\Sigma=(\rho^\left | j-i \right | )_{p\times p},求tr(\Sigma^2) ,tr(\Sigma^4)
时间: 2023-08-27 12:15:34 浏览: 46
根据矩阵乘法的定义,$\Sigma^2$ 中的第 $(i,j)$ 个元素为
$$
(\Sigma^2)_{i,j}=\sum_{k=1}^p \rho^{\left|j-k\right|}\rho^{\left|k-i\right|}=\sum_{k=1}^p \rho^{\left|j-k\right|+\left|k-i\right|}
$$
当 $j\neq i$ 时,上式等价于
$$
(\Sigma^2)_{i,j}=\sum_{k=i}^{j-1} \rho^{j-i}+\sum_{k=j}^{i-1} \rho^{i-j}=\begin{cases}
(j-i)\rho^{j-i}, & j>i \\
(i-j)\rho^{i-j}, & j<i
\end{cases}
$$
当 $i=j$ 时,$(\Sigma^2)_{i,i}=\sum_{k=1}^p \rho^{2\left|i-k\right|}=\sum_{k=0}^{i-1} \rho^{2(i-k)}+\sum_{k=i}^{p-1} \rho^{2(k-i)}=\frac{1-\rho^{2i}}{1-\rho^2}+\frac{1-\rho^{2(p-i)}}{1-\rho^2}$。因此,$\Sigma^2$ 的迹为
$$
tr(\Sigma^2)=\sum_{i=1}^p (\Sigma^2)_{i,i}+\sum_{i\neq j} (\Sigma^2)_{i,j}=p+\frac{p(p-1)}{2}(j-i)\rho^{j-i}
$$
类似地,$\Sigma^4$ 中的第 $(i,j)$ 个元素为
$$
(\Sigma^4)_{i,j}=\sum_{k=1}^p (\Sigma^2)_{i,k}(\Sigma^2)_{k,j}=\sum_{k=1}^p (\Sigma^2)_{i,k}(\Sigma^2)_{j,k}
$$
当 $j\neq i$ 时,上式等价于
$$
(\Sigma^4)_{i,j}=\sum_{k=i}^{j-1} (j-k)(k-i)\rho^{2(j-i)}+\sum_{k=j}^{i-1} (i-k)(k-j)\rho^{2(i-j)}=\frac{(j-i)^2}{2}(3-2\rho^{2(j-i)})
$$
当 $i=j$ 时,$(\Sigma^4)_{i,i}=\sum_{k=1}^p (\Sigma^2)_{i,k}^2=\left(\frac{1-\rho^{2i}}{1-\rho^2}+\frac{1-\rho^{2(p-i)}}{1-\rho^2}\right)^2+\sum_{k\neq i}\left[(\Sigma^2)_{i,k}\right]^2$。因此,$\Sigma^4$ 的迹为
$$
\begin{aligned}
tr(\Sigma^4)&=\sum_{i=1}^p (\Sigma^4)_{i,i}+\sum_{i\neq j} (\Sigma^4)_{i,j} \\
&=p\frac{(3-p)+(p-1)(1-\rho^2)}{2}+\frac{p(p-1)(j-i)^2}{2}(3-2\rho^{2(j-i)})
\end{aligned}
$$