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Jan
20
accepted Shift Invariance of Backward Martingales for tail trivial probability measures
Jan
20
comment Shift Invariance of Backward Martingales for tail trivial probability measures
I was just verifying that :)
Jan
20
comment Shift Invariance of Backward Martingales for tail trivial probability measures
@NateEldredge I agree that different versions of the conditional expectation give us different sets ${x:g(x)=g(σx)}$ but I guess that what I had in mind (and not written here) is that independently of the choice of the version of the conditional expectation, this set always have measure one ?
Jan
20
comment Shift Invariance of Backward Martingales for tail trivial probability measures
This construction is nice, but I did not understand why we can apply the zero-one law since the r.v. $(X_i)_{i\in\mathbb{N}}$ are not independent. Of course, the projections $Y_i\in \{0,1\}^2$ are independent and the theorem applies, but its tail $\sigma$-algebra is different from the tail $\sigma$-algebra generated by the $X_i$ variables (which seems richer).
Jan
20
awarded  Inquisitive
Jan
20
comment Shift Invariance of Backward Martingales for tail trivial probability measures
@NateEldredge I guess your last comment answer the question. But I need to digest it because it implies that any probability measure trivial on "this" tail $\sigma$-algebra is shift invariant (its restricition to the tail $\sigma$-algebra) which seems very strong claim. Anyway thanks a lot for this clarification.
Jan
19
revised Shift Invariance of Backward Martingales for tail trivial probability measures
edited body
Jan
19
comment Shift Invariance of Backward Martingales for tail trivial probability measures
Yes. I am fixing it. Thanks.
Jan
19
asked Shift Invariance of Backward Martingales for tail trivial probability measures
Jan
4
revised Isolated Eigenvalue of $T$ is also an isolated eigenvalue of $\overline{T}$?
Improved formatting and clarity of the question.
Dec
22
comment Isolated Eigenvalue of $T$ is also an isolated eigenvalue of $\overline{T}$?
Thanks again @ChristianRemling but I don't understand why. I will think about it more.
Dec
21
comment Isolated Eigenvalue of $T$ is also an isolated eigenvalue of $\overline{T}$?
Dear @ChristianRemling first of all thanks for your comment. From the hypothesis it follows that $T-(\lambda-\varepsilon)$ has bounded inverse and therefore $\exists \ c>0$ such that $c\|x\|_{\infty}\leq \| [T-(\lambda-\varepsilon)]x\|_{\infty}$. From your hint should I be able to conclude that there is a positive constant $d$ such that $d\|x\|_{L^1}\leq \| [T-(\lambda-\varepsilon)]x\|_{L^1}$ ? Could you elaborate a little bit more on your comment ?
Dec
18
asked Isolated Eigenvalue of $T$ is also an isolated eigenvalue of $\overline{T}$?
Jul
27
awarded  Electorate
Feb
20
awarded  Necromancer
Dec
8
awarded  Yearling
Jul
2
awarded  Curious
May
28
awarded  Popular Question
May
14
comment Is $\text{Bow}(X,T)$ a Banach Space?
@BenWillson in this context $C^0(X)$ is usually the space of all continuous functions taking values on $\mathbb{R}$ or $\mathbb{C}$.
May
8
awarded  Popular Question