89 lines
2.6 KiB
Markdown
89 lines
2.6 KiB
Markdown
# Kolmogorov-Smirnov test
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## KS
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Quantify distance between expected and observed CDF
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. . .
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:::: {.columns}
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::: {.column width=50% .c}
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KS statistic:
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$$
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D_N = \text{sup}_x |F_N(x) - F(x)|
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$$
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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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:::
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::: {.column width=50%}
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\setbeamercovered{}
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\begin{center}
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\begin{tikzpicture}
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% axes
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\draw [thick, ->] (-2.5,0) -- (0,0) -- (0,4.5);
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\draw [thick, ->] (0,0) -- (2.5,0);
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% empiric
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\draw [cyclamen, fill=cyclamen!20!white] (-2.5,0) rectangle (-1.5,0.5);
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\draw [cyclamen, fill=cyclamen!20!white] (-1.5,0) rectangle (-0.9,1);
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\draw [cyclamen, fill=cyclamen!20!white] (-0.9,0) rectangle (-0.6,1.5);
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\draw [cyclamen, fill=cyclamen!20!white] (-0.6,0) rectangle ( 0.2,2);
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\draw [cyclamen, fill=cyclamen!20!white] ( 0.2,0) rectangle ( 0.5,2.5);
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\draw [cyclamen, fill=cyclamen!20!white] ( 0.5,0) rectangle ( 0.8,3);
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\draw [cyclamen, fill=cyclamen!20!white] ( 0.8,0) rectangle ( 1.6,3.5);
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\draw [cyclamen, fill=cyclamen!20!white] ( 1.6,0) rectangle ( 2.3,4);
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\draw [cyclamen, fill=cyclamen!20!white] ( 2.3,0) rectangle ( 2.5,4.5);
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% points
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\draw [blue!50!black, fill=blue] (-2.6,-0.1) rectangle (-2.4,0.1); %-2.5
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\draw [blue!50!black, fill=blue] (-1.6,-0.1) rectangle (-1.4,0.1); %-1.5
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\draw [blue!50!black, fill=blue] (-1,-0.1) rectangle (-0.8,0.1); %-0.9
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\draw [blue!50!black, fill=blue] (-0.7,-0.1) rectangle (-0.5,0.1); %-0.6
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\draw [blue!50!black, fill=blue] (0.1,-0.1) rectangle (0.3,0.1); % 0.2
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\draw [blue!50!black, fill=blue] (0.4,-0.1) rectangle (0.6,0.1); % 0.5
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\draw [blue!50!black, fill=blue] (0.7,-0.1) rectangle (0.9,0.1); % 0.8
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\draw [blue!50!black, fill=blue] (1.5,-0.1) rectangle (1.7,0.1); % 1.6
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\draw [blue!50!black, fill=blue] (2.2,-0.1) rectangle (2.4,0.1); % 2.3
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% expected
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\pause
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\draw[domain=-2.5:2.5, yscale=5, smooth, variable=\x, blue, very thick]
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plot ({\x}, {((atan(\x)*pi/180) + pi/2)/pi});
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\pause
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\draw [very thick, cyclamen] (0.8,3.6) -- (0.8,4.05);
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\end{tikzpicture}
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\end{center}
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\setbeamercovered{transparent}
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:::
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::::
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## KS
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$H_0$: points sampled according to $F(x)$
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. . .
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If $H_0$ is true:
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- $\sqrt{N}D_N \xrightarrow{N \rightarrow + \infty} K$
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Kolmogorov distribution with CDF:
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$$
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P(K \leqslant K_0) = 1 - p = \frac{\sqrt{2 \pi}}{K_0}
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\sum_{j = 1}^{+ \infty} e^{-(2j - 1)^2 \pi^2 / 8 K_0^2}
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$$
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. . .
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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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