If X,Y and Z are independent, are X and YZ independent?

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?

2. 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.

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 $\mathcal G = \mathcal D$.

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