MATLAB 中两个矩阵之间的协方差
我有两个矩阵，X 和 Y，每列代表随机变量的多个实现；
X = [x_11 x_21 .... x_n1
x_12 x_22 .... x_n2
. . .... .
. . .... .
x_1m x_2m .... x_nm]
其中 Y 是 X 的函数： Y = f(X)
Y = [y_11 y_21 .... y_n1
y_12 y_22 .... y_n2
. . .... .
. . .... .
y_1m y_2m .... y_nm]
我想找到变量 x_n 和 y_n 之间的协方差矩阵；
E{(X - E{Y}) * (Y - E{Y})^H}
其中 ()^H 表示向量的 厄米转置
在 matlab 中，当我运行 cov(X,Y) 在矩阵上，（每 1000 次试验 20 个变量）我只得到一个 2x2 矩阵，这让我相信它将每个矩阵视为单个矩阵不知何故“变量”。如果我连接两个矩阵并对结果调用 cov：
cov( [X Y] )
我得到一个 40x40 矩阵，cov( X ) 的结果位于左上角，< code>cov( Y ) 在右下角，我想要的矩阵在右上角和左下角，但是有没有一种方法可以计算这个而不必求助于这个？
谢谢
I have two matricies, X and Y, with each column representing multiple realizations of a random variable;
X = [x_11 x_21 .... x_n1
x_12 x_22 .... x_n2
. . .... .
. . .... .
x_1m x_2m .... x_nm]
And where Y is a function of X: Y = f(X)
Y = [y_11 y_21 .... y_n1
y_12 y_22 .... y_n2
. . .... .
. . .... .
y_1m y_2m .... y_nm]
I want to find the covariance matrix between the variables x_n and y_n;
E{(X - E{Y}) * (Y - E{Y})^H}
Where ()^H denotes the Hermitian Transpose of the vector
In matlab, when I run cov(X,Y) on the matricies, (each 1000 trials of 20 variables) I only get a 2x2 matrix back, which leads me to believe that it is treating each matrix as a single "variable" somehow. If I concatenate the two matricies and call cov on the result:
cov( [X Y] )
I get a 40x40 matrix, with the result of cov( X ) in the top left, the result of cov( Y ) in the bottom right, and the matrix I want in the top right and bottom left, but is there a way to calculate this without having to resort to this?
Thanks
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cov(X,Y)相当于cov([x(:) y(:)])。但是[x(:) y(:)]对您来说是 20000 x 2，并且cov()将行视为观察值，将列视为维度，因此您得到 2通过 2 个协方差矩阵。我会根据定义自己实现它：
如果您有不支持
bsxfun()的旧版本 matlab，只需使用repmat()。cov(X,Y)is equivalent tocov([x(:) y(:)]). But[x(:) y(:)]is 20000 by 2 for you, andcov()treats rows as observations and columns as dimensions, so you get a 2 by 2 covariance matrix.I would just implement it myself by the definition:
If you have an older version of matlab that doesn't support
bsxfun(), just userepmat().