多項分布から多変量正規分布への近似について(8)

問題

\[ |\mathrm{A}| = \left| \begin{array}{cccc} 1 + \frac{p_{1}}{p_{5}} & \frac{\sqrt{p_{1}p_{2}}}{p_{5}} & \frac{\sqrt{p_{1}p_{3}}}{p_{5}} & \frac{\sqrt{p_{1}p_{4}}}{p_{5}} \\ \frac{\sqrt{p_{1}p_{2}}}{p_{5}} & 1 + \frac{p_{2}}{p_{5}} & \frac{\sqrt{p_{2}p_{3}}}{p_{5}} & \frac{\sqrt{p_{2}p_{4}}}{p_{5}} \\ \frac{\sqrt{p_{1}p_{3}}}{p_{5}} & \frac{\sqrt{p_{2}p_{3}}}{p_{5}} & 1 + \frac{p_{3}}{p_{5}} & \frac{\sqrt{p_{3}p_{4}}}{p_{5}} \\ \frac{\sqrt{p_{1}p_{4}}}{p_{5}} & \frac{\sqrt{p_{2}p_{4}}}{p_{5}} & \frac{\sqrt{p_{1}p_{3}}}{p_{5}} & 1 + \frac{p_{4}}{p_{5}} \\ \end{array} \right| = \frac{1}{p_{5}} \]
になることを示せ。

考えたこと

大きな行列式を機械的に計算する1つの方法に、行列式の値を変えない操作を繰り返して、三角行列を作るという方法がある。このときの行列式の値は、三角行列の対角成分の積になる。

今回の場合は2つの操作を合わせて1回と数えたとき、\(i\)回目の操作を、次のように定義する。

1. \(i+1\)列の\(\sqrt{\frac{p_{i}}{p_{i+1}}}\)倍を\(i\)列から引く
2. \(i\)行の\(\sqrt{\frac{p_{i}}{p_{i+1}}}\)倍を\(i+1\)行に足す

後はこのルールに従って、計算していけば、三角行列が得られる(最初の等式の右辺は計算を追いやすいように書き方を工夫しています)。

\[ \begin{align*} |\mathrm{A}|&= \left| \begin{array}{cccc} 1 + \frac{\sqrt{p_{1}}}{p_{5}}\left(\frac{p_{1}}{\sqrt{p_{1}}}\right) & \frac{\sqrt{p_{2}}}{p_{5}}\left(\frac{p_{1}}{\sqrt{p_{1}}}\right) & \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{p_{1}}{\sqrt{p_{1}}}\right) & \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{p_{1}}{\sqrt{p_{1}}}\right)\\ \frac{\sqrt{p_{1}}}{p_{5}}\left(\frac{p_{2}}{\sqrt{p_{2}}}\right) & 1 + \frac{\sqrt{p_{2}}}{p_{5}}\left(\frac{p_{2}}{\sqrt{p_{2}}}\right) & \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{p_{2}}{\sqrt{p_{2}}}\right) & \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{p_{2}}{\sqrt{p_{2}}}\right)\\ \frac{\sqrt{p_{1}}}{p_{5}}\left(\frac{p_{3}}{\sqrt{p_{3}}}\right) & \frac{\sqrt{p_{2}}}{p_{5}}\left(\frac{p_{3}}{\sqrt{p_{3}}}\right) & 1 + \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{p_{3}}{\sqrt{p_{3}}}\right) & \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{p_{3}}{\sqrt{p_{3}}}\right)\\ \frac{\sqrt{p_{1}}}{p_{5}}\left(\frac{p_{4}}{\sqrt{p_{4}}}\right) & \frac{\sqrt{p_{2}}}{p_{5}}\left(\frac{p_{4}}{\sqrt{p_{4}}}\right) & \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{p_{4}}{\sqrt{p_{4}}}\right) & 1 + \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{p_{4}}{\sqrt{p_{4}}}\right)\\ \end{array} \right|\\ &= \left| \begin{array}{cccc} 1 & \frac{\sqrt{p_{2}}}{p_{5}}\left(\frac{p_{1}}{\sqrt{p_{1}}}\right) & \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{p_{1}}{\sqrt{p_{1}}}\right) & \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{p_{1}}{\sqrt{p_{1}}}\right)\\ -\sqrt{\frac{p_{1}}{p_{2}}} & 1 + \frac{\sqrt{p_{2}}}{p_{5}}\left(\frac{p_{2}}{\sqrt{p_{2}}}\right) & \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{p_{2}}{\sqrt{p_{2}}}\right) & \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{p_{2}}{\sqrt{p_{2}}}\right)\\ 0 & \frac{\sqrt{p_{2}}}{p_{5}}\left(\frac{p_{3}}{\sqrt{p_{3}}}\right) & 1 + \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{p_{3}}{\sqrt{p_{3}}}\right) & \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{p_{3}}{\sqrt{p_{3}}}\right)\\ 0 & \frac{\sqrt{p_{2}}}{p_{5}}\left(\frac{p_{4}}{\sqrt{p_{4}}}\right) & \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{p_{4}}{\sqrt{p_{4}}}\right) & 1 + \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{p_{4}}{\sqrt{p_{4}}}\right)\\ \end{array} \right|\\ &= \left| \begin{array}{cccc} 1 & \frac{\sqrt{p_{2}}}{p_{5}}\left(\frac{p_{1}}{\sqrt{p_{1}}}\right) & \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{p_{1}}{\sqrt{p_{1}}}\right) & \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{p_{1}}{\sqrt{p_{1}}}\right)\\ 0 & 1 + \frac{\sqrt{p_{2}}}{p_{5}}\left(\frac{\sum_{l=1}^{2}p_{l}}{\sqrt{p_{2}}}\right) & \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{\sum_{l=1}^{2}p_{l}}{\sqrt{p_{2}}}\right) & \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{\sum_{l=1}^{2}p_{l}}{\sqrt{p_{2}}}\right)\\ 0 & \frac{\sqrt{p_{2}}}{p_{5}}\left(\frac{p_{3}}{\sqrt{p_{3}}}\right) & 1 + \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{p_{3}}{\sqrt{p_{3}}}\right) & \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{p_{3}}{\sqrt{p_{3}}}\right)\\ 0 & \frac{\sqrt{p_{2}}}{p_{5}}\left(\frac{p_{4}}{\sqrt{p_{4}}}\right) & \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{p_{4}}{\sqrt{p_{4}}}\right) & 1 + \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{p_{4}}{\sqrt{p_{4}}}\right)\\ \end{array} \right|\\ &= \left| \begin{array}{cccc} 1 & 0 & \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{p_{1}}{\sqrt{p_{1}}}\right) & \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{p_{1}}{\sqrt{p_{1}}}\right)\\ 0 & 1 & \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{\sum_{l=1}^{2}p_{l}}{\sqrt{p_{2}}}\right) & \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{\sum_{l=1}^{2}p_{l}}{\sqrt{p_{2}}}\right)\\ 0 & 0 & 1 + \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{\sum_{l=1}^{3}p_{l}}{\sqrt{p_{3}}}\right) & \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{\sum_{l=1}^{3}p_{l}}{\sqrt{p_{3}}}\right)\\ 0 & 0 & \frac{\sqrt{p_{3}}}{p_{5}}\left(\frac{p_{4}}{\sqrt{p_{4}}}\right) & 1 + \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{p_{4}}{\sqrt{p_{4}}}\right)\\ \end{array} \right|\\ &= \left| \begin{array}{cccc} 1 & 0 & 0 & \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{p_{1}}{\sqrt{p_{1}}}\right)\\ 0 & 1 & 0 & \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{\sum_{l=1}^{2}p_{l}}{\sqrt{p_{2}}}\right)\\ 0 & 0 & 1 & \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{\sum_{l=1}^{3}p_{l}}{\sqrt{p_{3}}}\right)\\ 0 & 0 & 0 & 1 + \frac{\sqrt{p_{4}}}{p_{5}}\left(\frac{\sum_{l=1}^{4}p_{l}}{\sqrt{p_{4}}}\right)\\ \end{array} \right| \end{align*} \]

三角行列の対角成分の積から行列式が\(\sum_{l=1}^{5}p_{l}/p_{5}\)だと分かる。更に\(\sum_{i=1}^{5} p_{i} = 1\)を使うと\(1/p_{5}\)が得られる。これが示したいことであった。

次回

次回、最終回は今回の計算を参考に一般の場合の計算を扱います。三角行列を得る手続きは、次元に依らず変わらないことから一般化できます。