slides: reshape a bit section 5-6
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@ -3,7 +3,9 @@
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## KS
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Quantify distance between expected and observed CDF. KS statistic:
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Quantify distance between expected and observed CDF.
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KS statistic:
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:::: {.columns}
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::: {.column width=50% .c}
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@ -11,9 +13,8 @@ Quantify distance between expected and observed CDF. KS statistic:
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D_N = \text{sup}_x |F_N(x) - F(x)|
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$$
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\vspace{20pt}
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- $F(x)$ is the expected CDF
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- $F_N(x)$ is the empirical CDF
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- sort points in ascending order
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- number of points preceding the point normalized by $N$
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@ -58,21 +59,18 @@ Quantify distance between expected and observed CDF. KS statistic:
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## KS
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$H_0$: points sampled according to $F(x)$
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$\bold{H_0}$: points sampled from $F$
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. . .
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::: incremental
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If $H_0$ is true: $\sqrt{N}D_N \xrightarrow{N \rightarrow + \infty} K$
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- $\sqrt{N}D_N \xrightarrow{N \rightarrow + \infty} K$, independent of $F$
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$K$ Kolmogorov variable with CDF:
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- Kolmogorov variable $K$ with CDF:
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$$
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P(K \leqslant K_0) = \frac{\sqrt{2 \pi}}{K_0}
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\sum_{j = 1}^{+ \infty} e^{-(2j - 1)^2 \pi^2 / 8 K_0^2}
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$$
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$$
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P(K \leqslant K_0) = \frac{\sqrt{2 \pi}}{K_0}
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\sum_{j = 1}^{+ \infty} e^{-(2j - 1)^2 \pi^2 / 8 K_0^2}
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$$
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- $p$-value given by: $p = 1 - P(K \leq K_0)$
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. . .
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A $p$-value can be computed
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- At 95% confidence level, $H_0$ cannot be disproved if $p > 0.05$
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:::
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@ -20,12 +20,12 @@
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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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- Sample $L$ from a Landau PDF
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- Sample $M$ from a Moyal PDF
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. . .
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\vspace{20pt}
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\vspace{2em}
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:::: {.columns}
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::: {.column width=50% .c}
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@ -50,24 +50,22 @@
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## Infinite moments
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- Previous tests: points sampled from Landau PDF?
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::: incremental
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. . .
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- Trapani test: check whether a moment is infinite
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- Trapani test: check whether a moment is finite or infinite
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- Consistency test:
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\begin{align*}
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\text{infinite} &\thus \text{Landau} \\
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\text{infinite} &\thus \text{may be 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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:::
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## Trapani test
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@ -163,7 +161,7 @@ $$
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\left[ \sum_{j} \zeta_j (u) - \frac{r}{2} \right]
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$$
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- If $a_j$ uniformly distributed, for the CLT:
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- If $a_j$ uniformly distributed, by the CLT:
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$$
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\sum_j \zeta_j (u) \hence
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G \left( \frac{r}{2}, \frac{\sqrt{r}}{2} \right)
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@ -205,18 +203,19 @@ $$
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## Trapani test
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According to L. Trapani [@trapani15]:
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MC simulations [@trapani15] gives:
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- $r = o(N) \hence r = N^{0.75}$
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- $\underbar{u} = -1 \quad \wedge \quad \bar{u} = 1$
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- $\psi(u) = \frac{1}{2} \, \chi_{[\underbar{u}, \bar{u}]}$
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- $\psi(u)$ uniform
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. . .
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$\mu_k$ must be scale invariant for $k > 1$:
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$$
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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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$$
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\Begin{alertblock}{Important}
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For $k > 1$, $\mu_k$ must be made scale-invariant:
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$$
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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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\End{alertblock}
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