How to find variance between multidimensional points? Suppose I have a matrix X which is n by p, i.e. it has n observations, with each observation in p-dimensional space.
How do I find the variance of these n observations?
In the case where p = 1, I just need to use the regular variance formula. What about the cases where p > 1.
 A: For  a $p$-dimensional random variable $X = {\left( {{X_1}, \ldots ,{X_p}} \right)^\intercal}$, we have the following definition of the variance:
$$Var\left( X \right) = E\left[ {\left( {X - EX} \right){{\left( {X - EX} \right)}^\intercal}} \right] = \left( {\begin{array}{*{20}{c}}
  {Var\left( {{X_1}} \right)}& \ldots &{Cov\left( {{X_1},{X_p}} \right)} \\ 
   \vdots & \ddots & \vdots  \\ 
  {Cov\left( {{X_p},{X_1}} \right)}& \ldots &{Var\left( {{X_p}} \right)} 
\end{array}} \right)$$
That is, the variance of a random vector is defined as the matrix which stores all the variances on the main diagonal and the covariances between the different components in the other elements. The sample $p \times p$ covariance matrix would then be calculated by plugging in the sample analogs for the population variables: 
$$\frac{1}{{n - 1}}\left( {\begin{array}{*{20}{c}}
  {\sum\limits_{i = 1}^n {{{\left( {{X_{i1}} - {{\bar X}_{\cdot1}}} \right)}^2}} }& \ldots &{\sum\limits_{i = 1}^n {\left( {{X_{i1}} - {{\bar X}_{\cdot1}}} \right)\left( {{X_{ip}} - {{\bar X}_{\cdot p}}} \right)} } \\ 
   \vdots & \ddots & \vdots  \\ 
  {\sum\limits_{i = 1}^n {\left( {{X_{ip}} - {{\bar X}_{\cdot p}}} \right)\left( {{X_{i1}} - {{\bar X}_{\cdot 1}}} \right)} }& \ldots &{\sum\limits_{i = 1}^n {{{\left( {{X_{ip}} - {{\bar X}_{\cdot p}}} \right)}^2}} } 
\end{array}} \right)$$
where ${X_{ij}}$ denotes the $i$th observation for feature $j$ and ${{\bar X}_{ \cdot j}}$ the sample mean of the $j$th feature. To sum up, the variance of a random vector is defined as the matrix containing the individual variances and covariances. It therefore suffices to calculate the sample variances and covariances for all the vector components individually.
