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The first question I have is: For conditionally independent (given $Z$), random variables $X$ and $Y$, does $\mathbb E [X | Y,Z] = \mathbb E [X | Z]$?

And I also wanted to know if when $X$ is independent of $Y$ and $X$ is independent of $Z$, does $\mathbb E [X | Y,Z] = \mathbb E [X]$? What about if $Y$ and $Z$ were also independent (so all $3$ random variables are pairwise independent)?

For the first question, I know that if $\sigma(Y)$ was independent of the smallest sigma algebra containing $\sigma(X)$ and $\sigma(Z)$ then it would be true, but conditional independence doesn't imply this. I still think it is true but not sure how to show or prove it. I showed it using the expectation equation with probability densities and that $p(x|y,z) = \frac{p(xy|z)}{p(y|z)}$. But I am not sure if this is the correct.

\begin{equation} \mathbb E [X | Y,Z] = \int x p(x|y,z)dx = \int x\frac{p(xy|z)}{p(y|z)}dx \\\text{ using conditional independence } = \int x\frac{p(x|z)p(y|z)}{p(y|z)}dx = \int x p(x|z)dx = \mathbb E [X | Z] \end{equation}

For the second question, I don't believe it is true since the smallest sigma algebra containing $\sigma(Y)$ and $\sigma(Z)$ contains more information than each of them alone. And I'm not sure about the expression when $Y$ and $Z$ are also independent.

Thanks in advance for the help!

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2 Answers 2

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Hint to 2nd question: Whenever I think about pairwise independence, the first thing I check is the famous example:

  • $X, Y$ are i.i.d. Bernoulli variables with $P(X=1) = P(X=0) = 1/2$.

  • $Z = 1$ if $X = Y$, otherwise $Z=0$.

As is well known, these $3$ are pairwise independent but not mutually independent - any two of them determine the third.

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For your first question, have you tried to prove this if $X$ has densities $p(x | y, z)$ and $p(x | z)$? In this case you can write $\mathbb E [ X | Y, Z] = \int x p(x | Y, Z) dx$ and proceed from there.

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