\[ \mathbb{P}\big(\sup_{t\in[0,1]} |Z^{(n)}(t)| \geq C\big) \leq \frac{4}{(C-1)^2} \] For $C\to\infty$: process exceeds $C$ with probability $\to 0$.
Proof: construction of related martingale; application of Doob's $L^p$-inequality.
\begin{align*} \mathbf{K}_{\mathbf{t}n} = \big( K_{\lfloor t_1 n\rfloor}^{(n)}, \dots, K_{\lfloor t_r n\rfloor}^{(n)} \big) \end{align*} \begin{align*} \frac{\mathbf{K}_{\mathbf{t}n} - \mathbb{E} \mathbf{K}_{\mathbf{t}n}}{\sqrt{n}} \xrightarrow[n\to\infty]{d} \mathcal{N} \end{align*}
Proof: construct p.g.f. from $E_n$, \[ P_n = \prod_{i=2}^{n-1} \Big( \frac{1}{n} + \frac{i}{n} z_i + \sum_{h=i+1}^{n-1} \frac{1}{n} z_h \Big), \] sum of multinomial variables! Then: multivariate central limit theorem.
\begin{align*} \mathcal{F} &= \mathcal{Z} \ast \mathrm{Set}(\mathcal{G}) + \mathcal{Z}\times \{U, V\} \ast \mathrm{Set}(\mathcal{F}) \\ \mathcal{G} &= (\mathcal{Z} + \mathcal{Z}\times \{U\}) \ast \mathrm{Set}(\mathcal{G}) \\ \end{align*}
\begin{align*} \mathcal{F} &= \mathcal{Z} \ast \mathrm{Set}(\mathcal{G}) + \mathcal{Z}\times \{U, V\} \ast \mathrm{Set}(\mathcal{F}) \\ \mathcal{G} &= (\mathcal{Z} + \mathcal{Z}\times \{U\}) \ast \mathrm{Set}(\mathcal{G}) \\ \end{align*}
\[ F = z e^G + zuv e^F \qquad G = z (1 + u) e^G \]
$k$ |
$k = o(n)$ |
$k \sim \alpha n$ |
$d=\omega(\sqrt{n})$ $k = n-d$ |
$d\sim c\sqrt{n}$ $k = n-d$ |
$d=o(\sqrt{n})$ $k = n-d$ |
$f(n)$ | $k/n$ | $\frac{\alpha}{(1 - \alpha)^2}$ | $n^2/d^2$ | $\kappa n$ | $n - d\sqrt{\pi n /2}$ |
\[ \mathbb{P}(R_n^{(k)} = r) = \begin{cases} 1 - k/n & \text{if } r = 0,\\ r^r (n-k) (n-r)^{k-r-1} \binom{k}{r} n^{-k} & \text{if } 1\leq r\leq k < n,\\ 1 & \text{if } r = k = n. \end{cases} \]