hulik

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	63 reputation	bio	website		visits	member for	1 year, 5 months
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Jan 4	revised	Properties about a certain martingale added 95 characters in body
Jan 4	comment	conditional expectation and conditional variance @ Henry: Thanks for your answer. Though I have an additional question. In fact I would like to bound $ Var(X\|\mathcal{F}) $ by $ \operatorname{const}\cdot E(X^2)$. Is this possible?
Jan 4	comment	conditional expectation and conditional variance @ Didier Piau: Thanks for your comment, but unfortunately the book is not free available or I didn't find it.
Jan 4	asked	conditional expectation and conditional variance
Dec 30	accepted	Cantor diagonalization method for subsequences
Dec 30	accepted	Convergence of series $ \frac1{n}\sum\limits_{k=1}^n a_k$
Dec 26	asked	Convergence of series $ \frac1{n}\sum\limits_{k=1}^n a_k$
Dec 23	comment	Boundedness in probability @ Nate Eldrege: Thanks for your help! But there still some question around: The case for just one such $X$ is clear. But how could I prove the general case? Suppose there are k such $ X_i $, then I know, that $ P(\|X_i\|>N) \to 0 $ as $ N\to \infty$ as in the case for just one such $ X $, right? Therefore the sup over all this is $0$. This is the whole arguement for the general case, right? Or am I wrong? How else should I prove the general case?
Dec 21	comment	Boundedness in probability Perhaps there's a element $ \tilde{X} $ of the sequence which is on a set (with large enough measure) as big as I want. If I can prove that it doesn't hold for 1) then using induction I can prove your claim. However, I do not see how to prove this for one, aso mentioned above. Again thanks for your help!
Dec 21	comment	Boundedness in probability Eldrege: Thanks for answering 2)! I have a question about that: We know it exists $ \epsilon $ such that for every $ N>0 $ there's a $ X_n $ with $ P(\|X_n\|>N) > \epsilon$. You claim that for every $ N $ there are infinitely many $ X_n$ with this property. So let $ N>0 $ and suppose there's just one of this $ X_n$. First it's clear: $ \mathbf1\{\|X_n\|>N\} \ge \mathbf1\{\|X_n\|>N-1\} \dots \mathbf1\{\|X_n\|>1\}$ So if I can prove, that there's no $ X_n $ such that for this fixed $ X_n$ : $ P(\|X_n\|> N ) > \epsilon $ for all $ N $ then I'm done. But I do not see why this could not be the case....
Dec 20	comment	Boundedness in probability Thanks Nate Eldrege for you answer! with " 1 is not correct" you mean : for every $ N >0 $ there exists an $ \epsilon >0$ and $ X \in M$ such that $ P(\|X\|> N)\ge \epsilon $, right? As you can see in the comments above, Srivatsan corrected this already. At this point I just have trouble with 2), see also my observations so far for 2) above.
Dec 19	comment	Boundedness in probability I deleted question 3. There was a mistake! I also updated my thoughts about 2. I would appreciat it much if somone could help me.
Dec 19	revised	Boundedness in probability deleted 801 characters in body
Dec 19	comment	Boundedness in probability I do not know that the $ (X_n) $ are independent. The reason for question 3 was $ (1) $. As I worte, I try to prove $ (1) $ and I do not see why this is true unless we know that for almost all $ \omega $ and infinitly man $ k$, it's true that $\|X_{n_k}(\omega)\| \ge k$. Since I can not assume independence, I have to find a different arguement why $ (1) $ is true.
Dec 19	awarded	Commentator
Dec 19	revised	Boundedness in probability deleted 2 characters in body
Dec 19	comment	Boundedness in probability You're right, I edited that. I'm very sorry but I don't know what you mean with: " but what does question 3 become?"
Dec 19	revised	Boundedness in probability added 46 characters in body
Dec 19	asked	Boundedness in probability
Dec 15	comment	Cantor diagonalization method for subsequences By your last sentence, do you mean, that the last inclusion in $ \mathbb{N}\supset \Omega_1 \supset \dots \supset \Omega_l \dots \supset \Omega $ is wrong?