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$\mathbb{P}(\limsup_{t\rightarrow\infty}\frac{\left|B_{t}\right|}{t}>\frac{1}{M})\leq\mathbb{P}(\bigl\{\sup_{t\in\left[n,n+1\right]}\frac{\left|B_{t}\right|}{n}>\frac{1}{M}\bigr\}\;\mbox{i.o.})$

In this question (Law of large numbers for Brownian Motion (Direct proof using L2-convergence)) in the second answer, I don't see how this inequality is true.

Indeed, it seems counter intuitive to me since the supremum of the right side is on a smaller set then that of the left side.

Did
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user405156
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    You might agree that it suffices to show that, for every positive $c$, $$\left{\limsup_{t\rightarrow\infty}\frac{\left|B_{t}\right|}{t}>c\right}\subseteq\left{\sup_{t\in\left[n,n+1\right]}\frac{\left|B_{t}\right|}{n}>c;\mbox{infinitely often}\right}$$ or, equivalently, that $$\left{\sup_{t\in\left[n,n+1\right]}\frac{\left|B_{t}\right|}{n}\leqslant c;\mbox{for every $n$ large enough}\right}\subseteq\left{\limsup_{t\rightarrow\infty}\frac{\left|B_{t}\right|}{t}\leqslant c\right}$$ Thus, let $\omega$ be in the set on the LHS, ... – Did May 11 '17 at 12:36
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    ... there exists $N(\omega)$ such that for every $n\geqslant N(\omega)$, ... Surely you can continue and arrive at the consequence that $\omega$ belong to the set on the RHS? (One sees that the result is purely real analysis, with not a whiff of stochasticity in it.) – Did May 11 '17 at 12:36
  • Ok thanks. Latter in the same post, $\frac{M^{4}}{n^{4}}\mathbb{E}\bigl|B_{n+1}\bigr|^{4}\asymp\frac{1}{n^{2}}$. Is there an easy way of calculating $E(X^4)$? I don't see how he gets that it behaves as $1/n^2$ – user405156 May 19 '17 at 13:41
  • If $Z$ is standard normal then $E(Z^4)=3$. This yields the estimate you cite. (One may find surprising to see someone engaged in the study of the properties of Brownian motion, the gaussian process par excellence, and being unaware of the values of the first moments of normal random variables...) – Did May 19 '17 at 15:41
  • but in our case $B_{n+1}$ is not standard normal and as here https://fr.wikipedia.org/wiki/Loi_normale#Moments, $E(B_{n+1}) = 3(n+1)^4$ so this does not work... – user405156 Jun 09 '17 at 09:02
  • No, $B_{n+1}$ is not standard normal but it is centered normal hence the computation in my comment is all you need to compute $E(B_{n+1}^4)$, so yes, this does work... (But no, $E(B_{n+1})$ is not $3(n+1)^4$.) – Did Jun 09 '17 at 10:24
  • $B_{n+1}$ follows $N(0,n+1)$. If I follow the formula's from https://en.wikipedia.org/wiki/Normal_distribution#Moments, I find that $E[\vert B_{n+1}\vert^4] = 3(n+1)^4$ – user405156 Jun 10 '17 at 08:59
  • Sorry but at the end of the day, you have to be much more serious about this... No, $E(B_t^4)$ is not $3t^4$. This new suggestion of yours makes slightly more sense than your previous one that $E(B_t)=3t^4$, but it is still wrong. – Did Jun 10 '17 at 09:17
  • I don't see how what I say can be wrong. I'm just following the formula. On wikipedia it states that for a $X$ following $N(\mu,\sigma^2), E[X^4] = \mu^4 + 6\mu^2\sigma^2 + 3\sigma^4$ which in the case of $B_{t}$ gives $3t^4$ – user405156 Jun 10 '17 at 09:22
  • take it all back $\sigma^4 = (\sigma^2)^2 = t^2$ – user405156 Jun 10 '17 at 09:24
  • My stupid mistake – user405156 Jun 10 '17 at 09:24

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