Impact
~177k
people reached
- 0 posts edited
- 0 helpful flags
- 1,189 votes cast
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 |