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For a standard Brownian motion define \begin{align}\mathcal{F}_{0+} &= \bigcap_{t>0} \mathcal{F_t},\\ \mathcal{F_t} &= \sigma(W_s, 0 \le s \le t)\end{align}

Which of the following events belong to $\mathcal{F_{0+}}$?

(i) $$\left\{ W_t > \sqrt{t} \text{ for infinitely many } t>0\} = \limsup \{W_t > \sqrt{t}\right\}$$

(ii) $$\limsup\left\{W_{\frac{1}{n^2}} > \sqrt{\frac{1}{n}}\right\}$$

(iii)$$\left\{W_{t_n} > \sqrt{t_n} \text{ for $(t_n)_n$ converging to $0$ monotonically}\right\}$$

First of all I guess that (i) does not belong to $\mathcal{F_{0+}}$, since for a realisation of W, where the "part-event" {$W_t > \sqrt{t}$} occurs infinitely often just for $t>s>0$, one cannot decide with just knowing W at 0 (with an infinitesimal view in the future).

On the other hand, for (ii) and (iii) the "part-events" have to occur infinitely often arbitrary close to 0. So I guess at least (ii) is in $\mathcal{F_{0+}}$ (for (iii) all "part-events" have to occur). This is anyway just my guess, how to show it properly? For sigma-algebras one usually has to argue with the generators of the algebras. The borel algebra of $\mathcal{R}$ is generated (for example) by $\{(a,\infty) , a \in \mathcal{R}\}$. The $\mathcal{F_t}$ are hence generated by $\{ W_s > a, 0 \le s \le t, a \in \mathcal{R}\}$.

So in the case of (ii) we have

$$\left\{W_{\frac{1}{n^2}} > \sqrt{\frac{1}{n}}\right\} \in \mathcal{F_t} \text{ for } \frac{1}{n^2} \le t$$

so

$$\bigcup_{m>n}\left\{W_{\frac{1}{m^2}} > \sqrt{\frac{1}{m}}\right\} \in \mathcal{F_t} \text{ for } \frac{1}{n^2} \le t$$

and

$$\bigcap_{i>n} \bigcup_{m>i}\left\{W_{\frac{1}{m^2}} > \sqrt{\frac{1}{m}}\right\} \in \mathcal{F_t} \text{ for } \frac{1}{n^2} \le t$$

For arbitrary $t>0$ chose $n$ big enough and since the intersection goes over monotonically decreasing sets it follows that

$$\bigcap_{i>0} \bigcup_{m>i} \left\{W_{\frac{1}{m^2}} > \sqrt{\frac{1}{m}}\right\} \in \mathcal{F_t} \text{ for } 0 < t.$$

In the case of (iii) I just know that for $t>0$,

$$\bigcap_{n \ge m} \left\{W_{t_n} > \sqrt{t_n}\right\} \in \mathcal{F_t} \text{ for } t_m \le t$$ since the sequence is monotone.

So $$\bigcap_{n \ge 1} \left\{W_{\hat{t}_n} > \sqrt{\hat{t}_n }\right\} \in \mathcal{F_t} \text{ for } \hat{t}_n = t_{n+m}$$ and $(\hat{t}_n)_n$ is also a monotone decreasing sequence smaller than $t$.

Since $t>0$ was arbitrary (iii) is also in the intersection sigma-algebra.

Is this argumentation okay? And how to show that (i) is not in the sigma-algebra?

Thanks in advance (I'm sorry for missspelling)


One more thing to (iii)

Is not $$\left\{W_{t_n} > \sqrt{t_n} \text{ for a sequence } t_n \to 0\right\} = \bigcup_{(t_n)_n} \bigcap_{n \ge 1} \left\{W_{t_n} > \sqrt{t_n}\right\},$$ so my argument is not working since the union is uncountable?

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  • $\begingroup$ I fixed the formatting on your post. Feel free to edit back if I messed anything up. $\endgroup$ – Math1000 Jun 19 '15 at 21:01
  • $\begingroup$ Another idea: {$W_n > \sqrt{t_n}$ for a monotone sequence $t_n \to 0$} = {inf {$t>0 : W_t > \sqrt{t}$} which is clearly in $ \mathcal{F}_{0+}$ but still, how to show an event is not in a sigma-algebra (for (i)) ? Since I don't know how the elements look like I have to do it implicitly (through contradiction maybe?). $\endgroup$ – crankk Jun 21 '15 at 11:50
  • $\begingroup$ it should say {$ W_n > \sqrt{t_n}$ for a monotone sequence $t_n \to 0$} = { inf {$ t>0 : W_t > \sqrt{t} $} = 0 } $\endgroup$ – crankk Jun 22 '15 at 6:00
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Set $$A := \{W_t < \sqrt{t} \, \text{for infinitely many $t>0$}\}.$$ Since, by the law of the iterated logarithm,

$$\limsup_{t \to \infty} \frac{W_t}{\sqrt{t \log \log t}} = 1 \qquad \text{almost surely}$$

we have $A = \Omega \backslash N$ for some null set $N$. So the trouble is basically: Is $N \in \mathcal{F}_{0+}$? Since $\mathcal{F}_{0+}$ is (in general) not complete, we cannot expect $N \in \mathcal{F}_{0+}$. On the other hand, there are processes for which $N \in \mathcal{F}_{0+}$; just consider the Brownian motion

$$\tilde{W}_t(\omega) := \begin{cases} W_t(\omega), & \omega \in \Omega \backslash N, \\ 2 \sqrt{t}, & \omega \in N \end{cases}.$$ Then $$\{\tilde{W}_t> \sqrt{t} \, \text{for infinitely many $t$}\} = \Omega \in \mathcal{F}_{0+}.$$


Next, consider $$B := \limsup_{n \to \infty} \left\{W_{\frac{1}{n^2}}> \sqrt{\frac{1}{n}} \right\}.$$ Well, by definition,

$$B = \bigcap_{n \in \mathbb{N}} \underbrace{\bigcup_{k \geq n} \left\{W_{\frac{1}{k^2}}> \sqrt{\frac{1}{k}} \right\}}_{=:B_n}.$$

Since $B_n \downarrow B$, we have

$$B = \bigcap_{n \geq N} B_n$$

for any (fixed) $N \in \mathbb{N}$. Since $B_n \in \mathcal{F}_{\frac{1}{n^2}}$, this implies $B \in \mathcal{F}_{\frac{1}{N^2}}$. Hence, $B \in \mathcal{F}_{0+}.$$


For $$C := \{\exists t_n \downarrow 0: W_{t_n}>\sqrt{t_n}\}$$

we note that

$$C = \bigcap_{k \in \mathbb{N}} \underbrace{ \left\{ \sup_{t \leq \frac{1}{k}} \frac{W_t}{\sqrt{t}}>1 \right\}}_{=:C_k}.$$

Since $C_k \downarrow C$ and $C_k \in \mathcal{F}_{\frac{1}{k}}$, this shows $C \in \mathcal{F}_{0+}$.

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  • $\begingroup$ Thank you, so basically what I said is right then. $\endgroup$ – crankk Jun 23 '15 at 20:04
  • $\begingroup$ @crankk Yes. Just as a remark: One possibility to prove that a set $A$ is not contained in $\mathcal{F}_{0+}$ is to show that $\mathbb{P}(A) \notin \{0,1\}$. $\endgroup$ – saz Jun 24 '15 at 5:24

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