1
$\begingroup$

Consider two random variables $X$ and $Y$ defined on the same probability space $(\Omega,\sigma,P)$. We know that they are equivalent in the sense that $P(\{X \ne Y\})=0$. Let $A_X$ and $A_Y$ be the sigma algebras generated by these random variables. What can we say about $A_X$ and $A_Y$? Are they the same?

PS: I had this doubt when looking at martingales, since if the two sigma algebras come out to be different it seems to me that we can modify the characteristic of a process of being adapted with respect to a given filtration simply by changing the process over a set of measure zero. Which I find counterintuitive.

$\endgroup$
2
  • $\begingroup$ Of course the sigma algebras can differ, particularly when X does not agree with Y on a set of measure zero. This need not cause you strife for martingales though- in fact we should expect for measure zero sets to make all the difference in the world for martingales. Consider what happens when we condition based upon two different paths of the process determined by the martingale, ie, if $S_0 = 1$ and $T_0=2$ but both were equal before conditioning at 0- we will still have different expectations for the future path of the processes, despite the fact the we wouldn't have before we conditioned. $\endgroup$
    – Micah
    Jul 5, 2016 at 15:20
  • $\begingroup$ Thanks. You had a typo maybe $T_0=2$ should be $S_0=2$? Otherwise what is $T$? I see that if $S_n$ is a martingale than $E[S_1|\{S_0=2\}]=2$ and $E[S_1|\{S_0=1\}]=1$ but I do not see the relationship with my doubt on martingale theory, probably for my poor understanding. Could you please explain more how this does show the role played by zero measure sets in martingale theory? $\endgroup$
    – Thomas
    Jul 5, 2016 at 15:51

2 Answers 2

Reset to default
2
$\begingroup$

Let $A$ be a non-trivial subset of $\Omega$ and let $\Omega$ be equipped with $\sigma$-algebra $\left\{ \varnothing,A,A^{c},\Omega\right\} $ and probability measure $P$ determined by $P\left(A\right)=1$.

Let $X:\Omega\to\mathbb{R}$ be prescribed by $\omega\mapsto1$

Let $Y:\Omega\to\mathbb{R}$ be prescribed by $\omega\mapsto1$ if $\omega\in A$ and $\omega\to0$ otherwise.

Then $X,Y$ are random variables with $P\left(X\neq Y\right)=P\left(A^{c}\right)=0$.

However the $\sigma$-algebra generated by $X$ is $\left\{ \varnothing,\Omega\right\} $ but the $\sigma$-algebra generated by $Y$ is $\left\{ \varnothing,A,A^{c},\Omega\right\} $.

$\endgroup$
2
  • $\begingroup$ Good! So they can be different! Now the doubt comes if they differ only by sets of measure $0$ and $1$... Do you also have some insight (as in previous comments) on any implication for martingale theory? $\endgroup$
    – Thomas
    Jul 5, 2016 at 15:53
  • $\begingroup$ Sorry, but I must dissapoint you here. I have no familiarity with martingales. $\endgroup$
    – drhab
    Jul 5, 2016 at 16:04
0
$\begingroup$

Something that is closely related can be a statement like that:

Claim: Let $X$ and $Y$ equivalent in the sense already described. If $U \in A_X$, than there exists $U' \in A_Y$ such that the symmetric difference has zero measure: $P(U \Delta U')=0$.

Trial proof: We first prove the statement for all sets of the type $U=\{X^{-1}(B)\}$ where $B$ is a Borel set. We temporarily call $F=\{U=\{X^{-1}(B)\},B \in Borel \}$ this subset of $A_X$. In this case let take $U'=\{Y^{-1}(B)\}$. We have the inclusion $U \Delta U' \subset \{X \ne Y\}$. Since $P(\{X \ne Y\})=0$ we also have $P(U \Delta U')=0$. To complete the proof and show that $F=A_X$, since $\sigma(F)=A_X$ ($F$ generates $A_X$), we need to show that the ensemble of sets respecting the claim indeed forms a sigma algebra, i.e. it is closed under countable union and complements. These are a consequence of the algebraic properties of the symmetric difference reported here https://en.wikipedia.org/wiki/Symmetric_difference .

I hope the statement is correct (can somebody check it?).

For the martingale relation of my question I found this interesting blog:

https://almostsure.wordpress.com/2009/11/08/filtrations-and-adapted-processes/

$\endgroup$
2
  • $\begingroup$ Actually the previous trial proof is not complete. I think I proved that all the sets of the type $U={X^{-1}(B)}$ respect that property, but they are only a generator of the sigma algebra $A_X$. One way to complete the proof would be to show now that all the sets respecting the statement form a sigma algebra, but it is not straightforward to me... $\endgroup$
    – Thomas
    Jul 13, 2016 at 15:41
  • $\begingroup$ I tried to sketch an argument for improving the trial proof $\endgroup$
    – Thomas
    Jul 13, 2016 at 16:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .