Suppose $\tilde{X}=(X_1,X_2,\dots)$ is a sequence of RVs on $(\Omega,\mathcal{B})$. Prove that $\sigma(\tilde{X})$ is generated by events of the form:

$$\bigcap_{i=1}^m \{X_i\leq x_i\}\mbox{ for }x_1,\dots,x_n \in\mathbb{R}, n\geq 1.$$


I do know that if $X:(\Omega,\mathcal{B})\rightarrow(\Omega^',\mathcal{B}^')$ then $\sigma(X) = \{\{X\in B\}:B\in\mathcal{B}^'\}$.

I also worked in the past to prove that if I take a $\pi$-class $\mathcal{P}=\{(-\infty,a]:a\in\mathbb{R}\}$ and the class of cylinder events in $\mathbb{R}^{\infty}$ determined by $\mathcal{P}$. That is:

$\mathcal{C}=\{\omega=(x_1,x_2,\dots):x_i\leq b_i,i\leq n\}$ for some $b_i\in\mathbb{R},i\leq n, n\geq 1$ is a $\pi$-class and $\sigma(\mathcal{C})=\mathcal{B}(\mathbb{R}^{\infty})$.

Attempted solution:

$\mathcal{C}=\{\bigcap_{i=1}^n\{X_i\le x_i\},x_1,\dots,x_n\in\mathbb{R},n\ge 1\};\quad \sigma(\tilde{X})=\{X_n\le a\},n\in\mathbb{N},a\in\mathbb{R}\\ \{X_n\le a\}=\bigcup_{k=1}^\infty \left ( (\{X_n\le a\}) \cap(\bigcap_{i=1}^{n-1}\{X_n\le k\})\right )\\ \\ k\in\mathbb{N}:k>\max_{1\le i\le n-1}X_i(\omega)\Rightarrow\omega\in \{X_n\le k\},\forall i\in\{1,...,n-1\}\\ \Rightarrow\{X_n\le a\}\in\sigma(\mathcal{C})\\ \Rightarrow\{X_n\le a\}\subset\sigma(\tilde{X})\Rightarrow \sigma(\tilde{X})\subset\sigma(\mathcal{C})$

Problems, etc.:

A critique of my solution: You have the general idea but it seems you are confusing classes of events.  1st, you have a sigma-field on the left and a single event on the right.  But even if you meant the class of all such events, it would not come anywhere close to being a sigma-field.  For example, it does not include $\{\limsup X_n \le a\}$. 2nd, the object is to show that inverse images of all sets in $B(R^\infty)$ are in the sigma-field generated by $C$.  However, it suffices to consider inverse images of cylinder sets.Finally, cylinder sets are not finite dimensional; they are subsets of $R^\infty$.  They do have obvious finite-dimensional counterparts, but don't confuse the two kinds of sets.

If you can help me clean this up and correct it, I would be greatly appreciated.

  • 1
    $\begingroup$ By the way, it is different from saying that we have a random variable $\tilde{X}:\Omega\to\mathbb{R}^{\infty}$ given by $\tilde{X}(\omega)=(X_{1}(\omega),X_{2}(\omega),...)$ than saying that $\tilde{X}=(X_{i})_{i=1}^{\infty}$ is a sequence of random variables. When looking at the $\sigma$-algebras they generate, in first case you need to consider preimages of Borel sets in $\mathbb{R}^{\infty}$ and in latter you need to consider preimages or Borel sets in $\mathbb{R}$ for each $X_{n}$. And whether these two agree depends on the topology of $\mathbb{R}^{\infty}$. Which one did you mean? $\endgroup$ – T. Eskin Jun 17 '12 at 8:56
  • $\begingroup$ @ThomasE., the case I am considering is the former example you mention. $\endgroup$ – Justin Jun 17 '12 at 18:07
  • $\begingroup$ Okey. Just as I had assumed too. $\endgroup$ – T. Eskin Jun 17 '12 at 18:51
  • $\begingroup$ @THomasE., thank you for all of your help. You might be able to help me with another question (and not that it matters, but I did set up a bounty for that one).math.stackexchange.com/questions/155446/… $\endgroup$ – Justin Jun 17 '12 at 19:09

The second last implication $\{X_{n}\leq a\}\in \sigma(\mathscr{C})\Rightarrow \{X_{n}\leq a\}\subset \sigma(\tilde{X})$ is false, and I don't see how it is relevant here. One reason why this makes no sense is that $\{X_{n}\leq a\}$ is a subset of $\Omega$ and $\sigma(\tilde{X})$ is a subset of the power-set of $\Omega$. This also means that in start you should correct the definition of $\sigma(\tilde{X})$, the right way to write it would be $\sigma(\tilde{X})=\sigma(\{\{X_{n}\leq a\}:a\in\mathbb{R},\,\,n\in\mathbb{N}\})$. The last implications should instead be just: \begin{align*} &\{X_{n}\leq a\}\in\sigma(\mathscr{C}) \\ &\Rightarrow \sigma(\tilde{X})\subset \sigma(\mathscr{C}), \end{align*} otherwise it's correct. And this is the case because you just showed that $\{X_{n}\leq a\}\in \sigma(\mathscr{C})$ for all $a\in\mathbb{R}$ and $n\in\mathbb{N}$. Since $\sigma(\tilde{X})$ is the smallest $\sigma$-algebra that contains such sets and $\sigma(\mathscr{C})$ is some $\sigma$-algebra containing them, then $\sigma(\tilde{X})\subset \sigma(\mathscr{C})$.

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