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 Sep6 comment A problem about strong law of large numbers of Shiryaev's Probability @DavideGiraudo I think you have followed mike's hint, am I right? If so, how to prove his hint 3? Thanks! Aug15 comment Prove this group $G$ is abelian @MartinBrandenburg Thank you! Aug15 asked Prove this group $G$ is abelian Aug14 comment subgroups of finitely generated groups with a finite index @mez It can be proved this $t$ is exactly $t_1=1$ Aug12 comment Is group in which every $a$ satisfies $a^3=e$ abelian? @ZhenLin Thank you! Aug12 asked Is group in which every $a$ satisfies $a^3=e$ abelian? Jul31 comment Manifold has uncountable many smooth stuctures if it has one @AnthonyCarapetis I think you are right. Would you please coppy your comment as an anwser so I can accept it. Thank you! Jul30 asked Manifold has uncountable many smooth stuctures if it has one Jul30 comment the equivalence between paracompactness and second countablity in a locally Euclidean and $T_2$ space Have you got the answer? I am also curious about this problem. Jul30 revised exercise 1.21 of chapter 1 of Revuz and Yor's edited tags Jul29 asked exercise 1.21 of chapter 1 of Revuz and Yor's Jul26 accepted Exercise 1.13 of chapter 1 of Revuz and Yor's Jul26 comment Exercise 1.13 of chapter 1 of Revuz and Yor's It can also use Fatou's lemma to prove $P(\varlimsup_{t\to+\infty}B_t/\sqrt{t}>0)\geq \varlimsup P(B_t/\sqrt{t}>0)=1/2>0$ and use Kolmogorov or HS 0-1 law. For 2, using the same reasoning we can prove $P(\varlimsup_{t\to+\infty}B_t/\sqrt{t}=+\infty)=P(\varliminf_{t\to-\infty}B_t/ \sqrt {t}=-\infty)=1$ and complet the proof. Jul25 comment Exercise 1.13 of chapter 1 of Revuz and Yor's Thank you! I found my proof of 1 is wrong: $t_0$ should depend on $\omega$. Jul25 asked Exercise 1.13 of chapter 1 of Revuz and Yor's Jul17 comment How do you prove a CW complex is locally path connected Dear @RyanBudney I saw this problem in Lee's book where he haven't introduce the conception of contractible. Can you tell me how to prove it directly? Thank you! Jul11 revised Product of independent continuous local martingales is local martingale edited tags Jul11 comment Product of independent continuous local martingales is local martingale @saz The book only said the independent means the natural fields generated by $M$ $N$ respectively are independent. I think $M$ $N$ are with respect to a same filtration wich is larger than the natural filtrations they generated respectively. Jul11 asked Product of independent continuous local martingales is local martingale Jul2 awarded Curious