# hulik

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 Feb10 comment Tightness of distributionAh....and it's also true that $E(|X|)=E(|Y|)$ and then everything is clear. Thanks a lot! Feb10 asked Diagonalization method by Cantor (2) Feb10 comment Tightness of distribution@ Byron Schmuland: I know the equivalent definition of weak convergence described in Chris' answer. But I do not see why in this case $Y_n = X_n$. My situation is different: If we look at $(|X_n|)$ we can use Fatou as you mentioned and we denote $|X|:=\lim_n |X_n|\in [0,\infty]$. But I don't see why $X$ should be a r.v. and why $\mu$ should be the distribution of $X$. From the definition of tightness we know that there exists $(Y_n)$ as Chris said, but the sequence $(Y_n)$ is different to the sequence $(X_n)$, isn't it? Feb10 asked Tightness of distribution Jan14 comment Probabilistic statementThanks for your answer! Jan14 accepted Probabilistic statement Jan14 revised Probabilistic statementadded 4 characters in body Jan14 comment Probabilistic statementit's true that for any $l\ge 1$ and $N\ge0$! I edited my question. But I do not quite understand your comment. Do you mean $\{X_n=l\}$? $P(\cup_N\cap_{n\ge N}\{X_n=l\})=0$ follows from sigma additivity, right? Jan14 asked Probabilistic statement Jan11 awarded Benefactor Jan11 comment Properties about a certain martingale@ math: Thank you for your answer! Now I do understand much better the answer of John Dawkins, too. I accepted your answer since 4 was quite important for me and you explanations were clear. Perhaps there's a mistake in 3. I will again check this. Jan5 comment Properties about a certain martingaleand why is $E(X_i)^2 -2E(X_iE(X_i|\mathcal{F}_{i-1})) + E((E(X_i|\mathcal{F}_{i-1}))^2) = E(X_i^2)-E((E(X_i|\mathcal{F}_{i-1}))^2)$? I would agree if $E(X_i|\mathcal{F}_{i-1})$ is constant, otherwise I do not see why this sould be true. Jan5 comment Properties about a certain martingale@ John Dawkins: Thanks a lot for your patience, but there still some arguements, which I do not understand: 1. In your second last comment do you mean $E(M_n^2) \le M_\infty$ instead of $E(M_n^2) \le E(M_\infty)$? It would be appreciated a lot, if you would update your answer and show how you get this. 2. Unfortunately there's no assumption as $|X_i-E(X_i|\mathcal{F}_{i-1})|\le Z$. Jan4 comment Properties about a certain martingale@ John Dawkins: Thanks for your answer, but there are some question around before accepting your answer. First question: How should I compute $E(M_n^2)$ ? This looks complicated, since I square the whole sum, and why is $E(M_n^2) \le M_\infty$ for all $n$. Second question: Why is $E((X_i-E(X_i|\mathcal{F}_{i-1}))^2|\mathcal{F}_{i-1})=E((X_i-E(X_i|\mathcal{F}_{‌​i-1}))^2)= E(X_i^2)-E(E(X_i|\mathcal{F}_{i-1})^2)$ ? Jan4 awarded Promoter Jan4 revised Properties about a certain martingaleadded 2 characters in body Jan4 accepted Boundedness in probability Jan4 accepted conditional expectation and conditional variance Jan4 comment conditional expectation and conditional variance@ Henry: Thanks for your quick answer. As mentioned below (see comment after Didier Piau's answer) the motiviation behind this question can be found in link. I wondered if in a more general setting something would be true. Because then I would be able to prove 4.) in the related question Jan4 comment conditional expectation and conditional varianceThanks for your answer. See link it's a related question, in fact it's 4. in the other question (see link). This was the reason for asking this question in a more general setting.