$\mathbb{E}[f(X_1+X_2+\cdots+X_{n}) \mid \sigma(X_1,X_2,\cdots,X_{n-1})] =\mathbb{E}[f(X_1+X_2+\cdots+X_{n}) \mid X_1+X_2+\cdots+X_{n-1} ]$

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$f$ is a bounded measurable functions and the $X_i$'s are independent

to prove the above one must show the following equality : $$I = J$$ where : $$I=\int_{A \in \sigma(X_1,X_2,\cdots,X_{n-1})} \mathbb{E}[f(X_1+X_2+\cdots+X_n) \mid X_1+X_2+\cdots+X_{n-1} ]d \mathbb{P}$$

$$J=\int_{A \in \sigma(X_1,X_2,\cdots,X_{n-1})} f(X_1+X_2+\cdots+X_n) d \mathbb{P}$$

clearly if $\sigma(X_1,X_2,\cdots,X_{n-1}) \subset\sigma(X_1+X_2+ \cdots+X_{n-1})$

then the equality holds

and that inclusion is actually correct for the following argument,

elements of $\sigma(X_1,X_2,\cdots,X_{n-1})$ are all intersections of elements of $\sigma(X_1+X_2+ \cdots+X_{n-1})$

is this right ? can someone correct any falseness I said ?