統計検定 1級 過去問 解答/解答例と解説
2014年11月30日 (日) 試験

$$\begin{equation} T \sim \varGamma(n+1) \end{equation}$$

$$\begin{equation} f_T(t) = \frac{1}{\varGamma(n+1)}t^{n}e^{-t} \end{equation}$$

$$\begin{align} f_{X_1, \ldots, X_{n+1}}(x_1, \ldots, x_{n+1}) &= \prod_{i=1}^{n+1}\frac{1}{\varGamma(1)}x^{1-1}e^{-x_i} \\ &= \prod_{i=1}^{n+1}e^{-x_i} \\ &= e^{-(x_1 + \cdots + x_{n+1})} \\ &= e^{-t} \end{align}$$

$$\begin{equation} t = x_1 + \cdots + x_{n+1} \end{equation}$$

$$\begin{align} x_1 &= y_1t \\ x_2 &= (y_2 - y_1)t \\ x_3 &= (y_3 - y_2)t \\ &\vdots \\ x_n &= (y_{n} - y_{n-1})t \end{align}$$

$$\begin{equation} x_{n+1} = t - (x_1 + \cdots + x_{n}) = t - t y_n = (1-y_n)t \end{equation}$$

$$\begin{align} J &= \begin{vmatrix} \ds\frac{\partial x_1}{\partial y_1} & \ds\frac{\partial x_1}{\partial y_2} & \ds\frac{\partial x_1}{\partial y_3} & \ds\frac{\partial x_1}{\partial t} \\ \ds\frac{\partial x_2}{\partial y_1} & \ds\frac{\partial x_2}{\partial y_2} & \ds\frac{\partial x_2}{\partial y_3} & \ds\frac{\partial x_2}{\partial t} \\ \ds\frac{\partial x_3}{\partial y_1} & \ds\frac{\partial x_3}{\partial y_2} & \ds\frac{\partial x_3}{\partial y_3} & \ds\frac{\partial x_3}{\partial t} \\ \ds\frac{\partial x_4}{\partial y_1} & \ds\frac{\partial x_4}{\partial y_2} & \ds\frac{\partial x_4}{\partial y_3} & \ds\frac{\partial x_4}{\partial t} \\ \end{vmatrix} \\ &= \begin{vmatrix} t & 0 & 0 & y_1 \\ -t & t & 0 & y_2 - y_1 \\ 0 & -t & t & y_3 - y_2 \\ 0 & 0 & -t & 1 - y_3 \end{vmatrix} \\ &= \begin{vmatrix} t & 0 & 0 & y_1 \\ 0 & t & 0 & y_2 \\ 0 & -t & t & y_3 - y_2 \\ 0 & 0 & -t & 1 - y_3 \end{vmatrix}~~\cdots\cdots~~\text{第2行目 + 第1行目} \\ &= \begin{vmatrix} t & 0 & 0 & y_1 \\ 0 & t & 0 & y_2 \\ 0 & 0 & t & y_3 \\ 0 & 0 & -t & 1 - y_3 \end{vmatrix}~~\cdots\cdots~~\text{第3行目 + 第2行目} \\ &= \begin{vmatrix} t & 0 & 0 & y_1 \\ 0 & t & 0 & y_2 \\ 0 & 0 & t & y_3 \\ 0 & 0 & 0 & 1 \end{vmatrix}~~\cdots\cdots~~\text{第4行目 + 第3行目} \\ &= t^3~~\cdots\cdots~~\text{三角行列の行列式は対角要素の積} \end{align}$$

$$\begin{equation} |J| = t^n \end{equation}$$

$$\begin{align} f_{Y_1, \ldots, Y_n, T}(y_1, \ldots, y_n, t) &= f_{X_1, \ldots, X_{n+1}}(y_1t,~(y_2 - y_1)t,~\ldots,~(y_{n} - y_{n-1})t,~(1-y_n)t)|J| \\ &= e^{-\left\{y_1t + (y_{2} - y_{1})t + \ldots + (y_{n} - y_{n-1})t + (1-y_n)t\right\}} \cdot t^n \\ &= e^{-t}\,t^n \end{align}$$