This is a more succinct question from a previous post, but I have arrived at two different answers, and need help determining which - if either - is correct.
I start with a 4*3 matrix:
5 4 -1
2 3 -3
3 4 -4
1 3 -2
and I need to compute the variance-covariance matrix. I had originally tried:
For the non-diagonals, calculate the covariance values using the formula:
$cov(x,y)=\frac{1}{n-1} \sum_{i=1}^n~(x_i- \overline{x} )*(y_i- \overline{y} )$
For the diagonals, I caculate the variance values using the formula:
$var(x)=\frac{1}{n-1} \sum_{i=1}^n~(x_i- \overline{x} )^2$
which lead me to an "answer" of:
2.916 4.042 -1.458
4.042 0.333 -2.917
-1.458 -2.917 1.666
However, I tried to double-check this hand-done work using R, and wrote:
m = matrix(c(5,2,3,1,4,3,4,3,-1,-3,-4,-2), nrow=4, ncol=3)
covM = cov(m, method = "pearson")
which lead to a different "answer" of:
2.916 0.833 0.833
0.833 0.333 0.000
0.833 0.000 1.666
So, it makes me wonder if I am using the wrong R function, or if I am using the wrong equations and calculations. Any ideas are greatly appreciated!!
Edit: What I am asking is, notice that the diagonal is the same for my hand calculation and R calculation. Only the off-diagonal is incorrect. For my hand calculation of the off-diagonals, I used the covariance equation I listed above (for my hand calculation of the diagonals, I used the varaince equation I listed above). The off-diagonal use of the covariance equation is what differs with the R output. So, am I using the correct equation? Many thanks!
Here is an example of how I got an off-diagonal value, and it was different than the same cell of the R output:
cov(1,2)=((5-2.75)(4-3.5)+(2-2.75)(3-3.5)+(3-2.75)(4-3.5)+(1-2.75)(3-3.5))/(4-1)=4.042