215 lines
4.6 KiB
Markdown
215 lines
4.6 KiB
Markdown
# Trapani test
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## Trapani test
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::: incremental
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- Random variable $\left\{ x_i \right\}$ sampled from a distribution $f$
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- Sample moments estimate as $f$ moments
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- $H_0$: $\mu_k \longrightarrow + \infty$
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- Statistic with 1 dof $\chi^2$ distribution
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- $p$-value $\hence$ reject or accept $H_0$
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:::
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## Infinite moments
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- Generate a sample $L$ from a Landau PDF
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- Generate a sample $M$ from a Moyal PDF
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. . .
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\vspace{20pt}
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:::: {.columns}
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::: {.column width=50% .c}
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For the Landau PDF:
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\begin{align*}
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\mu_1 &= \text{E}\left[|x|\right] = + \infty \\
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\mu_2 &= \text{E}\left[|x|^2\right] = + \infty
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\end{align*}
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:::
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::: {.column width=50%}
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. . .
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For the Moyal PDF:
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\begin{align*}
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\mu_1 &= \text{E}\left[|x|\right] < + \infty \\
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\mu_2 &= \text{E}\left[|x|^2\right] < + \infty
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\end{align*}
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:::
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::::
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## Infinite moments
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- Previous tests: points sampled from Landau PDF?
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. . .
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- Trapani test: check whether a moment is finite or infinite
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\begin{align*}
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\text{infinite} &\thus \text{Landau} \\
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\text{finite} &\thus \text{not Landau}
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\end{align*}
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. . .
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- Compatibility test with $\mu_k = + \infty$
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. . .
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- If points were sampled from a Cauchy distribution...
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## Trapani test
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## Trapani test
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- Start with $\left\{ x_i \right\}^N$ and compute $\mu_k$ as:
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$$
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\mu_k = \frac{1}{N} \sum_{i = 1}^N |x_i|^k
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$$
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. . .
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- Generate $r$ points $\left\{ \xi_j\right\}^r$ according to $G(0, 1)$ and define
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$\left\{ a_j \right\}^r$ as:
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$$
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a_j = \sqrt{e^{\mu_k}} \cdot \xi_j
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\thus G\left( 0, \sqrt{e^{\mu_k}} \right)
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$$
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. . .
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The greater $\mu^k$, the 'larger' $G\left( 0, \sqrt{e^{\mu_k}} \right)$
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$$
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\begin{cases}
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\mu_k \longrightarrow + \infty \\
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r \longrightarrow + \infty
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\end{cases}
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\thus a_j \text{ distributed uniformly}
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$$
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## Trapani test
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- Define the sequence: $\left\{ \zeta_j (u) \right\}^r$ as:
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$$
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\zeta_j (u) = \theta( u - a_j) \with \theta - \text{Heaviside}
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$$
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. . .
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\begin{center}
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\begin{tikzpicture}[>=Stealth]
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% line
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\draw [line width=3, ->, cyclamen] (0,0) -- (10,0);
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\node [right] at (10,0) {$u$};
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% tic
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\draw [thick] (5,-0.3) -- (5,0.3);
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\node [above] at (5,0.3) {$u_0$};
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% aj tics
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\draw [thick, cyclamen] (1,-0.2) -- (1,0.2);
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\node [below right, cyclamen] at (1,-0.2) {$a_{j+2}$};
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\draw [thick, cyclamen] (2,-0.2) -- (2,0.2);
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\node [below right, cyclamen] at (2,-0.2) {$a_j$};
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\draw [thick, cyclamen] (5.2,-0.2) -- (5.2,0.2);
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\node [below right, cyclamen] at (5.2,-0.2) {$a_{j+2}$};
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\draw [thick, cyclamen] (6,-0.2) -- (6,0.2);
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\node [below right, cyclamen] at (6,-0.2) {$a_{j+3}$};
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\draw [thick, cyclamen] (8.5,-0.2) -- (8.5,0.2);
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\node [below right, cyclamen] at (8.5,-0.2) {$a_{j+4}$};
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% notes
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\node [below] at (1,-1) {0};
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\node [below] at (2,-1) {0};
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\node [below] at (5.2,-1) {1};
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\node [below] at (6,-1) {1};
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\node [below] at (8.5,-1) {1};
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\draw [thick, ->] (1,-0.5) -- (1,-1);
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\draw [thick, ->] (2,-0.5) -- (2,-1);
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\draw [thick, ->] (5.2,-0.5) -- (5.2,-1);
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\draw [thick, ->] (6,-0.5) -- (6,-1);
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\draw [thick, ->] (8.5,-0.5) -- (8.5,-1);
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\end{tikzpicture}
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\end{center}
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. . .
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If $a_j$ uniformly distributed:
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- $\zeta_j (u)$ Bernoulli PDF with $P\left( \zeta_j (u) = 1 \right) = \frac{1}{2}$
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$\hence \text{E}[\zeta_j]_j = \frac{1}{2}
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\quad \wedge \quad \text{Var}[\zeta_j]_j = \frac{1}{4}$
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## Trapani test
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- Define the function $\vartheta (u)$ as:
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$$
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\vartheta (u) = \frac{2}{\sqrt{r}}
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\left[ \sum_{j} \zeta_j (u) - \frac{r}{2} \right]
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$$
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. . .
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If $a_j$ uniformly distributed, for the CLT:
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$$
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\sum_j \zeta_j (u) \hence
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G \left( \frac{r}{2}, \frac{r}{4} \right)
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\thus \vartheta (u) \hence
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G \left( 0, 1 \right)
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$$
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. . .
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- Test statistic:
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$$
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\Theta = \int_{\underbar{u}}^{\bar{u}} du \, \vartheta^2 (u) \psi(u)
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$$
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## Trapani test
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According to L. Trapani [@trapani15]:
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- $r = o(N) \hence r = N^{0.75}$
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- $\underbar{u} = -1 \quad \wedge \quad \bar{u} = 1$
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- $\psi(u) = \frac{1}{\bar{u} - \underbar{u}} \, \chi_{[\underbar{u}, \bar{u}]}$
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. . .
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$\mu_k$ must be scale invariant for $k > 1$:
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$$
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\mu_k^* = \frac{\mu_k}{ \left( \mu_{\phi} \right)^{k/\phi} }
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\with \phi \in (0, k)
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$$
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## Trapani test
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If $\mu_k \ne + \infty \hence \left\{ a_j \right\}$ are not uniformly distributed
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\vspace{20pt}
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Rewriting:
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$$
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\vartheta (u) = \frac{2}{\sqrt{r}}
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\left[ \sum_{j} \zeta_j (u) - \frac{r}{2} \right]
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= \frac{2}{\sqrt{r}}
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\sum_{j} \left[ \zeta_j (u) - \frac{1}{2} \right]
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$$
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\vspace{20pt}
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. . .
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Residues become very large $\hence$ $p$-values decreases.
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