Vtyjkyuk

Given three $sigma$-algebras $mathcalA, mathcalB, mathcalC$ defined on the same space $Omega$ with $mathcalA$ being independent of both $mathcalB$ and $mathcalC$ is it true that $mathcalA$ is independent of $sigma(mathcalB, mathcalC)$?

This looks like something that must be true but I have neither a proof nor a counterexample.

$mathcalB cap mathcalC := X subset Omega mid X = B cap C, B in mathcalB, C in mathcalC $ is a $pi$-system and it generates $sigma(mathcalB, mathcalC)$ but I don't see how I can show that

$$mathcalA perp mathcalB, mathcalA perp mathcalC implies mathcalA perp mathcalB cap mathcalC$$

What brought me to this question is the following assertion.

Let $Y_0, Y_1, ldots$ be iid random variables with $PY_0 = 1 = PY_0 = -1 = frac12$. Define $mathcalY = sigma(Y_1,Y_2,ldots), X_n = Y_0Y_1ldots Y_n$ and $mathcalT_n = sigma(X_r, r > n)$ for $n = 1,2,ldots$. Then $Y_0$ is independent of $sigma(mathcalY, cap_n mathcalT_n)$.

Obviously, $Y_0$ is independent of $mathcalY$. It is also not hard to show that it is independent of $mathcalT_1$, hence of the tail $sigma$-algebra. Here I could note that $cap_n mathcalT_n$ contains either null-measure or full-measure events since $X_1,X_2,ldots$ is an independent sequence, which I can also show. Then defining the $pi$-system $mathcalZ := mathcalY bigcap cap_n mathcalT_n$ I could write

$$PY_0 = 1 cap Z = PY_0 = 1 cap (Y cap T)$$
for some $mathcalZ ni Z = Y cap T$ with $Y in mathcalY$ and $T in cap_n mathcalT_n$. Continuing the expression above

$$PY_0 = 1 cap (Y cap T) = Pleft(Y_0 = 1 cap Yright) cap T)= PY_0 = 1 PY cap T$$

I did not show all the steps above but I considered the cases $PT = 1$ and $PT = 0$ separately. So this proves independence of $Y_0$ and $sigma(mathcalY, cap_n mathcalT_n)$. I was hoping (for whatever reason) that this would be true regardless of $cap_n mathcalT_n$ being a trivial $sigma$-algebra.

If my first claim is not true, is there a result that gives necessary and sufficient conditions under which it is true? A counterexample would also be appreciated.

The answer to your question is: no.

Let $X,Y,Z$ be random variables on the same probability space.

It can happen that random variables $X$ and $Y$ are independent and also $X$ and $Z$ are independent, but that not $X$ and $(Y,Z)$ are independent.

E.g. let $X$ and $Y$ be iid that both only take values in $0,1$ with $P(X=1)=P(Y=1)=0.5$ and let $Z$ take value $1$ if $X=Y$ and let $Z$ take value $0$ otherwise.

This can be modeled like this:

Let $Omega=Omega_1cupOmega_2cupOmega_3cupOmega_4$ where the $Omega_i$ are non-empty and disjoint.

Let $mathcal D=sigma(Omega_imid i=1,2,3,4)$ and let $P$ be a probability measure on $(Omega,mathcal D)$ that is determined by $P(Omega_i)=frac14$ for $i=1,2,3,4$.

Let $X=1_Omega_1+1_Omega_2$, $Y=1_Omega_1+1_Omega_3$ and $Z=1_Omega_1+1_Omega_4$.

Then it can be proved that $X$ and $Y$ are independent and that $X$ and $Z$ are independent.

Setting $mathcal A=sigma(X)$, $mathcal B=sigma(Y)$ and $mathcal C=sigma(Z)$ that means that $mathcal A$ is independent of both $mathcal B$ and $mathcal C$.

However $mathcal Asubseteqmathcal D=sigma(mathcal B,mathcal C)$ so $mathcal A$ and $sigma(mathcal B,mathcal C)$ are not independent.