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If there isn't a name for this like Doob-Dynkin-Brown-Markov Tower Lemma / Theorem, then at least what's going on here so that I can describe this proposition in words?

(I guess the ff is in probability space $(\Omega, \mathcal F, \mathbb P)$.)

Let $\mathcal{G}$ be a sub-$\sigma$-field of $\mathcal{F}, X, Y$ be two random variables such that $X$ is independent of $\mathcal{G}$ and $Y$ is $\mathcal{G}$-measurable, and let $\varphi: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a Borel-measurable function such that $\mathbb{E}(|\varphi(X, Y)|)<$ $+\infty$. Then $$ \mathbb{E}(\varphi(X, Y) | \mathcal{G}) = \psi(Y) \quad \text { a.s., } \quad \text { where } \psi(y)=\mathbb{E}(\varphi(X, y)). $$

From here:

  1. If $X$ is independent of $\mathcal{G}$ and $Y$ is $\mathcal{G}$-measurable, then $\mathbb{E}(\varphi(X, Y) | \mathcal{G}) = \mathbb{E}(\varphi(X, Y))$

  2. Why is Borel measurability needed here?

  3. Example 4.1.7 of Durrett PTE

Also I do recall it is in my class notes from before:

enter image description here

BCLC
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    For some people it's the Freezing lemma. There does not seem to be consensus on it. See for instance https://math.stackexchange.com/questions/3607972/do-you-know-this-result-about-conditional-expectation – Will Jan 15 '23 at 10:12
  • @Will ok thanks. in this case what is 'frozen' here? $Y$? I mean like instead of 'tower rule' I can say 'in a double expectation, the smaller $\sigma-$algebra wins.' What about this? – BCLC Jan 28 '23 at 02:26
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    Yes the thing frozen here is $Y$, because the expected value for a certain outcome $\omega$ is the same as if we computed it with $Y$ inside frozen equal to $y=Y(\omega)$. But that's just a point of view, which maybe does not make sense for everyone, which would explain why there is not a strong consensus on this lemma's name. For me it makes sense and I am happy to call this the freezing lemma. – Will Jan 28 '23 at 22:06
  • @Will oh very nice. thanks. post as answer? – BCLC Feb 07 '23 at 08:41
  • See also https://math.stackexchange.com/q/4618609/66096 – Gabriel Romon Feb 07 '23 at 08:53
  • the statement on the title is wrong, is not true that $\operatorname{E}[\varphi (X,Y)|\mathcal{G}]=\operatorname{E}[\varphi (X,Y)]$, what is true is that $\operatorname{E}\varphi (X,Y)|\mathcal{G}=\operatorname{E}[\varphi (X,Y(\omega ))]$ for almost every $\omega \in \Omega $ – Masacroso Feb 07 '23 at 08:56
  • @Masacroso it's like how several youtube titles are 'wrong'. There isn't enough space? Or feel free to make an edit proposal. – BCLC Feb 12 '23 at 08:28

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