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there are many questions on Mathexchange, e.g. Definition of conditional expectation, discussing the definition of conditional expectation. In most cases the random variable which is later named "conditional expectation" is assumed to be a measurable $\Omega\rightarrow\mathbb R$ function. In some lecture notes (that are maybe too advanced for me) conditional expectation is defined to be a $\Omega\rightarrow\mathbb R^n$ measurable function -- but imposes in the definition that this function is a.e. unqiue. For me it feels that we do not need a.e. uniqueness in the definition but that it can be proven (I failed though). So how is uniqueness of conditional expectation proven in $\mathbb R^n$ or even more general, in Banach spaces?

Edit: Interestingly, this five-year-old question receives some attention now. The conditional expectation $Z$ of a random variable $X:\Omega\rightarrow S$ given a sub-$\sigma$-algebra $\mathcal G$ satisfies $$\int_G Z\,\mathrm dP = \int_G X\,\mathrm dP$$ for all $G\in\mathcal G$. The common notation is $\mathbb E[X\mid\mathcal G] := Z$.

It can be shown in the case $S = \mathbb R$ that if such a random variable exists, it is unique. The proof can be found e.g. here: Proof that conditional expectation is defined uniquely almost everywhere? This proof does not work for $S = \mathbb R^n$, or more general $S = $ some Banach space. The question is: how can I prove uniqueness in these cases?

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  • $\begingroup$ Can you post a reference that define conditional expectation for random vector/element in a non-trivial way? To the best of my knowledge, even for the concept "(unconditional) expectation", it is normally defined for random variables -- and is generalized to random vector/element by collecting their component scalar expected values. For example, it is customary, for fixed $s < t$, to talk about $E(W_t | \mathscr{F}_s)$ for a Brownian motion $\{W_t\}$, but I have never seen any author discussed $E(\{W_t\}|\mathscr{F}_s)$ for the whole Brownian motion. $\endgroup$
    – Zhanxiong
    Nov 9, 2022 at 0:52
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    $\begingroup$ Regarding the proof you linked in the edit. What prevents you from applying the same arguments as in there to a vector-valued r.v.'s components or to its norm? I believe, all we need is to show $\pi_{i}:(\mathbb{R}^{n},\mathfrak{B}(\mathbb{R}^{n})\to(\mathbb{R},\mathfrak{B}(\mathbb{R})$ and $\|\cdot\|:(\mathbb{R}^{n},\mathfrak{B}(\mathbb{R}^{n})\to(\mathbb{R},\mathfrak{B}(\mathbb{R})$ are measurable, right? $\endgroup$
    – paperskilltrees
    Nov 9, 2022 at 16:46
  • $\begingroup$ You are right, I can apply the same reasoning componentwise. Thank you! $\endgroup$
    – Syd Amerikaner
    Nov 9, 2022 at 18:17

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The additional imposed condition (of the uniqueness up to a set of measure zero) in the definition for $\mathbb{R}^n$ might be just a way of phrasing it and condensing several statements into one. I am not aware of any principal difference between conditional expectations of variables taking values in $\mathbb{R}$ and $\mathbb{R}^n$. There could be some room for a formal subtlety of whether you define the expectation as a random variable or as a class of equivalence of random variables. I believe it is most convenient to define the conditional expectation as any representative of the equivalence class, and then proceed to establishing that if the measurable spaces are nice, then we can choose representatives from the classes of equivalence in such a way that the family of functions $$\Sigma\ni A \mapsto P(A|t):=\mathbb{E}[1_{A}| T=t],\quad \forall t\in \mathrm{range}(T)$$ has the properties we expect from a conditional probability measure $P(\cdot|t)$. Here $\mathbb{E}[1_{A}| T=t]$ is a measurable function from $\mathrm{range}(T)$, which are defined using Doob-Dynkim lemma, as described in this Wikipedia article. Note that $\mathbb{E}[1_{A}| T]$ and $\mathbb{E}[1_{A}| T=t]$ are different objects. See e.g. Chapter 2.5 in Testing Statistical Hypotheses by Lehmann and Romano for details on choosing the right representatives from the equivalence classes.

Your question about proving the uniqueness up to a set of measure zero is answered here. [Edit]

I don't feel qualified to talk about general Banach spaces, but the starting point is the Bochner Integral.

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    $\begingroup$ Thank you for your reply, but I don't see how the linked question proves uniqueness $\endgroup$
    – Syd Amerikaner
    Nov 9, 2022 at 9:52

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