Help us understand the problem. What is going on with this article?

メモ:変数ベクトルを定数行列倍した時の共分散行列

結論

転置は後ろ。
$Ax \in G(0,A\Sigma A^\top)$

確率変数がスカラの場合

元の変数が定数倍された時の分散がどうなるかはわかりやすいので覚えていることでしょう。

$x \in G(0,\sigma)$
とするならば
$ax \in G(0,a^2\sigma)$

となるわけです。

確率変数が縦ベクトルの場合

この場合のベクトルとはn個の要素を持つ1次元のベクトルとします。

$x \in G(0,\Sigma)$

スカラ倍された場合

これも簡単
$ax \in G(0,a^2\Sigma)$

線形行列をかけた場合

では座標変換をかけた時はどうなるか。

二次形式なのは確定としてどっちに転置のせるか迷いません?

結果はこれのはず。
$Ax \in G(0,A\Sigma A^\top)$

紛らわしい箇所と導出

LQとかだったら$x^\top Q x$という形式を見がちなので前半に転置のせそうってちょっと思っちゃいます。

  • 実際には,

$\Sigma=E[(X-\mu)(X-\mu)^\top]=E[XX^\top] \hspace{1cm}(while\ \mu=0)$
なので

$\Sigma_A=E[(AX)(AX^\top)] = A \Sigma A^\top \hspace{1cm}(while\ \mu=0)$

となるわけですねぇ。

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