方差，协方差及协方差矩阵的计算
1.方差
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方差是用来衡量一组数据的离散程度，数序表达式如下:
σ2=1N∑i=1N(xi−μ)2 \sigma^2=\frac1N\sum_{i=1}^N(x_i-\mu)^2 σ2=N1i=1∑N(xi−μ)2- σ2σ^2σ2表示样本的总体方差，
- NNN 表示样本总数，
- xix _ixi是第 i 个样本，
- μμμ 是数据集的平均值(μ=1N∑i=1Nxi)(\mu=\frac1N\sum_{i=1}^Nx_i)(μ=N1∑i=1Nxi)
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假设某班级有 5 名学生的数学成绩为：85,74,63,95,9985, 74, 63, 95, 9985,74,63,95,99, 计算平均值为 μ=85+74+63+95+995=83.2μ = \frac{85 +74 + 63 + 95 + 99} 5 =83.2μ=585+74+63+95+99=83.2 ,方差的计算如下：
σ成绩2=(85−83.5)2+(74−83.5)2+(63−83.5)2+(95−83.5)2+(99−83.5)25=177.05 {σ_{成绩}}^2 = \frac{(85-83.5)^2 +(74-83.5)^2 + (63-83.5)^2 + (95-83.5)^2 + (99 - 83.5)^2}{5} = 177.05 σ成绩2=5(85−83.5)2+(74−83.5)2+(63−83.5)2+(95−83.5)2+(99−83.5)2=177.05
2.协方差
- 协方差是用来衡量两组样本之间的关系，数序表达式如下:
σxσy=1N∑i=1N(xi−μx)(yi−μy) \sigma_x\sigma_y=\frac{1}{N}\sum_{i=1}^N(x_i-\mu_x)(y_i-\mu_y) σxσy=N1i=1∑N(xi−μx)(yi−μy)- σxσyσ_xσ_yσxσy表示两组样本的总体协方差，
- NNN 是样本对的总数，
- xi,yix _i, y_ixi,yi 是第 i 对样本，
- μx,μyμ_x,μ_yμx,μy分别表示两组样本的均值。
- 例如，要计算如下 x,yx, yx,y 两个样本的协方差：
| x | y |
|---|---|
| 500 | 200 |
| 600 | 250 |
| 451 | 180 |
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- 计算xxx的平均值 ux=500+600+4513=517u_x = \frac{500 +600 +451}3 = 517ux=3500+600+451=517
- 计算yyy的平均值 uy=200+250+1803=210u_y = \frac{200 +250 +180}3 = 210uy=3200+250+180=210
σxσy=(500−517)(200−210)+(600−517)(250−210)+(451−517)(180−210)3=1823.33... σ_xσ_y = \frac{(500-517)(200 - 210) + (600-517)(250-210) + (451-517)(180-210) }{3} = 1823.33... σxσy=3(500−517)(200−210)+(600−517)(250−210)+(451−517)(180−210)=1823.33...
3.协方差矩阵
协方差矩阵就是将多组数据的方差和协方差用矩阵的形式表达出来，例如如下三组样本,其协方差的矩阵排布如下，从下图可以看出，对角线上是各个样本的方差，两边则是样本之间的协方差
| x | y | z |
|---|---|---|
| x1x_1x1 | y1y_1y1 | z1z_1z1 |
| x2x_2x2 | y2y_2y2 | z2z_2z2 |
| x3x_3x3 | y3y_3y3 | z3z_3z3 |
cov=[σx2σxσyσxσzσyσxσy2σyσzσzσxσzσyσx2]
cov = \left[\begin{array}{ccc}σ_x^2&σ_xσ_y&σ_xσ_z\\
σ_yσ_x&σ_y^2&σ_yσ_z\\
σ_zσ_x&σ_zσ_y&σ_x^2\end{array}\right]
cov=σx2σyσxσzσxσxσyσy2σzσyσxσzσyσzσx2
4.通过矩阵运算求解一个矩阵的协方差矩阵
- 将第三节使用的样本写成矩阵的形式,本节主要求解这个矩阵的协方差矩阵
[x1y1z1x2y2z2x3y3z3] \left[\begin{array}{ccc}x_1&y_1&z_1\\ x_2&y_2&z_2\\ x_3&y_3&z_3\end{array}\right] x1x2x3y1y2y3z1z2z3 - 求出过度矩阵 aaa,根据下面计算可知，过度矩阵 aaa 其实是求解了每个样本的 标准差
a=[x1y1z1x2y2z2x3y3z3]−13[111111111][x1y1z1x2y2z2x3y3z3]−−−−a=[x1y1z1x2y2z2x3y3z3]−13[x1+x2+x3y1+y2+y3z1+z2+z3x1+x2+x3y1+y2+y3z1+z2+z3x1+x2+x3y1+y2+y3z1+z2+z3]−−−−a=[x1−13(x1+x2+x3)y1−13(y1+y2+y3)z1−13(z1+z2+z3)x2−13(x1+x2+x3)y2−13(y1+y2+y3)z2−13(z1+z2+z3)x3−13(x1+x2+x3)y3−13(y1+y2+y3)z3−13(z1+z2+z3)] a = \left[\begin{array}{ccc}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{array}\right] - \frac{1}{3} \left[\begin{array}{ccc}1&1&1\\1&1&1\\1&1&1\end{array}\right] \left[\begin{array}{ccc}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{array}\right] \\ ----\\ a = \left[\begin{array}{ccc}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{array}\right] - \frac{1}{3} \left[\begin{array}{ccc}x_1+x_2+x_3&y_1+y_2+y_3&z_1+z_2+z_3\\x_1+x_2+x_3&y_1+y_2+y_3&z_1+z_2+z_3\\x_1+x_2+x_3&y_1+y_2+y_3&z_1+z_2+z_3\end{array}\right] \\ ---- \\ a = \left[\begin{array}{ccc}x_1 - \frac{1}{3} (x_1+x_2+x_3)&y_1 - \frac{1}{3} (y_1+y_2+y_3)&z_1 -\frac{1}{3} (z_1+z_2+z_3) \\x_2 - \frac{1}{3} (x_1+x_2+x_3)&y_2 - \frac{1}{3} (y_1+y_2+y_3)&z_2 -\frac{1}{3} (z_1+z_2+z_3)\\x_3 - \frac{1}{3} (x_1+x_2+x_3)&y_3 - \frac{1}{3} (y_1+y_2+y_3)&z_3 -\frac{1}{3} (z_1+z_2+z_3)\end{array}\right] a=x1x2x3y1y2y3z1z2z3−31111111111x1x2x3y1y2y3z1z2z3−−−−a=x1x2x3y1y2y3z1z2z3−31x1+x2+x3x1+x2+x3x1+x2+x3y1+y2+y3y1+y2+y3y1+y2+y3z1+z2+z3z1+z2+z3z1+z2+z3−−−−a=x1−31(x1+x2+x3)x2−31(x1+x2+x3)x3−31(x1+x2+x3)y1−31(y1+y2+y3)y2−31(y1+y2+y3)y3−31(y1+y2+y3)z1−31(z1+z2+z3)z2−31(z1+z2+z3)z3−31(z1+z2+z3) - 最后求解协方差矩阵P
P=13∗aT∗a P = \frac{1}{3} * a^T *a P=31∗aT∗a
aT=[x1−13(x1+x2+x3)x2−13(x1+x2+x3)x3−13(x1+x2+x3)y1−13(y1+y2+y3)y2−13(y1+y2+y3)y3−13(y1+y2+y3)z1−13(z1+z2+z3)z2−13(z1+z2+z3)z3−13(z1+z2+z3)] a^T = \left[\begin{array}{ccc}x_1 - \frac{1}{3} (x_1+x_2+x_3)&x_2 - \frac{1}{3} (x_1+x_2+x_3)&x_3 - \frac{1}{3} (x_1+x_2+x_3) \\ y_1 - \frac{1}{3} (y_1+y_2+y_3)&y_2 - \frac{1}{3} (y_1+y_2+y_3)&y_3 - \frac{1}{3} (y_1+y_2+y_3)\\ z_1 - \frac{1}{3} (z_1+z_2+z_3)&z_2 -\frac{1}{3} (z_1+z_2+z_3)&z_3 -\frac{1}{3}(z_1+z_2+z_3)\end{array}\right] aT=x1−31(x1+x2+x3)y1−31(y1+y2+y3)z1−31(z1+z2+z3)x2−31(x1+x2+x3)y2−31(y1+y2+y3)z2−31(z1+z2+z3)x3−31(x1+x2+x3)y3−31(y1+y2+y3)z3−31(z1+z2+z3)
P=13∗[x1−13(x1+x2+x3)x2−13(x1+x2+x3)x3−13(x1+x2+x3)y1−13(y1+y2+y3)y2−13(y1+y2+y3)y3−13(y1+y2+y3)z1−13(z1+z2+z3)z2−13(z1+z2+z3)z3−13∗(z1+z2+z3)]∗[x1−13(x1+x2+x3)y1−13(y1+y2+y3)z1−13(z1+z2+z3)x2−13(x1+x2+x3)y2−13(y1+y2+y3)z2−13(z1+z2+z3)x3−13(x1+x2+x3)y3−13(y1+y2+y3)z3−13(z1+z2+z3)] P = \frac{1}{3} * \left[\begin{array}{ccc}x_1 - \frac{1}{3} (x_1+x_2+x_3)&x_2 - \frac{1}{3} (x_1+x_2+x_3)&x_3 - \frac{1}{3} (x_1+x_2+x_3) \\ y_1 - \frac{1}{3} (y_1+y_2+y_3)&y_2 - \frac{1}{3} (y_1+y_2+y_3)&y_3 - \frac{1}{3} (y_1+y_2+y_3)\\ z_1 - \frac{1}{3} (z_1+z_2+z_3)&z_2 -\frac{1}{3} (z_1+z_2+z_3)&z_3 -\frac{1}{3} *(z_1+z_2+z_3)\end{array}\right] * \left[\begin{array}{ccc}x_1 - \frac{1}{3} (x_1+x_2+x_3)&y_1 - \frac{1}{3} (y_1+y_2+y_3)&z_1 -\frac{1}{3} (z_1+z_2+z_3) \\x_2 - \frac{1}{3} (x_1+x_2+x_3)&y_2 - \frac{1}{3} (y_1+y_2+y_3)&z_2 -\frac{1}{3} (z_1+z_2+z_3)\\x_3 - \frac{1}{3} (x_1+x_2+x_3)&y_3 - \frac{1}{3} (y_1+y_2+y_3)&z_3 -\frac{1}{3} (z_1+z_2+z_3)\end{array}\right] P=31∗x1−31(x1+x2+x3)y1−31(y1+y2+y3)z1−31(z1+z2+z3)x2−31(x1+x2+x3)y2−31(y1+y2+y3)z2−31(z1+z2+z3)x3−31(x1+x2+x3)y3−31(y1+y2+y3)z3−31∗(z1+z2+z3)∗x1−31(x1+x2+x3)x2−31(x1+x2+x3)x3−31(x1+x2+x3)y1−31(y1+y2+y3)y2−31(y1+y2+y3)y3−31(y1+y2+y3)z1−31(z1+z2+z3)z2−31(z1+z2+z3)z3−31(z1+z2+z3)
- 如此运算便可通过矩阵运算求出协方差矩阵
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