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  1. 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?

  1. If no:

Counterexample please? ^-^

Possibly related:

How do you prove that if $ X_t \sim^{iid} (0,1) $, then $ E(X_t^{2}X_{t-j}^{2}) = E(X_t^{2})E(X_{t-j}^{2})$?

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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    $\begingroup$ How is this not a duplicate of the two questions linked last, applied to $\mathbf X=X$, $\mathbf Y=(Y,Z)$, $f$ the identity, $g:(y,z)\mapsto yz$. $\endgroup$ – Did Jul 24 '15 at 9:43
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    $\begingroup$ @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? $\endgroup$ – BCLC Jul 24 '15 at 10:05
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    $\begingroup$ There 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. $\endgroup$ – Did Jul 24 '15 at 10:08
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    $\begingroup$ @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 $\endgroup$ – BCLC Jul 24 '15 at 14:25
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    $\begingroup$ Posts 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. $\endgroup$ – quid Jul 24 '15 at 16:38
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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 $\mathcal G = \mathcal D$.

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  • $\begingroup$ Thanks Siméon. It's about time I learned about pi and lambda systems. Do you think @Did's answer is right? $\endgroup$ – BCLC Jul 24 '15 at 17:13
  • $\begingroup$ @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. $\endgroup$ – Siméon Jul 24 '15 at 17:19
  • $\begingroup$ So your and Conrado Costa's are, um, inelegant? $\endgroup$ – BCLC Jul 24 '15 at 17:21
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    $\begingroup$ @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. $\endgroup$ – Siméon Jul 24 '15 at 17:26
  • $\begingroup$ Good 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. $\endgroup$ – BCLC Jul 25 '15 at 9:08
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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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    $\begingroup$ I 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. $\endgroup$ – user940 Jul 2 '15 at 20:16
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    $\begingroup$ wright, If $D$ is of the form $B \times C$ then it is possible $\endgroup$ – Conrado Costa Jul 2 '15 at 20:30
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    $\begingroup$ Oops. $D$ is a subset of $\mathbb R$, not ${\mathbb R}^2$. $\endgroup$ – user940 Jul 2 '15 at 20:37
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    $\begingroup$ This 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 $\endgroup$ – Conrado Costa Jul 4 '15 at 14:32
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    $\begingroup$ @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!). $\endgroup$ – user940 Jul 4 '15 at 19:28

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