Standard Brownian motion, limit, square of expectation bound 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 $X \cap [0, 1]$.


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*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...
 A: (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.  
