$\chi^2(k)$

いし habe nur die Dinge geschrieben, die ich nicht vollständig verstand. Vielleicht wusstest du das schon, aber es muss einige Buchstaben geben, die du nicht ㄋㄠ.
você não consegue obter a resposta de 111x111 em 0,1 segundo mas eu posso. mas não é importante. porque eu já dominava isso.

\[\chi^2(k) \overset{\text{def}}{=} \sum_{i=1}^k Z_i^2,\quad Z_i \overset{\text{i.i.d.}}{\sim} N(0,1) \]

\[f_{\chi^2(k)}(x) = \frac{1}{2^{k/2}\Gamma(k/2)}x^{k/2-1}e^{-x/2},\quad x>0 \]

\[\phi_{\chi^2(k)}(t) = (1-2it)^{-k/2} \]

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\[X_1,\dots,X_n \overset{\text{i.i.d.}}{\sim} N(\mu,\sigma^2),\quad \bar{X} = \frac{1}{n}\sum_{i=1}^n X_i \]

\[Z_i = \frac{X_i-\mu}{\sigma} \sim N(0,1) \]

\[\exists\ \mathbf{H}\in\mathbb{R}^{n\times n}:\ \mathbf{H}\mathbf{H}^\top = \mathbf{I},\quad \mathbf{Y} = \mathbf{H}\mathbf{Z} \]

\[Y_1 = \sqrt{n}\,\bar{Z},\quad Y_2,\dots,Y_n \perp Y_1,\quad Y_i \overset{\text{i.i.d.}}{\sim} N(0,1) \]

\[\sum_{i=1}^n (X_i-\bar{X})^2 = \sigma^2\sum_{i=1}^n (Z_i-\bar{Z})^2 = \sigma^2\sum_{i=2}^n Y_i^2 \]

\[\frac{1}{\sigma^2}\sum_{i=1}^n (X_i-\bar{X})^2 \sim \chi^2(n-1) \]

\[\phi_{\frac{1}{\sigma^2}\sum (X_i-\bar{X})^2}(t) = (1-2it)^{-(n-1)/2} \]

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\[(O_1,\dots,O_k) \sim \text{Multinomial}(n;p_1,\dots,p_k),\quad \sum_{i=1}^k p_i = 1 \]

\[E_i = np_i,\quad Q = \sum_{i=1}^k \frac{(O_i-E_i)^2}{E_i} \]

\[\sqrt{n}\left(\frac{O_i}{n}-p_i\right) \xrightarrow{d} N\left(0,\ \text{diag}(p_i)-\mathbf{p}\mathbf{p}^\top\right) \]

\[\text{rank}\left(\text{diag}(p_i)-\mathbf{p}\mathbf{p}^\top\right) = k-1 \]

\[Q \xrightarrow{d} \sum_{i=1}^{k-1} Z_i^2 \sim \chi^2(k-1) \]

\[\lim_{n\to\infty}\phi_Q(t) = (1-2it)^{-(k-1)/2} \]

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\[X_1,\dots,X_9 \overset{\text{i.i.d.}}{\sim} N(2,5),\quad \sigma^2 = 5,\ n = 9 \]

\[S^2 = \frac{1}{8}\sum_{i=1}^9 (X_i-\bar{X})^2 \]

\[\frac{8S^2}{5} \sim \chi^2(8) \]

\[aS^2 \sim \chi^2(b) \implies a = \frac{8}{5},\ b = 8 \]

\[ab = \frac{8}{5}\times 8 = \frac{64}{5} \]