analistica/slides/sections/6.md

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# Trapani test
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## Trapani test
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::: incremental
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- Random variables $\left\{ x_i \right\}$ sampled from a distribution $f$
- Sample moments converge to $f$ 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 $\rightarrow$ reject or accept $H_0$
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:::
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## Infinite moments
- Generate a sample $L$ from a Landau PDF
- Generate a sample $M$ from a Moyal PDF
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. . .
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\vspace{20pt}
:::: {.columns}
::: {.column width=50% .c}
For the Landau PDF:
\begin{align*}
\mu_1 &= \text{E}\left[|x|\right] = + \infty \\
\mu_2 &= \text{E}\left[|x|^2\right] = + \infty
\end{align*}
:::
::: {.column width=50%}
. . .
For the Moyal PDF:
\begin{align*}
\mu_1 &= \text{E}\left[|x|\right] < + \infty \\
\mu_2 &= \text{E}\left[|x|^2\right] < + \infty
\end{align*}
:::
::::
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## Infinite moments
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- Previous tests: points sampled from Landau PDF?
. . .
- 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} \\
\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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![](images/cauchy-pdf.pdf)
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## Trapani test
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::: incremental
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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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- Generate $r$ points $\left\{ \xi_j\right\}^r$ according to $G(0, 1)$ and define
$\left\{ a_j \right\}^r$ as:
$$
a_j = \sqrt{e^{\mu_k}} \cdot \xi_j
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\thus G\left( 0, \sqrt{e^{\mu_k}} \right)
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$$
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- The greater $\mu^k$, the 'larger' $G\left( 0, \sqrt{e^{\mu_k}} \right)$
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$$
\begin{cases}
\mu_k \longrightarrow + \infty \\
r \longrightarrow + \infty
\end{cases}
\thus a_j \text{ distributed uniformly}
$$
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:::
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## Trapani test
- 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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. . .
\begin{center}
\begin{tikzpicture}[>=Stealth]
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% line
\draw [line width=3, ->, cyclamen] (0,0) -- (10,0);
\node [right] at (10,0) {$u$};
% tic
\draw [thick] (5,-0.3) -- (5,0.3);
\node [above] at (5,0.3) {$u_0$};
% aj tics
\draw [thick, cyclamen] (1,-0.2) -- (1,0.2);
\node [below right, cyclamen] at (1,-0.2) {$a_{j+2}$};
\draw [thick, cyclamen] (2,-0.2) -- (2,0.2);
\node [below right, cyclamen] at (2,-0.2) {$a_j$};
\draw [thick, cyclamen] (5.2,-0.2) -- (5.2,0.2);
\node [below right, cyclamen] at (5.2,-0.2) {$a_{j+2}$};
\draw [thick, cyclamen] (6,-0.2) -- (6,0.2);
\node [below right, cyclamen] at (6,-0.2) {$a_{j+3}$};
\draw [thick, cyclamen] (8.5,-0.2) -- (8.5,0.2);
\node [below right, cyclamen] at (8.5,-0.2) {$a_{j+4}$};
% notes
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\node [below] at (1,-1) {1};
\node [below] at (2,-1) {1};
\node [below] at (5.2,-1) {0};
\node [below] at (6,-1) {0};
\node [below] at (8.5,-1) {0};
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\draw [thick, ->] (1,-0.5) -- (1,-1);
\draw [thick, ->] (2,-0.5) -- (2,-1);
\draw [thick, ->] (5.2,-0.5) -- (5.2,-1);
\draw [thick, ->] (6,-0.5) -- (6,-1);
\draw [thick, ->] (8.5,-0.5) -- (8.5,-1);
\end{tikzpicture}
\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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$\hence \text{E}[\zeta_j]_j = \frac{1}{2}
\quad \wedge \quad \text{Var}[\zeta_j]_j = \frac{1}{4}$
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## Trapani test
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::: incremental
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- Define the function $\vartheta (u)$ as:
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$$
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\vartheta (u) = \frac{2}{\sqrt{r}}
\left[ \sum_{j} \zeta_j (u) - \frac{r}{2} \right]
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$$
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- If $a_j$ uniformly distributed, for the CLT:
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$$
\sum_j \zeta_j (u) \hence
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G \left( \frac{r}{2}, \frac{\sqrt{r}}{2} \right)
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\thus \vartheta (u) \hence
G \left( 0, 1 \right)
$$
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- Test statistic:
$$
\Theta = \int_{\underbar{u}}^{\bar{u}} du \, \vartheta^2 (u) \psi(u)
$$
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:::
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## Trapani test
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**Under** $\bold{H_0}$: $\mu_k \to +\infty$
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- $\Theta \to \chi^2$ as $n,r \to +\infty$
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. . .
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**Under** $\bold{H_a}$: $\mu_k < + \infty$
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::: incremental
- $\text{E}[\zeta_j] \neq \frac{1}{2}$
- the residues
$\vartheta (u) = \frac{2}{\sqrt{r}}
\sum_{j} \left[ \zeta_j (u) - \frac{1}{2} \right]$
become large
- $\Theta \to +\infty$ as $n,r \to +\infty$
:::
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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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- $\psi(u) = \frac{1}{2} \, \chi_{[\underbar{u}, \bar{u}]}$
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. . .
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$\mu_k$ must be scale invariant for $k > 1$:
$$
\mu_k^* = \frac{\mu_k}{ \left( \mu_{\phi} \right)^{k/\phi} }
\with \phi \in (0, k)
$$