1
$\begingroup$

I'm trying to prove the following statement:

If \begin{equation*} \left( X_1, X_2, \ldots, X_n\right) \stackrel{D}{=}\left( X_1, Y_2, \ldots, Y_n\right) \end{equation*} then \begin{equation*} \mathbb{E}[X_1|X_2, \ldots, X_n]\stackrel{D}{=}\mathbb{E}[X_1|Y_2, \ldots, Y_n] \end{equation*}

I have tried to use the definition of conditional expectation but this has gotten me nowhere. Does anyone have any idea?

$\endgroup$
2
$\begingroup$

Notice that there exists a measurable function $g\colon \mathbb R^{n-1}\to\mathbb R$ such that $$\mathbb{E}\left[X_1|X_2, \ldots, X_n\right]=g\left(X_2,\dots,X_n\right)\mbox{a.s.}, $$ by Doob-Dynkin's lemma. Similarly, there exists a measurable function $h\colon \mathbb R^{n-1}\to\mathbb R$ such that $$\mathbb{E}\left[X_1|Y_2, \ldots, Y_n\right]=h\left(Y_2,\dots,Y_n\right)\mbox{a.s.} $$ Let $B$ be a Borel subset of $\mathbb R^{n-1}$. Then, by assumption, $$\mathbb E\left[X_1\mathbb 1_{(X_2,\dots,X_n)\in B }\right]=\mathbb E\left[X_1\mathbb 1_{(Y_2,\dots,Y_n)\in B }\right],$$ and using the definition of conditional expectation $$\mathbb E\left[g\left(X_2,\dots,X_n\right)\mathbb 1_{(X_2,\dots,X_n)\in B }\right]=\mathbb E\left[h\left(Y_2,\dots,Y_n\right) \mathbb 1_{(Y_2,\dots,Y_n)\in B }\right].$$ Conclude by using the fact that $\left(X_2,\dots,X_n\right)$ has the same distribution as $\left(Y_2,\dots,Y_n\right)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.