$\mathbb{E}(\varphi(X, Y) | \mathcal{G}) = \mathbb{E}(\varphi(X, Y))$ What's the name for this proposition? Or alternatively what's happening?

Question

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:

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

Edit 1 for searchability: It's called freezing lemma. I like to call this independence-stability generalisation.

Edit 2: Btw I'm not sure the remark (2) is right unless $X$ & $Y$ are independent, in w/c case obviously you choose $\mathcal G = \sigma(Y)$.

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
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
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 $ — , 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
For the generalisation that does not assume independence, see https://math.stackexchange.com/questions/1366543 , https://math.stackexchange.com/questions/3451609, https://math.stackexchange.com/questions/4294255/, https://math.stackexchange.com/questions/4444874/ — paperskilltrees, Jun 04 '25 at 22:09

BCLC · Accepted Answer · 2024-08-22T23:41:42.510

No idea what I was/wasn't thinking before, but this is very obviously a generalisation of a combination of the ff 2 properties:

Pulling out independent factors
Stability

Perhaps this could be called the 'independence-stability generalisation'

Consider the ff 2 examples:

Example 1 - generalisation is not needed

$$\varphi(X, Y) = X\sin(Y); \mathcal{G} = \sigma(Y)$$

Then

$$\mathbb{E}(\varphi(X, Y) | \mathcal{G})$$

$$ = \mathbb{E}(X\sin(Y) | Y)$$

$$ = \sin(Y)\mathbb{E}(X | Y) \ \text{, by stability}$$

$$ = \sin(Y)\mathbb{E}(X) \ \text{, by pulling out independent factors}$$

Example 2 - needed

$$\varphi(X, Y) = \sin(XY); \mathcal{G} = \sigma(Y)$$

Then

$$\mathbb{E}(\varphi(X, Y) | \mathcal{G})$$

$$ = \mathbb{E}(\sin(XY) | Y)$$

$$ = \mathbb{E}(\sin(X)y)|_{y=Y} \ \text{, by freezing lemma or independence-stability generalisation}$$

Freezing Lemma in Example 1 would look like

$$ = \mathbb{E}(X\sin(Y) | Y)$$

$$ = [\sin(y)\mathbb{E}(X)]|_{y=Y} \ \text{by freezing lemma}$$

$$ = [\sin(y)|_{y=Y}\mathbb{E}(X)]$$

$$ = \sin(Y)\mathbb{E}(X)$$

$\mathbb{E}(\varphi(X, Y) | \mathcal{G}) = \mathbb{E}(\varphi(X, Y))$ What's the name for this proposition? Or alternatively what's happening?

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