# Trapani test


## Trapani test

::: incremental

  - Random variables $\left\{ x_i \right\}$ sampled from a distribution $f$

  - Sample moments converge to $f$ moments

  - $H_0$: $\mu_k \longrightarrow + \infty$

  - Statistic with 1 dof $\chi^2$ distribution

  - $p$-value $\rightarrow$ reject or accept $H_0$

:::


## Infinite moments

- Sample $L$ from a Landau PDF
- Sample $M$ from a Moyal PDF

. . .

\vspace{2em}

:::: {.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

::: incremental

- Trapani test: check whether a moment is infinite

- Consistency test:
\begin{align*}
  \text{infinite} &\thus \text{may be 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

![](images/cauchy-pdf.pdf)


## Trapani test

::: incremental

- 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}[>=Stealth]
  % 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)   {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};
  \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 $P\left( \zeta_j (u) = 1 \right) = \frac{1}{2}$
  $\hence \text{E}[\zeta_j]_j = \frac{1}{2}
   \quad \wedge \quad \text{Var}[\zeta_j]_j = \frac{1}{4}$


## Trapani test

::: incremental

- 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, by the CLT:
$$
  \sum_j \zeta_j (u) \hence
    G \left( \frac{r}{2}, \frac{\sqrt{r}}{2} \right)
  \thus \vartheta (u) \hence
    G \left( 0, 1 \right)
$$

- Test statistic:
$$
  \Theta = \int_{\underbar{u}}^{\bar{u}} du \, \vartheta^2 (u) \psi(u)
$$

:::


## Trapani test

**Under** $\bold{H_0}$: $\mu_k \to +\infty$

- $\Theta \to \chi^2$ as $n,r \to +\infty$

. . .

**Under** $\bold{H_a}$: $\mu_k < + \infty$

::: 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 $n,r \to +\infty$

:::


## Trapani test

MC simulations [@trapani15] gives:

- $r = o(N) \hence r = N^{0.75}$
- $\underbar{u} = -1 \quad \wedge \quad \bar{u} = 1$
- $\psi(u)$ uniform

. . .

\Begin{alertblock}{Important}

  For $k > 1$, $\mu_k$ must be made scale-invariant:
  $$
    \mu_k^* = \frac{\mu_k}{ \left( \mu_{\phi} \right)^{k/\phi} }
    \with \phi \in (0, k)
  $$
\End{alertblock}