# hulik

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# 457 Actions

 Jan4 revised Properties about a certain martingaleadded 95 characters in body Jan4 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? Jan4 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. Jan4 asked conditional expectation and conditional variance Dec30 accepted Cantor diagonalization method for subsequences Dec30 accepted Convergence of series $\frac1{n}\sum\limits_{k=1}^n a_k$ Dec26 asked Convergence of series $\frac1{n}\sum\limits_{k=1}^n a_k$ Dec23 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? Dec21 comment Boundedness in probabilityPerhaps 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! Dec21 comment Boundedness in probabilityEldrege: 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.... Dec20 comment Boundedness in probabilityThanks 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. Dec19 comment Boundedness in probabilityI deleted question 3. There was a mistake! I also updated my thoughts about 2. I would appreciat it much if somone could help me. Dec19 revised Boundedness in probabilitydeleted 801 characters in body Dec19 comment Boundedness in probabilityI 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. Dec19 awarded Commentator Dec19 revised Boundedness in probabilitydeleted 2 characters in body Dec19 comment Boundedness in probabilityYou're right, I edited that. I'm very sorry but I don't know what you mean with: " but what does question 3 become?" Dec19 revised Boundedness in probabilityadded 46 characters in body Dec19 asked Boundedness in probability Dec15 comment Cantor diagonalization method for subsequencesBy 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?