Sample/Population Covariance Calculator

Find the sample covariance between $\left\{4, 6, 1, 2, 3\right\}$ and $\left\{1, 4, 5, 3, 2\right\}$.

The sample covariance of data is given by the formula $cov(x,y) = \frac{\sum_{i=1}^{n} \left(x_{i} - \mu_{x}\right)\cdot \left(y_{i} - \mu_{y}\right)}{n - 1}$, where $n$ is the number of values, $x_i, i=\overline{1..n}$ and $y_i, i=\overline{1..n}$ are the values themselves, $\mu_{x}$ is the mean of the x-values, and $\mu_{y}$ is the mean of the y-values.

The mean of the x-values is $\mu_{x} = \frac{16}{5}$ (for calculating it, see mean calculator).

The mean of the y-values is $\mu_{y} = 3$ (for calculating it, see mean calculator).

Since we have $n$ points, $n = 5$.

The sum of $\left(x_{i} - \mu_{x}\right)\cdot \left(y_{i} - \mu_{y}\right)$ is $\left(4 - \frac{16}{5}\right)\cdot \left(1 - 3\right) + \left(6 - \frac{16}{5}\right)\cdot \left(4 - 3\right) + \left(1 - \frac{16}{5}\right)\cdot \left(5 - 3\right) + \left(2 - \frac{16}{5}\right)\cdot \left(3 - 3\right) + \left(3 - \frac{16}{5}\right)\cdot \left(2 - 3\right) = -3.$

Thus, $cov(x,y) = \frac{\sum_{i=1}^{n} \left(x_{i} - \mu_{x}\right)\cdot \left(y_{i} - \mu_{y}\right)}{n - 1} = \frac{-3}{4} = - \frac{3}{4}$.

The sample covariance is $cov(x,y) = - \frac{3}{4} = -0.75$A.