統計検定 準1級 例題集 解答/解答例と解説

問題の要約

1. $\begin{pmatrix} X + Y \\ X - Y \end{pmatrix}$ が従う 2 変量正規分布を求めよ．

2. $X$ を与えたときの $Y$ の条件つき分布を求めよ．

解答

1. 答 : ③

   $$\begin{align} E[X + Y] &= E[X] + E[Y] \\ &= 1.0 + 2.0 \\ &= 3.0 \end{align}$$
   $$\begin{align} E[X - Y] &= E[X] - E[Y] \\ &= 1.0 - 2.0 \\ &= -1.0 \end{align}$$
   $$\begin{align} V[X + Y] &= V[X] + 2Cov[X, Y] + V[Y] \\ &= 3.0 + 2 \times 2.0 + 4.0 \\ &= 11.0 \end{align}$$
   $$\begin{align} V[X - Y] &= V[X] - 2Cov[X, Y] + V[Y] \\ &= 3.0 - 2 \times 2.0 + 4.0 \\ &= 3.0 \end{align}$$
   $$\begin{align} &Cov[X + Y, X - Y] \\ &= E[(X + Y - E[X + Y])(X - Y - E[X - Y])] \\ &= E[(X + Y - E[X] - E[Y])(X - Y - E[X] + E[Y])] \\ &= E[\{(X - E[X]) + (Y - E[Y])\}\{(X - E[X]) - (Y - E[Y])\}] \\ &= E[(X - E[X])^2 - (Y - E[Y])^2] \\ &= E[(X - E[X])^2] - E[(Y - E[Y])^2] \\ &= V[X] + V[Y] \\ &= 3.0 - 4.0 \\ &= -1.0 \end{align}$$
   よって，
   $$\begin{equation} \begin{pmatrix} X ＋ Y \\ X - Y \end{pmatrix} ~\sim~ N\left( \begin{pmatrix} 3.0 \\ -1.0 \end{pmatrix}, \begin{pmatrix} 11.0 & -1.0 \\ -1.0 & 3.0 \end{pmatrix} \right) \end{equation}$$

2. 答 : ④

   $$\begin{equation} (X, Y)~\sim~ N\left( \begin{pmatrix} \mu_X \\ \mu_Y \end{pmatrix}, \begin{pmatrix} {\sigma_X}^2 & {\sigma_{XY}} \\ {\sigma_{XY}} & {\sigma_Y}^2 \end{pmatrix} \right) \end{equation}$$
   のとき，
   $$\begin{equation} Y \mid X~\sim~ N\left( \mu_Y + \frac{\sigma_{XY}}{{\sigma_X}^2}(X - \mu_X),~ {\sigma_Y}^2 - \frac{{\sigma_{XY}}^2}{{\sigma_X}^2} \right) \end{equation}$$
   である．
   $$\begin{align} Y \mid X &~\sim~ N\left( 2.0 + \frac{2.0}{3.0}(X - 1.0),~ 4.0 - \frac{{2.0}^2}{3.0} \right) \\ &~\sim~ N\left( 1.33 +0.67X,~ 2.67 \right) \end{align}$$