Quadratic Form

二次型

假设 $ \boldsymbol{X}=\left(X_{1}, \cdots, X_{n}\right)^{\prime} $ 为$ n \times 1 $ 随机向量, $ \boldsymbol{A}=\left(a_{i j}\right)$ 为 $ n \times n $ 对称矩阵, 则称随机变量 $$ \boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}=\sum_{i=1}^{n} \sum_{j=1}^{n} a_{i j} X_{i} X_{j} $$ 为 $ \boldsymbol{X} $ 的二次型, 称 $ \boldsymbol{A} $ 为此二次型的二次型矩阵.

期望

定理1：设 $\mathrm{E}(\boldsymbol{X})=\boldsymbol{\mu}, \operatorname{Cov}(\boldsymbol{X})=\boldsymbol{\Sigma}$，则 $$ \mathrm{E}\left(\boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}\right)=\boldsymbol{\mu}^{\prime} \boldsymbol{A} \boldsymbol{\mu}+\operatorname{tr}(\boldsymbol{A} \boldsymbol{\Sigma}) $$

证明：因为 $$ \begin{aligned} \boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}=&(\boldsymbol{X}-\boldsymbol{\mu}+\boldsymbol{\mu})^{\prime} \boldsymbol{A}(\boldsymbol{X}-\boldsymbol{\mu}+\boldsymbol{\mu}) \ =&(\boldsymbol{X}-\boldsymbol{\mu})^{\prime} \boldsymbol{A}(\boldsymbol{X}-\boldsymbol{\mu})+\boldsymbol{\mu}^{\prime} \boldsymbol{A}(\boldsymbol{X}-\boldsymbol{\mu}) \ &+(\boldsymbol{X}-\boldsymbol{\mu})^{\prime} \boldsymbol{A} \boldsymbol{\mu}+\boldsymbol{\mu}^{\prime} \boldsymbol{A} \boldsymbol{\mu} . \end{aligned} $$ 上式第二项与第三项的数学期望为零. 因此，只需证明 $$ \mathrm{E}\left[(\boldsymbol{X}-\boldsymbol{\mu})^{\prime} \boldsymbol{A}(\boldsymbol{X}-\boldsymbol{\mu})\right]=\operatorname{tr}(\boldsymbol{A} \boldsymbol{\Sigma}) $$ 利用迹的性质 $\operatorname{tr}(\boldsymbol{A B})=\operatorname{tr}(\boldsymbol{B} \boldsymbol{A})$ 以及求迹和求期望可交换次序 , 可知 $$ \begin{aligned} \mathrm{E}\left[(\boldsymbol{X}-\boldsymbol{\mu})^{\prime} \boldsymbol{A}(\boldsymbol{X}-\boldsymbol{\mu})\right] &=\mathrm{E}\left[\operatorname{tr}\left((\boldsymbol{X}-\boldsymbol{\mu})^{\prime} \boldsymbol{A}(\boldsymbol{X}-\boldsymbol{\mu})\right)\right] \ &=\mathrm{E}\left{\operatorname{tr}\left[\boldsymbol{A}(\boldsymbol{X}-\boldsymbol{\mu})(\boldsymbol{X}-\boldsymbol{\mu})^{\prime}\right]\right} \ &=\operatorname{tr}\left{\boldsymbol{A E}\left[(\boldsymbol{X}-\boldsymbol{\mu})(\boldsymbol{X}-\boldsymbol{\mu})^{\prime}\right]\right} \ &=\operatorname{tr}(\boldsymbol{A} \boldsymbol{\Sigma}) \end{aligned} $$

方差

定理2：设随机变量 $X_{i}, i=1, \cdots, n$ 相互独立, 且 $\mathrm{E}\left(X_{i}\right)=\mu_{i}$ . 假设 $X_{i}-\mu_{i}, i=1, \cdots, n$ 是同分布的且 $$ \operatorname{Var}\left(X_{i}\right)=\sigma^{2}, m_{r}=E\left(X_{i}-\mu_{i}\right)^{r}, r=3,4 $$ $ \boldsymbol{A}=\left(a_{i j}\right)$ 为 $n \times n$ 对称矩阵. 记 $$ \boldsymbol{X}=\left(X_{1}, \cdots, X_{n}\right)^{\prime}, \quad \boldsymbol{\mu}=\left(\mu_{1}, \cdots, \mu_{n}\right)^{\prime} $$

则 $$ \operatorname{Var}\left(\boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}\right)=\left(m_{4}-3 \sigma^{4}\right) \boldsymbol{a}^{\prime} \boldsymbol{a}+2 \sigma^{4} \operatorname{tr}\left(\boldsymbol{A}^{2}\right)+4 \sigma^{2} \boldsymbol{\mu}^{\prime} \boldsymbol{A}^{2} \boldsymbol{\mu}+4 m_{3} \boldsymbol{\mu}^{\prime} \boldsymbol{A} \boldsymbol{a} $$

其中 $\boldsymbol{a}=\left(a_{11}, \cdots, a_{n n}\right)^{\prime}$ ，即 $\boldsymbol{A}$ 的对角元组成的列向量。

证明：首先注意到 $$ \operatorname{Var}\left(\boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}\right)=\mathrm{E}\left(\boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}\right)^{{2}-\left[\mathrm{E}\left(\boldsymbol{X}}{\prime} \boldsymbol{A} \boldsymbol{X}\right)\right]^{2}, $$

由定理1以及 $\mathrm{E}(\boldsymbol{X})=\boldsymbol{\mu}$ 和 $\operatorname{Cov}(\boldsymbol{X})=\sigma^{2} \boldsymbol{I}_{n}$ 可推得

$$\mathrm{E}\left(\boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}\right)=\boldsymbol{\mu}^{\prime} \boldsymbol{A} \boldsymbol{\mu}+\sigma^{2} \operatorname{tr}(\boldsymbol{A})$$

所以我们的问题主要是计算第一项（平方的期望）. 将 $\boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}$ 改写为

$$\boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}=(\boldsymbol{X}-\boldsymbol{\mu})^{\prime} \boldsymbol{A}(\boldsymbol{X}-\boldsymbol{\mu})+2 \boldsymbol{\mu}^{\prime} \boldsymbol{A}(\boldsymbol{X}-\boldsymbol{\mu})+\boldsymbol{\mu}^{\prime} \boldsymbol{A} \boldsymbol{\mu}$$

将其平方得到

$$\begin{aligned} \left(\boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}\right)^{2}=& {\left[(\boldsymbol{X}-\boldsymbol{\mu})^{\prime} \boldsymbol{A}(\boldsymbol{X}-\boldsymbol{\mu})\right]^{2}+4\left[\boldsymbol{\mu}^{\prime} \boldsymbol{A}(\boldsymbol{X}-\boldsymbol{\mu})\right]^{2}+\left(\boldsymbol{\mu}^{\prime} \boldsymbol{A} \boldsymbol{\mu}\right)^{2} } \\ &+2 \boldsymbol{\mu}^{\prime} \boldsymbol{A} \boldsymbol{\mu}\left[(\boldsymbol{X}-\boldsymbol{\mu})^{\prime} \boldsymbol{A}(\boldsymbol{X}-\boldsymbol{\mu})+2 \boldsymbol{\mu}^{\prime} \boldsymbol{A}(\boldsymbol{X}-\boldsymbol{\mu})\right] \\ &+4 \boldsymbol{\mu}^{\prime} \boldsymbol{A}(\boldsymbol{X}-\boldsymbol{\mu})(\boldsymbol{X}-\boldsymbol{\mu})^{\prime} \boldsymbol{A}(\boldsymbol{X}-\boldsymbol{\mu}) . \end{aligned}$$

令 $ \boldsymbol{Z}=\boldsymbol{X}-\boldsymbol{\mu}$ , 则 $\mathrm{E}(\boldsymbol{Z})=\mathbf{0}$ . 再次利用定理1得 $$ \begin{aligned} \mathrm{E}\left(\boldsymbol{X}^{\prime} \boldsymbol{A} \boldsymbol{X}\right)^{{2}=\mathrm{E}\left(\boldsymbol{Z}}{\prime} \boldsymbol{A} \boldsymbol{Z}\right)^{2}+4 \mathrm{E}\left(\boldsymbol{\mu}^{\prime} \boldsymbol{A} \boldsymbol{Z}\right)^{{2}+\left(\boldsymbol{\mu}}{\prime} \boldsymbol{A} \boldsymbol{\mu}\right)^{2} \ +2 \boldsymbol{\mu}^{\prime} \boldsymbol{A} \boldsymbol{\mu}\left(\sigma^{2} \operatorname{tr}(\boldsymbol{A})\right)+4 \mathrm{E}\left(\boldsymbol{\mu}^{\prime} \boldsymbol{A} \boldsymbol{Z} \boldsymbol{Z}^{\prime} \boldsymbol{A} \boldsymbol{Z}\right) \end{aligned} $$ 下面逐个计算上式所含的每个均值. 由 $$ \left(\boldsymbol{Z}^{\prime} \boldsymbol{A} \boldsymbol{Z}\right)^{2}=\sum_{i} \sum_{j} \sum_{k} \sum_{l} a_{i j} a_{k l} Z_{i} Z_{j} Z_{k} Z_{l} $$ 及 $ Z_{i} $ 的独立性可得 $$ \mathrm{E}\left(Z_{i} Z_{j} Z_{k} Z_{l}\right)=\left{\begin{array}{ll} m_{4}, & \text { 若 } i=j=k=l, \ \sigma^{4}, & \text { 若 } i=j \neq k=l ; i=k \neq j=l ; i=l \neq j=k, \ 0, & \text { 其它, } \end{array}\right. $$ 便有下列结果: $$ \mathrm{E}\left(\boldsymbol{Z}^{\prime} \boldsymbol{A} \boldsymbol{Z}\right)^{2} = m_{4}\left(\sum_{i = 1}^{n} a_{i i}^{{2}\right)+\sigma}{4}\left(\sum_{i \neq k} a_{i i} a_{k k}+\sum_{i \neq j} a_{i j}^{2}+\sum_{i \neq j} a_{i j} a_{j i}\right) \ = m_{4} \boldsymbol{a}^{\prime} \boldsymbol{a}+\sigma^{{4}\left{[\operatorname{tr}(\boldsymbol{A})]}{2}-\boldsymbol{a}^{\prime} \boldsymbol{a}+2\left[\operatorname{tr}\left(\boldsymbol{A}^{{2}\right)-\boldsymbol{a}}{\prime} \boldsymbol{a}\right]\right} \ = \left(m_{4}-3 \sigma^{4}\right) \boldsymbol{a}^{\prime} \boldsymbol{a}+\sigma^{{4}\left{[\operatorname{tr}(\boldsymbol{A})]}{2}+2 \operatorname{tr}\left(\boldsymbol{A}^{2}\right)\right}, \quad(2.2 .5) \ $$ 而 $$ \begin{aligned} \mathrm{E}\left(\boldsymbol{\mu}^{\prime} \boldsymbol{A} \boldsymbol{Z}\right)^{2} = \mathrm{E}\left(\boldsymbol{Z}^{\prime} \boldsymbol{A} \boldsymbol{\mu} \boldsymbol{\mu}^{\prime} \boldsymbol{A} \boldsymbol{Z}\right) \ = \sigma^{2} \cdot \operatorname{tr}\left(\boldsymbol{A} \boldsymbol{\mu} \boldsymbol{\mu}^{\prime} \boldsymbol{A}\right) \ = \sigma^{2} \cdot \boldsymbol{\mu}^{\prime} \boldsymbol{A}^{2} \boldsymbol{\mu} . \end{aligned} $$ 最后, 若记 $b=A \mu$ , 则 $$ \mathrm{E}\left(\boldsymbol{\mu}^{\prime} \boldsymbol{A} \boldsymbol{Z} \boldsymbol{Z}^{\prime} \boldsymbol{A} \boldsymbol{Z}\right)=\sum_{i} \sum_{j} \sum_{k} b_{i} a_{j k} \mathrm{E}\left(Z_{i} Z_{j} Z_{k}\right) $$ 因为 $$ \mathrm{E}\left(Z_{i} Z_{j} Z_{k}\right)=\left{\begin{array}{ll} m_{3}, & \text { 若 } i=j=k, \ 0, & \text { 其它, } \end{array}\right. $$ 所以 $$ \mathrm{E}\left(\boldsymbol{\mu}^{\prime} \boldsymbol{A} \boldsymbol{Z} \boldsymbol{Z}^{\prime} \boldsymbol{A} \boldsymbol{Z}\right)=m_{3} \sum_{i} b_{i} a_{i i}=m_{3} \boldsymbol{b}^{\prime} \boldsymbol{a}=m_{3} \boldsymbol{\mu}^{\prime} \boldsymbol{A} \boldsymbol{a} $$ 代回，便可得到需要证明的结果.

整理自王松桂老师《线性模型引论》和庞天晓老师课件，使用 https://www.latexlive.com/home 进行 Latex OCR。

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