In my physics class, we've been doing uncertainty problems with the General Error Propagation Equation. However, I felt uncomfortable using the equation without understanding its derivation. So I was reading this webpage (https://www.deanza.edu/faculty/marshburnthomas/pdf/ErrorPropagation.pdf) which was fine until the following portion:
$$\sigma^2 = \frac{(\partial f)} {(\partial x)}^2 \Sigma \frac{(\Delta x)^2}{n} + \frac{(\partial f)} {(\partial y)}^2 \Sigma \frac{(\Delta y)^2}{n} + \frac{(\partial f)} {(\partial z)}^2 \Sigma \frac{(\Delta z)^2}{n} + 2\frac{(\partial f)} {(\partial x)} \frac{(\partial f)} {(\partial y)} \frac{1}{n}\Sigma \Delta x\Delta y + 2\frac{(\partial f)} {(\partial x)} \frac{(\partial f)} {(\partial z)} \frac{1}{n}\Sigma \Delta x\Delta z + 2\frac{(\partial f)} {(\partial y)} \frac{(\partial f)} {(\partial z)}\frac{1}{n}\Sigma \Delta y\Delta z$$
"Since the uncertainties in the measurement x, y and z are random and independent, then $$\Sigma\Delta x\Delta y = 0$$ $$\Sigma\Delta x\Delta z = 0$$ $$\Sigma\Delta y\Delta z = 0$$ Therefore, $$\sigma^2 = \frac{(\partial f)} {(\partial x)}^2 (\sigma _x)^2 + \frac{(\partial f)} {(\partial y)}^2 (\sigma _y)^2 + \frac{(\partial f)} {(\partial z)}^2 (\sigma _z)^2$$
What I'm having trouble grasping is why the sum of the products of the uncertainties is 0. Thank you for your help.