A question about covariance I know this question might get marked as one that can easily be answered via textbook but it is something that I checked in a few of them and still can't get my head around. Basically what I'm trying to understand is how can $$Cov(X,Y) = E[(X - m_X)(Y - m_Y)] = E[(X - m_X)Y].$$
How I'm seeing the problem is that $(Y - m_Y)$ can give a negative result if the random observation that I just took is smaller than the mean of the $Y$ variable, however $Y$ could be positive. So let's say $(X - m_X)$ is a positive number, if we multiply it by the negative result that $(Y - m_Y)$ gives we would get a negative covariance between the two variables. However, as I said, if $Y$ is a smaller than the mean of $Y$ positive number and we multiply that positive number by the supposed positive result that $(X - m_X)$ gives us we would get a positive covariance. 
I hope your answers would be conceptual so I can get the general idea behind those equations. 
 A: This always rubbed me the wrong way. It helps to draw some rectangles, though. Suppose you start with two identical point clouds, one centered on x and y and the other only centered on y.

When drawing from the empirical distribution of these data, the contributions to $E[(X-\mu_x) (Y - \mu_Y)]$ can be shown as the saturated rectangles, and the contributions to $E[X(Y - \mu_Y)]$ appears as the less saturated rectangles (the blue one is partially hidden by its vivid counterpart).

The contributions are positive if in quadrants 1 or 3, negative if in 2 or 4 (numbering CCW from top right). By failing to center the x coordinate, we have increased the blue contribution but decreased the red one. 
The exact amount of the increase is $|y_1|*\mu_x$ and the exact amount of the decrease is $|y_2|*\mu_x$, where $y_1$ ($y_2$) is the height of the top right (bottom left) point. These are not the same (!) but if you average over all the points, you get $\mu_x$ times the mean of y -- which is zero.
This is what @whuber is trying to get you to prove algebraically -- sorry for spoiling the punch line, but what's a Q&A site without the A?
