If X,Y and Z are independent, are X and YZ independent?
Solution 1
Let $\mathcal A,\mathcal B,\mathcal C$ be three independent $\sigma$-algebras and let $\mathcal{D}$ be the $\sigma$-algebra generated by $\mathcal{B}$ and $\mathcal{C}$. We want to show that $\mathcal{A}$ and $\mathcal{D}$ are independent.
Indeed: if $\mathcal A,\mathcal B,\mathcal C$ are the natural $\sigma$-algebras of $X,Y,Z$ the product $YZ$ is a measurable function of $Y$ and $Z$, hence is $\mathcal D$-measurable. Therefore, the independence of $\mathcal A$ and $\mathcal D$ implies the independence of $X$ and $YZ$.
Sketch of proof for the independence.
This is standard application of the $\lambda-\pi$ theorem.
Let $\mathcal{G} = \{D \in \mathcal{D} : \forall A \in \mathcal{A},\; P(A\cap D) = P(A)P(D)\}$ and $\mathcal S = \{B\cap C ; B \in \mathcal{B}, C \in \mathcal{C}\}$.
Fact 1. $\mathcal S$ is a $\pi$-system (it is stable by finite intersections).
Comment. This is quite obvious since $\mathcal B$ and $\mathcal C$ are $\sigma$-algebras.
Fact 2. $\mathcal G$ is a $\lambda$-system.
Comment. It should be checked but there is no difficulty.
Fact 3. $\mathcal S \subset \mathcal G$.
Comment. This is a direct consequence of the independence of $\mathcal A,\mathcal B,\mathcal C$.
An application of the $\lambda$-$\pi$ theorem now yields $\sigma(\mathcal{S}) \subset \mathcal{G}$. Since the inclusion $\mathcal G \subset \mathcal D$ is obvious and $\sigma(\mathcal{S}) = \mathcal D$ (by definition), it follows that $\mathcal G = \mathcal D$.
Solution 2
If $X,Y,Z$ are independent then:
$$\mathbb{P}[X \in A, Y \in B Z \in C] =\mathbb{P}[X \in A] \mathbb{P}[Y \in B] \mathbb{P}[Z \in C] $$
consider now
$s^1(x) = \sum_{i =1}^{k_1} 1_{A_i}(x) \alpha_i$
$s^2(y) = \sum_{j =1}^{k_2} 1_{B_j}(y) \beta_j$
$s^3(z) = \sum_{l =1}^{k_3} 1_{C_l}(z) \gamma_l$
Now compute $$\mathbb{E}[s^1(X) s^2(Y) s^3(Z)] = \sum_{i,j,l} \alpha_i \beta_j \gamma_l \mathbb{E}[1_{A_i}(X) 1_{B_j}(Y) 1_{C_l}(Z) ]= \sum_{i,j,l} \alpha_i \beta_j \gamma_l \mathbb{P}[X \in A_i, Y \in B_j Z \in C_l] =\sum_{i,j,l} \alpha_i \beta_j \gamma_l\mathbb{P}[X \in A_i] \mathbb{P}[Y \in B_j] \mathbb{P}[Z \in C_l] = \mathbb{E}[s^1(X)]\mathbb{E}[ s^2(Y)]\mathbb{E}[ s^3(Z)]$$
Edit: Note that by the same argument we have $$ \mathbb{E}[ s^2(Y)]\mathbb{E}[ s^3(Z)] = \mathbb{E}[ s^2(Y) s^3(Z)]$$
Now take $\mathcal{F}_A= \{D \in \mathcal{B}^2: \Bbb{E}[1_A(X) 1_D(Y,Z)] = \Bbb{E}[1_A(X)]\Bbb{E}[1_D(Y,Z)]\}$
Note now that
1) $\emptyset \in \mathcal{F}_A$ 2) $D\in \mathcal{F} \Rightarrow D^c \in \mathcal{F}_A$
$$\Bbb{E}[1_A(X) 1_{D^c}(Y,Z)] = \Bbb{E}[1_A(X) (1 - 1_D(Y,Z)] -\Bbb{E}[1_A(X)] = \Bbb{E}[1_A(X)]-\Bbb{E}[1_A(X) 1_D(Y,Z)] = \Bbb{E}[1_A(X)]-\Bbb{E}[1_A(X)]\Bbb{E}[1_D(Y,Z)] = \Bbb{E}[1_A(X)] (1-\Bbb{E}[1_D(Y,Z)]) = \Bbb{E}[1_A(X)] (\Bbb{E}[1 - 1_D(Y,Z)]) = \Bbb{E}[1_A(X)] (\Bbb{E}[ 1_{D^c}(Y,Z)])$$
3) $D_1,\ldots, D_n ,\ldots \in \mathcal{F}$ disjoints sets then $\cup_i D_i \in \mathcal{F}_A$ $$\Bbb{E}[1_A(X) 1_{\cup_iD_i}(Y,Z)] = \Bbb{E}[1_A(X) \sum_i 1_{D_i}(Y,Z)] = \sum_i \Bbb{E}[1_A(X) 1_{D_i}(Y,Z)] = \sum_i\Bbb{E}[1_A(X)] \Bbb{E}[1_{D_i}(Y,Z)] =\Bbb{E}[1_A(X)] \Bbb{E}[\sum_i 1_{D_i}(Y,Z)] = \Bbb{E}[1_A(X)] \Bbb{E}[ 1_{\cup_i D_i}(Y,Z)] $$
4) $B \times C \in \mathcal{F}_A$
as $B\times C \cap B'\times C' = B\cap B'\times C \cap C'$ we conclude that $\mathcal{F}_A$ is a $\lambda$-system that contains a $\pi$-system. Therefore it contains the $\sigma$- algebra generated by the sets $B\times C$ which is the $\mathcal{B}^2$ (Borel algebra of $\Bbb{R}^2$) (see https://en.wikipedia.org/wiki/Dynkin_system)
Now we now that for every $A \in \mathcal{B}$
$$ \Bbb{E}[1_A(X) 1_D(Y,Z)] = \Bbb{E}[1_A(X)]\Bbb{E}[1_D(Y,Z)]$$
To conclude, denote by $\hat{D} = \{(y,z)\in \Bbb{R}^2 \; \vert \;1_B(yz)=1\} $. Note that $\hat{D} \in \mathcal{B}^2$ therefore $$ \Bbb{E}[1_A(X) 1_B(YZ)] = \Bbb{E}[1_A(X) 1_{\hat{D}}(Y,Z)]= \Bbb{E}[1_A(X)]\Bbb{E}[1_{\hat{D}}(Y,Z)] = \Bbb{E}[1_A(X)]\Bbb{E}[1_{B}(YZ)] $$
So we see that $X$ and $YZ$ are independent
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BCLC
Updated on July 28, 2020Comments
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BCLC over 3 years
- If yes:
I know that f(X) and g(Y) are independent if X and Y are independent and f and g are "measurable".*
If that is to be used, is g(Y) = YZ measurable?
If not, how else to approach this?
- If no:
Counterexample please? ^-^
Possibly related:
Prove that $f(X)$ and $g(Y)$ are independent if $X$ and $Y$ are independent
If $X$ and $Y$ are independent then $f(X)$ and $g(Y)$ are also independent.
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Did over 8 yearsHow is this not a duplicate of the two questions linked last, applied to $\mathbf X=X$, $\mathbf Y=(Y,Z)$, $f$ the identity, $g:(y,z)\mapsto yz$.
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BCLC over 8 years@Did In case you didn't read, I did suggest g(y,z). I think there was this comment (that I guess was deleted) that explains why such is wrong. Could you explain further? If it was that simple, what's the point of Conrado's complicated answer?
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Did over 8 yearsThere is no point (and no, you did not suggest this function $g$ but the absurd $g(Y)=YZ$, not even a function). Simply read the other answers.
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BCLC over 8 years@Did Oh sorry I meant I was close. What I suggested is much closer (in terms of strategy not correctness) to your answer than the answer provided below. I had a feeling there was a short solution. Again, it seems so simple yet there's this extremely complicated answer below. Where are these 'other answers' ? I only see 1 below
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quid over 8 yearsI assume @Did meant answers on the duplicate target and the near identical post you link to.
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BCLC over 8 years@quid In what way are my post and that post near-identical? It just seems e is using a fact asked about elsewhere to answer this question.
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BCLC over 8 yearsUn-duplicated! Hooray!
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quid over 8 yearsPosts do not have to be identical to be duplicates. As I told you on meta the point is that it is a special case. Now, you can also express this as saying it is an application of a general result to a special situation. One may or may not think this justifies a separate question, but once again, the reason it was marked as duplicate is that it is a simple special case of the target.
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BCLC over 8 years@quid I think simplicity is relative.
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quid over 8 yearsWell, yes, which is why I wrote: "One may or may not think this justifies a separate question[...]" Still it does not alter the fact that it is a special case. Do you at least realize now how it is a special case? It is not really my business but at the end of the day I would consider this as a lot more relevant than if this question was marked as duplicate or not.
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BCLC over 8 yearsThanks! What are those alpha, beta, gammas?
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Conrado Costa over 8 yearsthey are the constants that you choose for your simple function.
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Admin over 8 yearsI think your solution needs more work. It is not possible to find functions $s^2_n$ and $s^3_n$ so that $s^2_n(y)s^3_n(z)\to 1_D(yz)$ pointwise.
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Conrado Costa over 8 yearswright, If $D$ is of the form $B \times C$ then it is possible
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Admin over 8 yearsOops. $D$ is a subset of $\mathbb R$, not ${\mathbb R}^2$.
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BCLC over 8 yearsWhat simple function? Is that something you use to 'approximate' X, Y and Z ('Approximate' in the sense that you use MCT on them or something).
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BCLC over 8 yearsWhat is $s_n^1$ ?
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BCLC over 8 yearsHow do you get $\lim_n \mathbb{E}[s_n^1(X) s_n^2(Y) s_n^3(Z)] = \lim_n \mathbb{E}[s_n^1(X)]\mathbb{E}[ s_n^2(Y) s_n^3(Z)]$ ?
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Conrado Costa over 8 yearsThis step goes by linearity. use the following $$\Bbb{P}[X \in A, Y \in B]= \Bbb{E} [1_A(X) 1_B(Y)] $$ note that $\alpha_1\beta_1\Bbb{P}[X \in A_1, Y \in B_1]+ \alpha_2\beta_1\Bbb{P}[X \in A_2, Y \in B_1] = \alpha_1\beta_1\Bbb{E}[1_{A_1}(X) 1_{ B_1}(Y)]+ \alpha_2\beta_1\Bbb{E}[1_{A_2}(X) 1_{ B_1}(Y)] =\Bbb{E}[(\alpha_1 + \alpha_2)1_{A_1}(X)\beta_1 1_{ B_1}(Y)] $ and so on
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Admin over 8 years@ConradoCosta This solution is still completely wrong. I suggest you work through it very carefully to try and fix it. I have temporarily downvoted (sorry!).
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Conrado Costa over 8 yearsYou are a careful reader, thank you. Is it ok now?
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Admin over 8 years@ConradoCosta Much better, I have upvoted.
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BCLC over 8 yearsThanks Siméon. It's about time I learned about pi and lambda systems. Do you think @Did's answer is right?
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Siméon over 8 years@BCLC: Did's comment makes sense if the lemma about $\sigma$-algebras is known. His comment actually corresponds to the first part of my answer.
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BCLC over 8 yearsSo your and Conrado Costa's are, um, inelegant?
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Siméon over 8 years@BCLC: It depends on what tools are at your disposal, but I think that you cannot totally escape this proof at some point in the theory. Moreover, it is important to learn how to use the $\lambda-\pi$ machinery for proving results in measure theory.
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BCLC over 8 yearsGood point. I guess my probability and stochastic calculus classes likely would've continued initially more towards what you were doing rather than what Conrad or Did were doing.
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BCLC over 8 yearsBtw did you mean $\subseteq$ somewhere there?
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Siméon over 8 yearsAs far as I know, there is no difference between $\subset$ and $\subseteq$.
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BCLC over 8 yearsSiméon, I think some people (and you?) use $\subset$ to indicate the possibility of equality meanwhile using $\subsetneq$ to indicate strict inclusion or something. Meanwhile some authors use $\subset$ to indicate strict inclusion. Anyway, I guess you mean inclusion is not strict.
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BCLC almost 8 yearsWhy is this a sketch only?