analistica/slides/sections/6.md

Trapani test

Trapani test

::: incremental

• Random variable \left\{ x_i \right\} sampled from a distribution f
• Sample moments estimate as f moments
• H_0: \mu_k \longrightarrow + \infty
• Statistic with 1 dof \chi^2 distribution
• $p$-value \hence reject or accept H_0

:::

Infinite moments

• Generate a sample L from a Landau PDF
• Generate a sample M from a Moyal PDF

. . .

\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*} ::: ::::

Infinite moments

• Previous tests: points sampled from Landau PDF?

. . .

• Trapani test: check whether a moment is finite or infinite \begin{align*} \text{infinite} &\thus \text{Landau} \ \text{finite} &\thus \text{not Landau} \end{align*}

. . .

• Compatibility test with \mu_k = + \infty

. . .

• If points were sampled from a Cauchy distribution...

Trapani test

Trapani test

• Start with \left\{ x_i \right\}^N and compute \mu_k as:

  
    \mu_k = \frac{1}{N} \sum_{i = 1}^N |x_i|^k
  

. . .

• 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
    \thus G\left( 0, \sqrt{e^{\mu_k}} \right)
  

. . .

The greater \mu^k, the 'larger' G\left( 0, \sqrt{e^{\mu_k}} \right)

\begin{cases}
  \mu_k \longrightarrow + \infty \\
  r \longrightarrow + \infty
\end{cases}
\thus a_j \text{ distributed uniformly}

Trapani test

• Define the sequence: \left\{ \zeta_j (u) \right\}^r as:

  
    \zeta_j (u) = \theta( u -  a_j) \with \theta - \text{Heaviside}
  

. . .

\begin{center} \begin{tikzpicture} % 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 \node [below] at (1,-1) {0}; \node [below] at (2,-1) {0}; \node [below] at (5.2,-1) {1}; \node [below] at (6,-1) {1}; \node [below] at (8.5,-1) {1}; \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}

. . .

If a_j uniformly distributed:

• \zeta_j (u) Bernoulli PDF with $P\left( \zeta_j (u) = 1 \right) = \frac{1}{2}$ \hence E[\zeta_j]_j = \frac{1}{2} \quad \wedge \quad V[\zeta_j]_j = \frac{1}{4}

Trapani test

• Define the function \vartheta (u) as:

  \vartheta (u) = \frac{2}{\sqrt{r}}
                  \left[ \sum_{j} \zeta_j (u) - \frac{r}{2} \right]

. . .

If a_j uniformly distributed, for the CLT:

  \sum_j \zeta_j (u) \hence
    G \left( \frac{r}{2}, \frac{r}{4} \right)
  \thus \vartheta (u) \hence
    G \left( 0, 1 \right)

. . .

• Test statistic:

  \Theta = \int_{\underbar{u}}^{\bar{u}} du \, \vartheta^2 (u) \psi(u)

Trapani test

According to L. Trapani [@trapani15]:

• r = o(N) \hence r = N^{0.75}
• \underbar{u} = -1 \quad \wedge \quad \bar{u} = 1
• \psi(u) = \frac{1}{\bar{u} - \underbar{u}} \, \chi_{[\underbar{u}, \bar{u}]}

. . .

\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)

Trapani test

If \mu_k \ne + \infty \hence \left\{ a_j \right\} are not uniformly distributed

\vspace{20pt}

Rewriting:

  \vartheta (u) = \frac{2}{\sqrt{r}}
                  \left[ \sum_{j} \zeta_j (u) - \frac{r}{2} \right]
                = \frac{2}{\sqrt{r}}
                  \sum_{j} \left[ \zeta_j (u) - \frac{1}{2} \right]

\vspace{20pt}

. . .

Residues become very large \hence $p$-values decreases.