• 卡方分佈
  • 自由度為1的卡方分佈
  • 自由度為n的卡方分佈
• t分佈
• F分佈
• 樣本平均數 & 樣本變異數

卡方分佈

令 $Z \sim N(0, 1)$ 為常態隨機變量

平均數：$E(V) = n$
變異數：$Var(V) = 2n$

自由度為1的卡方分佈

• 定義： $\bar{U} = Z^2$ 為自由度為1的卡方分佈，記為 $\chi^2(1)$
• cdf：$F(u) = P(\bar{U} \leq u) = P(Z^2 \leq u) = P(-\sqrt{u} \leq Z \leq \sqrt{u}) = \Phi(\sqrt{u}) - \Phi(-\sqrt{u})$
• pdf：$\displaystyle f(u) = \phi(\sqrt{u}) \frac{1}{2} u^{-1/2} = \phi(-\sqrt{u}) \frac{1}{2} u^{-1/2}$ = $\frac{1}{\sqrt{2\pi}} e^{-u/2} u^{-1/2}$, $u > 0$

自由度為n的卡方分佈

• 定義：$\bar{U_1}, \bar{U_2}, ..., \bar{U_n}$ 為獨立且服從 $\chi^2(1)$ 的隨機變量，則 $V = \sum_{i=1}^{n} \bar{U_i}$ 稱為自由度為n的卡方分佈，記為 $\chi^2(n)$
• pdf：$\displaystyle f(v) = \frac{1}{2^{n/2} \Gamma(n/2)} v^{n/2 - 1} e^{-v/2}$, $v > 0$

如果 $V$ 和 $U$ 獨立，且 $V \sim \chi^2(m)$, $U \sim \chi^2(n)$，則 $V + U \sim \chi^2(m + n)$

• 證明：
  $\displaystyle M_{\sum_{i=1}^{m} \bar{U_i}}(t) = \prod_{i=1}^{m} M_{\bar{U_i}}(t) \Rightarrow V \sim P(\frac{n}{2}, \frac{1}{2})$,

  一般化：$\displaystyle V \sim P(\frac{n}{2}, \frac{1}{2})$, $\displaystyle U \sim P(\frac{m}{2}, \frac{1}{2})$，則 $\displaystyle V + U \sim \chi^2(m + n)$

t分佈

• 定義：$Z \sim N(0, 1)$, $V \sim \chi^2(n)$, $Z, V$ 獨立，則 $T = \frac{Z}{\sqrt{V/n}}$ 稱為自由度為n的t分佈，記為 $t(n)$
• pdf：$\displaystyle f(t) = \frac{\Gamma(\frac{n+1}{2})}{\sqrt{n\pi} \Gamma(\frac{n}{2})} (1 + \frac{t^2}{n})^{-\frac{n+1}{2}}$, $-\infty < t < \infty$

  證明：

  1. Derive the pdf of $\bar{U}/n = Y$
     $\begin{aligned} F_Y(y) &= P(\sqrt{\bar{U}/n} \leq y) = P(\bar{U} \leq ny^2) = F_{\bar{U}}(ny^2) \\ f_Y(y) &= \frac{d}{dy} F_Y(y) = \frac{d}{dy} F_{\bar{U}}(ny^2) = (2ny) f_{\bar{U}}(ny^2) \\ &= (2ny) \frac{1}{2^{(n/2)} \Gamma(n/2)} (ny^2)^{(n/2) - 1} e^{-ny^2/2} = \frac{1}{2^{(n/2)-1}\Gamma(n/2)}n^{n/2} y^{n-1} e^{-ny^2/2} \end{aligned}$
  2. The pdf of $T = Z/Y$
     $\displaystyle \begin{aligned} f_T(t) &= \int_{-\infty}^{\infty} |y| f_{Z, Y}(y, ty) dy \\[10pt] &= \int_{-\infty}^{\infty} y f_Y(y) f_Z(ty) dy \\[10pt] &= \int_{-\infty}^{\infty} \left[ y \frac{1}{2^{(n/2)-1}\Gamma(n/2)}n^{n/2} y^{n-1} e^{-ny^2/2}\right] \left[\frac{1}{\sqrt{2\pi}} e^{-t^2y^2/2}\right] dy \\[10pt] &= \left[ \frac{n^{n/2}}{2^{(n/2)-1}\Gamma(n/2) \sqrt{\pi}} \right] \int_{0}^{\infty} y^{n} e^{-(n+t^2)y^2/2} dy \end{aligned}$
  3. Derive the integral
     $\begin{aligned} \int_{0}^{\infty} y^{n} e^{-(n+t^2)y^2/2} dy &= \int_{0}^{\infty} (y^2)^{n/2} e^{-(n+t^2)y^2/2} dy \\[10pt] &= \frac{1}{2} \int_{0}^{\infty} (y^2)^{(n-1)/2} e^{-(n+t^2)y^2/2} (2y) dy \\[10pt] &= \frac{1}{2} s^{(n-1)/2} e^{-(n+t^2)s/2} ds \\[10pt] &= \frac{1}{2} \frac{\Gamma((n+1)/2)}{[(n+t^2)/2]^{(n+1)/2}} \int_{0}^{\infty} \frac{((n+t^2)/2)^{(n+1)/2}}{\Gamma((n+1)/2)} s^{(n+1)/2-1} e^{-(n+t^2)s/2} ds \\[10pt] &= \frac{1}{2} \frac{\Gamma((n+1)/2)}{[(n+t^2)/2]^{(n+1)/2}} \end{aligned}$
  4. Derive the pdf of $T$
     $\displaystyle f_T(t) = \frac{n^{n/2}}{2^{(n/2)-1}\Gamma(n/2) \sqrt{\pi}} \left[\frac{1}{2} \frac{\Gamma((n+1)/2)}{[(n+t^2)/2]^{(n+1)/2}}\right] = \frac{\Gamma((n+1)/2)}{\sqrt{(n\pi)} \Gamma(n/2)} \left[1 + \frac{t^2}{n}\right]^{-(n+1)/2}$
     $-\infty < t < \infty$
     FACT:
     $t_n \rightarrow Z$ as $n \rightarrow \infty$
     $\displaystyle \lim_{n \rightarrow \infty} \left[1 + \frac{t^2}{n}\right]^{(n+1)/2} = \lim_{n \rightarrow \infty} \left[1 + \frac{t^2/2}{n/2}\right]^{n/2}\left[1 + \frac{t^2/2}{n}\right]^{1/2} = e^{t^2/2}$

F分佈

• 定義：$\bar{U} \sim \chi^2(m)$, $\bar{V} \sim \chi^2(n)$, $\bar{U}, \bar{V}$ 獨立，則 $\displaystyle F = \frac{\bar{U}/m}{\bar{V}/n}$ 稱為自由度為m, n的F分佈，記為 $F(m, n)$
• pdf：$\displaystyle f(f) = \frac{\Gamma(\frac{m+n}{2})}{\Gamma(\frac{m}{2}) \Gamma(\frac{n}{2})} \left(\frac{m}{n}\right)^{m/2} f^{m/2 - 1} (1 + \frac{m}{n} f)^{-(m+n)/2}$, $f > 0$

  證明：

  • $\bar{U} \sim \chi^2(m)$, $\bar{V} \sim \chi^2(n)$, $\bar{U}, \bar{V}$ 獨立，則 $\displaystyle \frac{\bar{U}/m}{\bar{V}/n} \sim F(m, n)$
    $\begin{aligned} \displaystyle f_{\bar{U}}(u) &= \frac{1}{2^{m/2} \Gamma(m/2)} u^{m/2 - 1} e^{-u/2} \\[10pt] f_{\bar{V}}(v) &= \frac{1}{2^{n/2} \Gamma(n/2)} v^{n/2 - 1} e^{-v/2} \end{aligned}$

  1. $\displaystyle F = \frac{\bar{U}/m}{\bar{V}/n} \Rightarrow \begin{cases} \bar{U} = u \\ \bar{V} = (m/n) u \end{cases}$

     Jacobian : $\displaystyle |J| = \begin{vmatrix} \partial \bar{V}/\partial F & \partial \bar{V}/\partial u \\ \partial \bar{U}/\partial F & \partial \bar{U}/\partial u \end{vmatrix} = \begin{vmatrix} (m/n) u & (m/n) F \\ 0 & 1 \end{vmatrix} = -\frac{m}{n} u$

  2. $\displaystyle f(F, u) = f_{\bar{U}}(u) f_{\bar{V}}((m/n)u) |J| = \frac{(m/n)^{m/2}}{2^{(m+n)/2} \Gamma(m/2) \Gamma(n/2)} u^{(m+n)/2 - 1} e^{-u(1+(m/n)F)} F^{m/2 - 1}$

  3. The pdf of $F$
     $\displaystyle \begin{aligned} f(F) = \int_{0}^{\infty} f(F, u) du &= \frac{(m/n)^{m/2}F^{m/2 - 1}}{2^{(m+n)/2} \Gamma(m/2) \Gamma(n/2)} \int_{0}^{\infty} u^{(m+n)/2 - 1} e^{-u(1+(m/n)F)/2} du \\[10pt] &= \frac{(m/n)^{m/2}F^{m/2 - 1}}{2^{(m+n)/2} \Gamma(m/2) \Gamma(n/2)} \frac{\Gamma((m+n)/2)}{(1+(m/n)F)^{(m+n)/2}} \\[10pt] &= \frac{\Gamma((m+n)/2)}{\Gamma(m/2) \Gamma(n/2)} \left(\frac{m}{n}\right)^{m/2} F^{m/2 - 1} (1 + \frac{m}{n} F)^{-(m+n)/2} \end{aligned}$

樣本平均數 & 樣本變異數

令 $X_1, X_2, ..., X_n \xrightarrow{iid} N(\mu, \sigma^2)$

• 樣本平均數：$\displaystyle \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i$
• 樣本變異數：$\displaystyle S^2 = \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X})^2$

引理：$\bar{X}_n$ and $S^2_n$ are independent