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How can one prove that with probability 1 $X(t)=Y(t)$ for infinitely many values of $t\in [0, \infty)$ , where $\{X (t); t \geq 0\}$ and $\{Y (t); t \geq 0\}$ are independent, standard Brownian motions? In a previous step we are asked to show that $Z(t) = X(t) - Y(t)$ is a Brownian motions with zero drift and volatility parameter 2t.

This is just a practice problem as I prepare for my exam.

EM823823
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  • Since you showed $X-Y$ is a Brownian motion, all you need is the fact that Brownian motion is recurrent. Do you know how to show that? – Ian Dec 24 '20 at 01:41
  • @Ian Is that to show that the time it takes $Z(t)$ to hit zero is finite? If so, I think I found an answer I can follow at https://math.stackexchange.com/questions/3098741/proof-that-a-standard-brownian-motion-visits-zero-infinitely-often-at-the-beginn. Thank you! – EM823823 Dec 24 '20 at 01:45

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You know that $Z(t) = X(t) - Y(t)$ and hence $\hat{W}(t) = \frac{Z(t)}{\sqrt{2}}$ is a standard Brownian motion. It follows from the law of iterated logarithm that

$$\overline{lim}_{t \to \infty} \frac{ \hat{W}_t}{ \sqrt{2t \ln \ln t} } = 1$$ and $$\underline{lim}_{t \to \infty} \frac{\hat{W}_t}{ \sqrt{2t \ln \ln t} } = -1.$$ Hence $\hat{W}(t) = 0$ for infinitely many values of $t$, q.e.d.

Botnakov N.
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  • Thank you for this different approach. I'm not familiar with the law of iterated logarithms, however. – EM823823 Dec 25 '20 at 18:22