Covariance

The covariance matrix of an m×n matrix X = $\left[\vphantom{ x_{ij}}\right.$xij$\left.\vphantom{ x_{ij}}\right]$ is an n×n matrix with (i, j)th entry

$\displaystyle {\frac{{\dsum\limits_{k=1}^{m}\left( x_{ki}-\frac{\sum_{s=1}^{m}x...
...j}-\frac{\sum_{t=1}^{m}x_{tj}}{%
\stackrel{\vspace{1pt}}{m}}\right) }}{{m-1}}}$

Note that for each i, the (i, i)th entry is the variance of the data in the ith column, making the variances of the column vectors occur down the main diagonal of the covariance matrix. The definition of covariance matrix is symmetric in i and j, so the covariance matrix is always a symmetric matrix.

$\blacktriangleright$ Statistics + Mean, Statistics + Variance, Statistics + Covariance

$\left[\vphantom{
\begin{array}{cc}
1 & 2 \\
3 & 5 \\
4 & 3
\end{array}
}\right.$$\begin{array}{cc}
1 & 2 \\
3 & 5 \\
4 & 3
\end{array}$$\left.\vphantom{
\begin{array}{cc}
1 & 2 \\
3 & 5 \\
4 & 3
\end{array}
}\right]$, Mean(s): 2.7, 3.3, Variance(s): $\left[\vphantom{
2.3333,2.3333}\right.$2.3333, 2.3333$\left.\vphantom{
2.3333,2.3333}\right]$,

                Covariance matrix: $\left[\vphantom{
\begin{array}{cc}
2.3333 & 1.1667 \\
1.1667 & 2.3333
\end{array}
}\right.$$\begin{array}{cc}
2.3333 & 1.1667 \\
1.1667 & 2.3333
\end{array}$$\left.\vphantom{
\begin{array}{cc}
2.3333 & 1.1667 \\
1.1667 & 2.3333
\end{array}
}\right]$

$\left[\vphantom{
\begin{array}{rrr}
8.5 & -5.5 & -3.7 \\
-3.5 & 9.7 & 5.0 \\
7.9 & 5.6 & 4.9
\end{array}
}\right.$$\begin{array}{rrr}
8.5 & -5.5 & -3.7 \\
-3.5 & 9.7 & 5.0 \\
7.9 & 5.6 & 4.9
\end{array}$$\left.\vphantom{
\begin{array}{rrr}
8.5 & -5.5 & -3.7 \\
-3.5 & 9.7 & 5.0 \\
7.9 & 5.6 & 4.9
\end{array}
}\right]$, Mean(s): 4.3, 3.3, 2.1, Variance(s): $\left[\vphantom{ 45.72,61.843,24.943}\right.$45.72, 61.843, 24.943$\left.\vphantom{ 45.72,61.843,24.943}\right]$,

                Covariance matrix: $\left[\vphantom{
\begin{array}{ccc}
45.72 & -39.3 & -18.45 \\
-39.3 & 61.843 & 38.018 \\
-18.45 & 38.018 & 24.943
\end{array}
}\right.$$\begin{array}{ccc}
45.72 & -39.3 & -18.45 \\
-39.3 & 61.843 & 38.018 \\
-18.45 & 38.018 & 24.943
\end{array}$$\left.\vphantom{
\begin{array}{ccc}
45.72 & -39.3 & -18.45 \\
-39.3 & 61.843 & 38.018 \\
-18.45 & 38.018 & 24.943
\end{array}
}\right]$

You can also apply Covariance to a matrix with column headings.

$\blacktriangleright$ Statistics + Mean, Statistics + Variance, Statistics + Covariance

$\left[\vphantom{
\begin{array}{ccc}
\text{x} & \text{y} & \text{z} \\
1 & 1 & 4 \\
3 & 2 & 5 \\
5 & 3 & 6 \\
7 & 4 & 7
\end{array}
}\right.$$\begin{array}{ccc}
\text{x} & \text{y} & \text{z} \\
1 & 1 & 4 \\
3 & 2 & 5 \\
5 & 3 & 6 \\
7 & 4 & 7
\end{array}$$\left.\vphantom{
\begin{array}{ccc}
\text{x} & \text{y} & \text{z} \\
1 & 1 & 4 \\
3 & 2 & 5 \\
5 & 3 & 6 \\
7 & 4 & 7
\end{array}
}\right]$, Mean(s): 4.0, 2.5, 5.5, Variance(s): $\left[\vphantom{ 6.6667,1.6667,1.6667}\right.$6.6667, 1.6667, 1.6667$\left.\vphantom{ 6.6667,1.6667,1.6667}\right]$,

            Covariance matrix: $\left[\vphantom{
\begin{array}{ccc}
6.6667 & 3.3333 & 3.3333 \\
3.3333 & 1.6667 & 1.6667 \\
3.3333 & 1.6667 & 1.6667
\end{array}
}\right.$$\begin{array}{ccc}
6.6667 & 3.3333 & 3.3333 \\
3.3333 & 1.6667 & 1.6667 \\
3.3333 & 1.6667 & 1.6667
\end{array}$$\left.\vphantom{
\begin{array}{ccc}
6.6667 & 3.3333 & 3.3333 \\
3.3333 & 1.6667 & 1.6667 \\
3.3333 & 1.6667 & 1.6667
\end{array}
}\right]$

$\left[\vphantom{
\begin{array}{cc}
\text{x} & \text{y} \\
a & b \\
u & v
\end{array}
}\right.$$\begin{array}{cc}
\text{x} & \text{y} \\
a & b \\
u & v
\end{array}$$\left.\vphantom{
\begin{array}{cc}
\text{x} & \text{y} \\
a & b \\
u & v
\end{array}
}\right]$, Mean(s): .5a + .5u,.5b + .5v,

        Variance(s): $\left[\vphantom{ \left( .5a-.5u\right) ^{2}+\left( .5u-.5a\right)
^{2},\left( .5b-.5v\right) ^{2}+\left( .5v-.5b\right) ^{2}}\right.$$\left(\vphantom{ .5a-.5u}\right.$.5a - .5u$\left.\vphantom{ .5a-.5u}\right)^{{2}}_{}$ + $\left(\vphantom{ .5u-.5a}\right.$.5u - .5a$\left.\vphantom{ .5u-.5a}\right)^{{2}}_{}$,$\left(\vphantom{ .5b-.5v}\right.$.5b - .5v$\left.\vphantom{ .5b-.5v}\right)^{{2}}_{}$ + $\left(\vphantom{ .5v-.5b}\right.$.5v - .5b$\left.\vphantom{ .5v-.5b}\right)^{{2}}_{}$$\left.\vphantom{ \left( .5a-.5u\right) ^{2}+\left( .5u-.5a\right)
^{2},\left( .5b-.5v\right) ^{2}+\left( .5v-.5b\right) ^{2}}\right]$,

        Covariance matrix: $\left[\vphantom{
\begin{array}{cc}
.5a^{2}-1.0au+.5u^{2} & .5ab-.5av-.5ub+.5uv \\
.5ab-.5av-.5ub+.5uv & .5b^{2}-1.0bv+.5v^{2}
\end{array}
}\right.$$\begin{array}{cc}
.5a^{2}-1.0au+.5u^{2} & .5ab-.5av-.5ub+.5uv \\
.5ab-.5av-.5ub+.5uv & .5b^{2}-1.0bv+.5v^{2}
\end{array}$$\left.\vphantom{
\begin{array}{cc}
.5a^{2}-1.0au+.5u^{2} & .5ab-.5av-.5ub+.5uv \\
.5ab-.5av-.5ub+.5uv & .5b^{2}-1.0bv+.5v^{2}
\end{array}
}\right]$

Note    The final matrix was obtained by applying Simplify to the first covariance matrix produced. Notice that the first row was interpreted as labels and ignored in the computation.