{Q: [0, T] is spaced equally, as usual notation:$\Delta B_i=B_{t_i}-B_{t_{i-1}}$, $t_i-t_{i-1}=T/n$. Show $\sum_{i=1}^{n} \Delta B_i^2\stackrel{ c.c. } \longrightarrow T $}

$Proof$: for all $\varepsilon$,

$\mathbb{P}\left(\{\omega\in \Omega:|\sum_{i=1}^{n} \Delta B_i^2-t|>\varepsilon\}\right)$

$\leq {\frac{E(\sum_{i=1}^{n}\Delta B_i^2-t)^4}{\varepsilon^4}}$(by Markov inequality)

$=12T^4(1/n^2+4/n^4)=O(1/n^2)$ .

then $\sum \mathbb{P}\left(\{ \omega\in \Omega:|\sum_{i=1}^{n} \Delta B_i^2-T|>\varepsilon\}\right)$ is a convergent series, satisfying the definition of c.c. (complete convergence),

So $\sum_{i=1}^{n} \Delta B_i^2\stackrel{ c.c. } \longrightarrow T $, that also implies $a.s.$ (or by Borel-Cantelli lemma).

Is there any error I cannot find? so it is appreciated if someone can point it out. wellcome for any comments, thanks so much.

  • $\begingroup$ Welcome to Math.SE! I noticed you've also posted the body of this question as an answer to two related questions here and here (where it does not answer the questions asked). In future, you should stick to just asking a question, rather than asking people to confirm your work in answers of other questions. $\endgroup$ – Rhys Steele Feb 27 at 17:13
  • $\begingroup$ ... and why do you post the same question two times? This one is a clear duplicate $\endgroup$ – saz Feb 27 at 17:22
  • $\begingroup$ deleted the original already. thanks for pointing it out. $\endgroup$ – Xisheng Yu Feb 28 at 1:39

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