slides: reshape a bit section 5-6

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Michele Guerini Rocco 2020-06-13 17:47:37 +02:00
parent c9709c5a9c
commit 6ed0d04d6b
2 changed files with 34 additions and 37 deletions

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@ -3,7 +3,9 @@
## KS
Quantify distance between expected and observed CDF. KS statistic:
Quantify distance between expected and observed CDF.
KS statistic:
:::: {.columns}
::: {.column width=50% .c}
@ -11,9 +13,8 @@ Quantify distance between expected and observed CDF. KS statistic:
D_N = \text{sup}_x |F_N(x) - F(x)|
$$
\vspace{20pt}
- $F(x)$ is the expected CDF
- $F_N(x)$ is the empirical CDF
- sort points in ascending order
- number of points preceding the point normalized by $N$
@ -58,21 +59,18 @@ Quantify distance between expected and observed CDF. KS statistic:
## KS
$H_0$: points sampled according to $F(x)$
$\bold{H_0}$: points sampled from $F$
. . .
::: incremental
If $H_0$ is true: $\sqrt{N}D_N \xrightarrow{N \rightarrow + \infty} K$
- $\sqrt{N}D_N \xrightarrow{N \rightarrow + \infty} K$, independent of $F$
$K$ Kolmogorov variable with CDF:
$$
- Kolmogorov variable $K$ with CDF:
$$
P(K \leqslant K_0) = \frac{\sqrt{2 \pi}}{K_0}
\sum_{j = 1}^{+ \infty} e^{-(2j - 1)^2 \pi^2 / 8 K_0^2}
$$
$$
. . .
- $p$-value given by: $p = 1 - P(K \leq K_0)$
A $p$-value can be computed
- At 95% confidence level, $H_0$ cannot be disproved if $p > 0.05$
:::

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@ -20,12 +20,12 @@
## Infinite moments
- Generate a sample $L$ from a Landau PDF
- Generate a sample $M$ from a Moyal PDF
- Sample $L$ from a Landau PDF
- Sample $M$ from a Moyal PDF
. . .
\vspace{20pt}
\vspace{2em}
:::: {.columns}
::: {.column width=50% .c}
@ -50,24 +50,22 @@
## Infinite moments
- Previous tests: points sampled from Landau PDF?
::: incremental
. . .
- Trapani test: check whether a moment is infinite
- Trapani test: check whether a moment is finite or infinite
- Consistency test:
\begin{align*}
\text{infinite} &\thus \text{Landau} \\
\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
@ -163,7 +161,7 @@ $$
\left[ \sum_{j} \zeta_j (u) - \frac{r}{2} \right]
$$
- If $a_j$ uniformly distributed, for the CLT:
- If $a_j$ uniformly distributed, by the CLT:
$$
\sum_j \zeta_j (u) \hence
G \left( \frac{r}{2}, \frac{\sqrt{r}}{2} \right)
@ -205,18 +203,19 @@ $$
## Trapani test
According to L. Trapani [@trapani15]:
MC simulations [@trapani15] gives:
- $r = o(N) \hence r = N^{0.75}$
- $\underbar{u} = -1 \quad \wedge \quad \bar{u} = 1$
- $\psi(u) = \frac{1}{2} \, \chi_{[\underbar{u}, \bar{u}]}$
- $\psi(u)$ uniform
. . .
$\mu_k$ must be scale invariant for $k > 1$:
$$
\Begin{alertblock}{Important}
For $k > 1$, $\mu_k$ must be made scale-invariant:
$$
\mu_k^* = \frac{\mu_k}{ \left( \mu_{\phi} \right)^{k/\phi} }
\with \phi \in (0, k)
$$
$$
\End{alertblock}