え 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.
you can not get the answer of 111x111 in 0.1 second but I kan. but it's not important. because I already mastered that.
\[\chi^2(k) \overset{\text{def}}{=} \sum_{i=1}^k Z_i^2,\quad Z_i \overset{\text{i.i.d.}}{\sim} N(0,1)
\]
\[f_{\chi^2(k)}(x) = \frac{1}{2^{k/2}\Gamma(k/2)}x^{k/2-1}e^{-x/2},\quad x>0
\]
\[\phi_{\chi^2(k)}(t) = (1-2it)^{-k/2}
\]
\[X_1,\dots,X_n \overset{\text{i.i.d.}}{\sim} N(\mu,\sigma^2),\quad \bar{X} = \frac{1}{n}\sum_{i=1}^n X_i
\]
\[Z_i = \frac{X_i-\mu}{\sigma} \sim N(0,1)
\]
\[\exists\ \mathbf{H}\in\mathbb{R}^{n\times n}:\ \mathbf{H}\mathbf{H}^\top = \mathbf{I},\quad \mathbf{Y} = \mathbf{H}\mathbf{Z}
\]
\[Y_1 = \sqrt{n}\,\bar{Z},\quad Y_2,\dots,Y_n \perp Y_1,\quad Y_i \overset{\text{i.i.d.}}{\sim} N(0,1)
\]
\[\sum_{i=1}^n (X_i-\bar{X})^2 = \sigma^2\sum_{i=1}^n (Z_i-\bar{Z})^2 = \sigma^2\sum_{i=2}^n Y_i^2
\]
\[\frac{1}{\sigma^2}\sum_{i=1}^n (X_i-\bar{X})^2 \sim \chi^2(n-1)
\]
\[\phi_{\frac{1}{\sigma^2}\sum (X_i-\bar{X})^2}(t) = (1-2it)^{-(n-1)/2}
\]
\[(O_1,\dots,O_k) \sim \text{Multinomial}(n;p_1,\dots,p_k),\quad \sum_{i=1}^k p_i = 1
\]
\[E_i = np_i,\quad Q = \sum_{i=1}^k \frac{(O_i-E_i)^2}{E_i}
\]
\[\sqrt{n}\left(\frac{O_i}{n}-p_i\right) \xrightarrow{d} N\left(0,\ \text{diag}(p_i)-\mathbf{p}\mathbf{p}^\top\right)
\]
\[\text{rank}\left(\text{diag}(p_i)-\mathbf{p}\mathbf{p}^\top\right) = k-1
\]
\[Q \xrightarrow{d} \sum_{i=1}^{k-1} Z_i^2 \sim \chi^2(k-1)
\]
\[\lim_{n\to\infty}\phi_Q(t) = (1-2it)^{-(k-1)/2}
\]
\[X_1,\dots,X_9 \overset{\text{i.i.d.}}{\sim} N(2,5),\quad \sigma^2 = 5,\ n = 9
\]
\[S^2 = \frac{1}{8}\sum_{i=1}^9 (X_i-\bar{X})^2
\]
\[\frac{8S^2}{5} \sim \chi^2(8)
\]
\[aS^2 \sim \chi^2(b) \implies a = \frac{8}{5},\ b = 8
\]
\[ab = \frac{8}{5}\times 8 = \frac{64}{5}
\]