X\mid\mathcal{H}], $$ that is, we need to show that $E[X\mid\mathcal{H}]$ can serve as the conditional expectation of $X$ given $\sigma(\mathcal{G}\cup\mathcal{H})$, i.e. show that

  • $E[X\mid\mathcal{H}]$ is $\sigma(\mathcal{G}\cup\mathcal{H})$-measurable,
  • $E[X\mid\mathcal{H}]$ is integrable,
  • for all $A\in\sigma(\mathcal{G}\cup\mathcal{H})$: $$\int_A E[X\mid\mathcal{H}]\,\mathrm dP=\int_A X\,\mathrm dP.$$

The first two are obvious. For the third, note that (by linearity, we can assume that $X$ is non-negative) $$ \sigma(\mathcal{G}\cup\mathcal{H})\ni A\mapsto \int_AE[X\mid\mathcal{H}]\,\mathrm dP $$ and $$ \sigma(\mathcal{G}\cup\mathcal{H})\ni A\mapsto \int_A X\,\mathrm dP $$ are two measures defined on $\sigma(\mathcal{G}\cup\mathcal{H})$ with equal total mass being $E[X]$. Hence, it is enough to show that the two measures are identical on some $\cap$-stable generator of $\sigma(\mathcal{G}\cup\mathcal{H})$. Here, we use that $$ \{A\cap B\mid A\in\mathcal{G},\,B\in\mathcal{H}\} $$ is indeed a $\cap$-stable generator of $\sigma(\mathcal{G}\cup\mathcal{H})$ (why?) and hence it suffices to show that $$ \int_{A\cap B} E[X\mid \mathcal{H}]\,\mathrm dP=\int_{A\cap B} X\,\mathrm dP,\quad A\in\mathcal{G},\,B\in\mathcal{H}. $$ Try to show this using the independence assumption. I think it will be clear to you that we in fact need the stronger independence assumption.

A counterexample showing that we indeed need the stronger assumption is the following: Let $U$ and $V$ be i.i.d. symmetric Bernoulli variables (i.e. $P(U=-1)=P(U=1)=\tfrac12$), $\mathcal{G}=\sigma(U)$, $\mathcal{H}=\sigma(V)$ and $X=UV$. Now, one can show that $X$ and $U$ are independent by showing that $$ P(X=a,U=b)=P(X=a)P(U=b) $$ for every combination of $a,b\in \{0,1\}$. But $\sigma(\sigma(X)\cup\sigma(V))$ is not independent of $\sigma(U)$ since for example $$ P(X=1,V=1,U=1)\neq P(X=1,V=1)P(U=1), $$ and hence we do not have the stronger independence assumption. In this case, $$ E[X\mid \sigma(\mathcal{G}\cup\mathcal{H})]=UV\neq E[U]V=E[X\mid\mathcal{H}]. $$

Stefan Hansen
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  • Thank you for your very clear answer. I am willing to assume the stronger independence. It remains me two questions. The first one is: the set $$S = {A\cap B | A \in \mathcal{G}, B \in \mathcal{H}}$$ is a generator of $\sigma (\mathcal{G} \cup \mathcal{H})$ because $ \mathcal{G}\cup \mathcal{H} \subseteq S \subseteq \sigma (\mathcal{G} \cup \mathcal{H})$, isn't it ? And the second one: do we have the following equivalence, $\sigma(\sigma(X) \cup \mathcal{H})$ is independent of $\mathcal{G}$ iff $\sigma(X)$ and $\mathcal{H}$ are independent of $\mathcal{G}$ ? –  Apr 18 '13 at 12:43
  • To the first question: Yes! To the second: That is not equivalent. Clearly, the first implies the second, but the reverse isn't true. Just look at the counterexample in my answer. There, $\sigma(X)$ and $\mathcal{H}=\sigma(V)$ are independent of $\mathcal{G}=\sigma(U)$ but $\sigma(\sigma(X)\cup\sigma(V))$ is not. – Stefan Hansen Apr 18 '13 at 12:49
  • Thanks! But I found this lemma in 'PDE and martingale methods in option pricing' (Pascucci): for $I$ and $J$ two $\cap$-stable collections of sets, $\sigma(I)$ is independent of $\sigma(J)$ iff $I$ is independent of $J$. So, if I have $\mathcal{G},\mathcal{H},\mathcal{I} \subseteq \mathcal{F}$ three sub-$\sigma$-algebras such that $\mathcal{H}$ and $\mathcal{I}$ are independent of $\mathcal{G}$ and by taking $J = { H \cap I | H \in \mathcal{H}, I \in \mathcal{I} }$, $J$ is $\cap$-stable and $\sigma(J) = \sigma( \mathcal{H} \cup \mathcal{I})$ is independent of $\mathcal{G}$. What's wrong? –  Apr 18 '13 at 19:40
  • How do you argue that $J$ is independent of $\mathcal{G}$? – Stefan Hansen Apr 18 '13 at 19:46
  • Got it! If $A \in I$, then $A$ is not necessarily in either $\mathcal{H}$ or $\mathcal{I}$, isn't it? –  Apr 18 '13 at 20:03
  • Well, it's true when they are independent sigma-algebras (right?). I'm considering two processes $W$ (a Wiener process with respect to his standard filtration $(\mathcal{F}t^W){t \geq 0}$) and $X$ (with his standard filtration $(\mathcal{F}t^X){t \geq 0}$) which are independent. Do I have, for $0 \leq s < t$, that $\sigma(W_t-W_s), \mathcal{F}s^W$ and $ \mathcal{F}_s^X$ are independent? I would like to prove that $W$ is a Wiener process with respect to the new filtration $(\mathcal{F}_t = \sigma(\mathcal{F}_t^W \cup \mathcal{F}_t^X)){t \geq 0}$. Thank you for the help! –  Apr 18 '13 at 21:43
  • @user73191: That is a completely different question, and I think you should post it as a new question (I can't give you a quick answer to it). – Stefan Hansen Apr 19 '13 at 05:15
  • Thank you. You can find my new question here: http://math.stackexchange.com/questions/366265/sigma-algebras-and-independent-stochastic-processes. –  Apr 19 '13 at 08:36
  • @StefanHansen: I have a new question concerning the two measures you defined. I don't think they are measures because the integral is not necessarily positive. What do you think about that? Thank you. –  Apr 20 '13 at 13:51
  • But maybe if we use the same proof firstly for $X^+ = \max(X,0)$ and secondly for $X^- = max(-X,0)$, we could prove the equality. Do you agree or did I made a mistake? –  Apr 20 '13 at 14:17
  • @user73191: Yes, that was a mistake. Splitting up the integral in $X^+$ and $X^-$ will do the job. – Stefan Hansen Apr 20 '13 at 22:13
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For a general result, the equation holds under the assumption A2 below, and assumption A2 holds when A1 holds. A1 is the condition mentioned in the other answer.

(A1) $X \vee \mathcal H$ and $\mathcal G$ are independent (where we write $X \vee \mathcal H$ to mean the smallest sub sigma algebra containing both $\sigma(X)$ and $\mathcal H$)

(A2) $X$ and $\mathcal G$ are conditionally independent given $\mathcal H$

(A3) $E(X | \mathcal H \vee \mathcal G) = E(X | \mathcal H)$

In short, A1 ==> A2 ==> A3.

On the other hand, X and $\mathcal G$ being independent does not imply A2 and A2 does not imply independence of X and $\mathcal G$.

Jisang Yoo
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