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Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and let $X,Y: \Omega \to \mathbb{R}$ be i.i.d. random variables such that $\mathbb{P}(X=1)=\mathbb{P}(X=-1)=\frac{1}{2}$. Define $XY: \Omega \to \mathbb{R}$ by $XY(\omega) = X(\omega)Y(\omega)$ for all $\omega \in \Omega$. Now, I want to prove that the two random variables $X$ and $XY$ are independent.

I know that the two random variables are independent if $\sigma(X)$ and $\sigma(XY)$ are independent. I know that $\sigma(X)$ and $\sigma(XY)$ are independent if \begin{align} (*) \qquad \forall S \in \sigma(X), \forall T \in \sigma(XY): \mathbb{P}(S \cap T) = \mathbb{P}(S)\mathbb{P}(T). \end{align} Since $X$ and $Y$ are i.i.d. random variables, I know that the events $X^{-1}((\infty,a])$ and $Y^{-1}((\infty,b])$ are independent for all $a,b\in \mathbb{R}$. How to conclude that equation $(*)$ holds?

Subsequently, it could be shown that $Y$ and $XY$ are independent. However, the three random variables $X$, $Y$ and $XY$ are not independent. Why?

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    $\begingroup$ My suggestion would be to stop being drowned into a sea of irrelevant formalism and to try to think about the situation... Both $X$ and $XY$ are always $\pm1$ hence they are independent if and only if $$P(X=x,XY=z)=P(X=x)P(XY=z)$$ for only four couples $(x,z)$. Can you identify these couples and check the identity holds for all of them? Re the non independence of $(X,Y,XY)$, find $(x,y,z)$ such that $$[X=x,Y=y,XY=z]=\varnothing$$ although $$P(X=x)P(Y=y)P(XY=z)\ne0$$ $\endgroup$
    – Did
    Dec 24 '16 at 18:01
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I will first answer your last question: given $X$ and $Y$ (as a vector), the random variable $XY$ is completely determined. So, you can easily finda three sets $A,$ $B$ and $C$ such that $\mathbf{P}(XY \in A \mid X \in B, Y \in C) \neq \mathbf{P}(XY \in A).$

For your first question, I will give you some results, I am quite unsure which one is the one you are missing.

Suppose $X$ and $Y$ are two discrete random variables and let $\Sigma(X)$ and $\Sigma(Y)$ the corresponding sigma algebras they generate.

  • If $\Sigma(X)$ and $\Sigma(Y)$ are independent, every $A \in \Sigma(X)$ is independent of every $B \in \Sigma(Y);$ in particular when $A = \{X = x_n\}$ and $B = \{Y = y_m\}.$
  • If every $\{X = x_n\}$ is independent of every $\{Y = y_m\},$ then for every pair of intervals $I,J$ the events $\{X \in I\}$ and $\{Y \in J\}$ are independent for $\mathbf{P}(X \in I, Y \in J\} = \sum\limits_{x_n \in I, y_m \in J} \mathbb{P}(X = x_n, Y = y_m) = \sum\limits_{x_n \in I, y_m \in J} \mathbb{P}(X = x_n) \mathbb{P}(Y = y_m) = \sum\limits_{x_n \in I} \mathbb{P}(X = x_n) \sum\limits_{y_m \in J} \mathbb{P}(Y = y_m) = \mathbb{P}(X \in I) \mathbb{P}(Y \in J).$ Then, $\Sigma(X)$ and $\Sigma(Y)$ are independent (let me clarify this last claim below).

For a function $f$ to be measurable from the measurable space $(X, \mathbf{X})$ to the measurable space $(Y, \mathbf{Y})$ it is necessary and sufficient that the set $f^{-1}(\mathbf{Y}) := \{f^{-1}(B) \mid B \in \mathbf{Y}\}$ be contained in $\mathbf{X}.$ This is just restating the definition of measurability. Now, $f^{-1}(\mathbf{Y})$ is a sigma algebra because preimage respect operations and so on. If $\mathbf{Y}$ is the sigma algebra generated by some set $\mathsf{Y} \subset 2^Y$ then $f^{-1}(\mathbf{Y})$ is the sigma algebra generated by $f^{-1}(\mathsf{Y})$ (since the Borel sigma algebra is generated by intervals of the form $(-a, \infty]$, you can apply this to $\mathbf{Y} = \text{Borel sigma algebra}$ and $\mathsf{Y} = \text{Intervals of the form } (-\infty, a]$). Since $\mathsf{Y} \subset \mathbf{Y}$ it is true that $f^{-1}(\mathsf{Y}) \subset f^{-1}(\mathbf{Y})$ and hence, as the right hand side is a sigma algebra, the sigma algebra generated by the left hand side will be contained in $f^{-1}(\mathbf{Y}).$ Let $\mathcal{Y}$ be the sigma algebra generated by $f^{-1}(\mathsf{Y}),$ I just established that $\mathcal{Y} \subset f^{-1}(\mathbf{Y});$ to prove the reverse inclusion consider the set $\bar{\mathsf{Y}}$ of subsets $B$ of $Y$ such that $f^{-1}(B) \in \mathcal{Y};$ so $\mathsf{Y} \subset \bar{\mathsf{Y}}.$ Since preimage behaves well with set operations, $\bar{\mathsf{Y}}$ is a sigma algebra that contains $\mathsf{Y},$ thus it contains the sigma algebra generated by the latter set, that is $\mathbf{Y} \subset \bar{\mathsf{Y}}.$ Therefore, for every $B \in \mathbf{Y},$ $f^{-1}(B) \in \mathcal{Y},$ thus $\mathcal{Y} = f^{-1}(\mathbf{Y}).$

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    $\begingroup$ It seems the first sentence of my comment on main applies to more users than I expected when I wrote it... $\endgroup$
    – Did
    Dec 24 '16 at 20:47
  • $\begingroup$ @Did I am afraid I disagree enormously with your first sentence. Formality is what allows some mathematician's minds to think and create. Besides, some people can't think in terms of intuition only. I am one of those mathematicians that needs the underground formality and then the intuition. If you like one method, good for you (you can try to convice people why you think your method is better). Just don't be a smug and think your method is universally better. $\endgroup$
    – William M.
    Dec 24 '16 at 21:38
  • $\begingroup$ Yeah, and in the present case, what you call formality is not what this OP needs. Did you try to have a look at their other recent questions? $\endgroup$
    – Did
    Dec 24 '16 at 22:41
  • $\begingroup$ Of course, for example: math.stackexchange.com/questions/1665533/… Clearly the person wants to learn measure theoretic techniques in random variables. I am not pursuing further any discussion, but if the person wants help, I'll help if I can. You can criticise, you are free to do so. $\endgroup$
    – William M.
    Dec 24 '16 at 22:48
  • $\begingroup$ Merci beaucoup, I really do appreciate both type of answers. Last but not least, merry christmas! $\endgroup$
    – iJup
    Dec 25 '16 at 9:13

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