Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space and $\mathcal F_i\subseteq\mathcal A$ be a $\sigma$-algebra on $\Omega$.

If $(\mathcal F_1,\mathcal F_2,\mathcal F_3)$ is independent, does it follow that $\mathcal F_1\vee\mathcal F_2:=\sigma(\mathcal F_1\cup\mathcal F_2)$ is independent of $\mathcal F_3$?

In the special case, where $\mathcal F_i=\sigma(X_i)$ for some random variable $X_i$ taking values in a measurable space $(E_i,\mathcal E_i)$, we've got $$\mathcal F_1\vee\mathcal F_2=(X_1,X_2)^{-1}(\mathcal E_1\otimes\mathcal E_2)=(X_1,X_2)^{-1}(\sigma(\mathcal G_1\times\mathcal G_2)\tag1,$$ where $\mathcal G_i\subseteq\mathcal E_i$ is arbitrary with $\sigma(\mathcal G_i)=\mathcal E_i)$. From $(1)$ the desired claim immediately follows.

So, I wondered whether the same holds in general or why it breaks down.

  • $\begingroup$ Yes; you already received a proof below. Note also that much more is true: see this answer of mine for the general theorem. $\endgroup$
    – peek-a-boo
    Jan 3, 2022 at 14:18
  • $\begingroup$ @peek-a-boo Thank you for your comment. One of the things I actually used to know, but forgot. Note that your Theorem 1 is even an equivalence (and you only need that $\mathcal P_i\cup\{\emptyset\}$ is a $\pi$-system). $\endgroup$
    – 0xbadf00d
    Jan 3, 2022 at 15:32

1 Answer 1


Sets of the form $A \cap B$ with $A \in \mathcal F_1, B \in \mathcal F_2$ form a $\pi$ system which generates $\mathcal F_1 \vee \mathcal F_2$. The equation $P(C\cap D) =P(C)P(D)$ holds if $C$ is above type and $D \in \mathcal F_3$. Apply Dymkin's $\pi -\lambda$ Theorem (with $D \in \mathcal F_3$ fixed) to finish.


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