3
$\begingroup$

Let $J_t$ be a standard Brownian motion, let $X = \{t : J_t = 0\}$ denote the zero set, and let $I(j, n)$ denote the indicator function of the event$$\left\{\text{there exists }s \in \left[{{j-1}\over{n}}, {j\over{n}}\right] \text{ with }J_s = 0\right\}.$$Let$$K_n = \sum_{j=1}^n I(j, n).$$Observe that $K_n$ denotes the number of intervals of the form $\left[{{j-1}\over{n}}, {j\over{n}}\right]$ needed to cover $X \cap [0, 1]$.

  • What is the constant $C$ such that$$\lim_{n \to \infty} n^{-1/2} \textbf{E}(K_n) = C?$$
  • Is there a constant $C < \infty$ such that for all $n$,$$\textbf{E}[K_n^2] \le C(\textbf{E}[K_n])^2?$$

My apologies, I need these two results for my research, and I am not an expert at probability...

$\endgroup$
  • $\begingroup$ By standard arguments using the reflection principle and a Taylor series expansion of $\cos^{-1}(1-x)$ around zero, I believe you should get $C = 4\sqrt{2}/\pi$ for the first part. $\endgroup$ – cardinal Sep 15 '15 at 6:09
5
+50
$\begingroup$

(Since I get a slightly different constant than cardinal, I detail a bit the computation).

For the first part, you need to compute the probability of zero crossing in $[t,t+\Delta]$ for $\Delta=1/n$. This is (conditioning on the value $z$ at time $t$ and using the reflection principle) $$A=2\int_0^\infty dz \frac{e^{-z^2/2t}}{\sqrt{2\pi t}} \cdot 2\int_z^\infty dy \frac{e^{-y^2/2\Delta}}{\sqrt{2\pi\Delta}}$$ By change of variables this is the same as $$\frac{4}{2\pi \sqrt{t}}\int_0^\infty dz e^{-z^2/2t} \int_{z/\sqrt{\Delta}}^\infty dx e^{-x^2/2}$$ The main contribution comes (except for $t\sim \Delta$ or a bit more, which eventually only will contribute a constant/logarithmic term) from $z\sim \sqrt{\Delta}$ and so you can replace the first exponential by $1$ and get finally $A=\frac{2\sqrt{\Delta}}{\pi \sqrt{ t}}$. Summing over $t=i\Delta$ gives $C=\frac{4}{\pi}$.

For the second part, the answer is yes - the second moment involves computing, for $t>s$, $P(W_t\sim \sqrt{\Delta},W_s\sim \sqrt{\Delta}) $ which is asymptotic to $\Delta \frac{1}{\sqrt{t(t-s)}}$. Summing as before you get your claim. No need to keep track of constants.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ It appears that, while transcribing, I must have accidentally dropped a square root in the argument of the arccosine at the last step which led to my extra factor of $\sqrt{2}$. I should have instead been considering a series expansion of $\cos^{-1}(\sqrt{1-x})$. $\endgroup$ – cardinal Sep 16 '15 at 18:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy