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Kurt.W.X
Kurt.W.X
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Consider a Brownian motion $B$ and let $f(r)=\sqrt{2r \ln(\ln(r))}.$

Is it true that $\lim_{r \to +\infty}\frac{1}{f(r)}(B_r-B_{\left \lfloor{f(r)}\right \rfloor})= 0$ a.s. ?

If so, how to prove it? Otherwise what counter-example do you suggest  ?

Consider a Brownian motion $B$ and let $f(r)=\sqrt{2r \ln(\ln(r))}.$

Is it true that $\lim_{r \to +\infty}\frac{1}{f(r)}(B_r-B_{\left \lfloor{f(r)}\right \rfloor})= 0$ a.s. ?

If so, how to prove it? Otherwise what counter-example do you suggest?

Consider a Brownian motion $B$ and let $f(r)=\sqrt{2r \ln(\ln(r))}.$

Is it true that $\lim_{r \to +\infty}\frac{1}{f(r)}(B_r-B_{\left \lfloor{f(r)}\right \rfloor})= 0$ a.s. ?

If so, how to prove it? Otherwise what counter-example do you suggest  ?

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Kurt.W.X
Kurt.W.X
  • 249
  • 1
  • 7

$\lim_{r \to +\infty}\frac{1}{\sqrt{2r \ln(\ln(r))}}(B_r-B_{\left \lfloor{\sqrt{2r \ln(\ln(r))}}\right \rfloor})= 0$ a.s.?

Consider a Brownian motion $B$ and let $f(r)=\sqrt{2r \ln(\ln(r))}.$

Is it true that $\lim_{r \to +\infty}\frac{1}{f(r)}(B_r-B_{\left \lfloor{f(r)}\right \rfloor})= 0$ a.s. ?

If so, how to prove it? Otherwise what counter-example do you suggest?