[Math] A question about independence of sigma algebras (generated by random variables)

measure-theoryprobability theoryrandom variablesrandom walkstochastic-processes

Let $X_1, X_2, \ldots$ i.i.d random variables. Is it possible that

$$\{X_{n+1} \in B\} \in \sigma({X_1, \ldots, X_n})$$ for some $B$? Why yes/not?

I want to show that $\sigma(X_{n+1})$ and $\sigma(X_1, \ldots, X_n)$ are independent. Is it possible to have in $\sigma(X_1, \ldots, X_n)$ events that are not independent of $\{ X_{n+1} \in B \}$, for some $B$? Why yes/not?

Notation: $\sigma(X)$ = $\sigma$-algebra generated by $X$.

Thank you!

Best Answer

Hint: If the event $A=\{X_{n+1} \in B\}$ is in $\sigma(X_1, \ldots, X_n)$ then $A$ is independent of $A$ hence...

(The exercise uses only the independence of the random variables $X_n$, not that they are identically distributed.)

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[Math] Are random variables independent of their tail sigma-algebra

[This answers the original edition of the question, which did not assume $X_1,X_2,X_3,\ldots$ are independent.]

No. Suppose $R\sim\mathrm{Uniform}(0,1)$ and $(Y_1,Y_2,Y_3,\ldots)\mid R\sim\mathrm{i.i.d. Bernoulli}(R)$.

Then the strong law of large numbers implies that $$ \Pr\left( \lim_{n\to\infty} \frac{Y_1+\cdots+Y_n} n= R \mid R\right) = 1. $$ So \begin{align} & \Pr\left( \lim_{n\to\infty} \frac{Y_1+\cdots+Y_n} n= R \right) \\[10pt] = {} & \operatorname{E} \left( \Pr\left( \lim_{n\to\infty} \frac{Y_1+\cdots+Y_n} n= R \mid R\right) \right) = \operatorname{E}(1) = 1. \end{align}

Let $X_n= (Y_1+\cdots+Y_n)/n$. The event that $\lim_{n\to\infty} X_n = r$ is in the tail sigma-algebra of $X_1,X_2,X_3,\ldots$. But $X_1$ is not independent of that event, since $\Pr(X_1=1\mid R=r)=r$.

If you want a tail event whose probability is positive, observe that $\lim\limits_{n\to\infty} X_n \overset{\text{a.s.}}= R \sim\mathrm{Uniform}(0,1)$, so $\Pr(X_1=1) = \operatorname{E}(\Pr(X_1=1\mid \lim\limits_{n\to\infty} X_n)) = \operatorname{E}(\lim\limits_{n\to\infty} X_n) = 1/2$, and find $\Pr(X_1=1\mid \lim\limits_{n\to\infty} X_n>1/2)$.

[Math] I.i.d. random variables and independent $\sigma$-algebras

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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[Math] Are random variables independent of their tail sigma-algebra

[Math] I.i.d. random variables and independent $\sigma$-algebras

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