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As a motivating example, let $Z_i, i=1,...,n$, be iid random elements, T a real-valued function in $(Z_1,...,Z_n)$, and $X_i, i=1,...,n$ are iid that are independent with $Z_i$'s; consider $P(T(Z_1,...,Z_n)\leq t|Z_{(1)}=X_1,...,Z_{(n)}=X_n, X_1=x_1,...,X_n=x_n)$.

It appears intuitive that we can simplify the above expression by manipulating the conditional part: $P(T(Z_1,...,Z_n)\leq t|Z_{(1)}=x_1,...,Z_{(n)}=x_n, X_1=x_1,...,X_n=x_n)$ Then, by independence between $Z$ and $X$, this equals $P(T(Z_1,...,Z_n)\leq t|Z_{(1)}=x_1,...,Z_{(n)}=x_n)$.

However, if the random elements are non-discrete, so that the conditional part has probability zero, then the conditional probability is in general defined using the measure-theoretic version of conditional expectation (via Radon-Nikodym derivative).

Question:

What properties of conditional expectation allow us to perform this kind of operation (swap equivalent events within the conditional part)?

If this operation does not hold in general, then under what condition does it hold? (e.g. does the use of regular conditional probability help?)

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Let $f$ be Borel measurable s.t. $E[|f(Z)|]<\infty$. Let $X,Z$ be not necessarily independent. Let $\mathscr{G}=\sigma(\sigma(X=Z)\cup\sigma(X))$ where $\sigma(X=Z)=\{\emptyset,\{X=Z\},\{X\neq Z\},\Omega\}$. Note, by taking the conditional expectation on both sides, $$E[f(Z)|\mathscr{G}]=f(X)\mathbf{1}_{\{X=Z\}}+E[f(Z)|\mathscr{G}]\mathbf{1}_{\{X\neq Z\}}\implies E[f(Z)|\mathscr{G}]\mathbf{1}_{\{X=Z\}}=f(X)\mathbf{1}_{\{X=Z\}}$$ So if $\omega\in \{X=Z\}$ we have $E[f(Z)|\mathscr{G}](\omega)=f(X(\omega))$.

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  • $\begingroup$ Thank you for your proof. I'm wondering if we can show a more general statement that $E[f(Y)|I_{Z=X},X]$ and $E[f(Y)|I_{Z=X}]$ match a.s. over $w\in\{Z=X\}$ $\endgroup$
    – NXWang
    Jan 26, 2023 at 20:29
  • $\begingroup$ @NXWang would $Y$ you introduced be another rv? $\endgroup$
    – Snoop
    Jan 26, 2023 at 20:35
  • $\begingroup$ Yes, and it may or may not have a relation with X and Z. I just want to see if we can swap events in the conditional part regardless of the front part. Thanks $\endgroup$
    – NXWang
    Jan 26, 2023 at 20:43
  • $\begingroup$ @NXWang ok, I thought it was a typo. You may open another question for that then :) $\endgroup$
    – Snoop
    Jan 26, 2023 at 21:42

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