I wrote:

>I'm not sure what Nad is wondering about, but here's a guess. She might be puzzled that we're calling
$$\langle A B \rangle - \langle A \rangle \langle B \rangle$$
the covariance between $A$ and $B$. She might be more used to
$$\langle A - \langle A \rangle \rangle \; \langle B - \langle B \rangle \rangle$$
However, these are equal! (Fun exercise.)

Nad wrote:

> No I think they are not equal.

Whoops, you're right - I made a typo.

Obviously

$$\langle A - \langle A \rangle \rangle \; \langle B - \langle B \rangle \rangle = 0$$

since

$$\langle A - \langle A \rangle \rangle = \langle A \rangle - \langle A \rangle = 0$$

using the principle that $\langle \langle X \rangle \rangle = \langle X \rangle$ - you can pull a number out of a mean.

I meant to write

$$\langle (A - \langle A \rangle ) \; (B - \langle B \rangle) \rangle$$

This is what equals

$$\langle A B \rangle - \langle A \rangle \langle B \rangle$$

Showing this really is a fun exercise:

$$\langle (A - \langle A \rangle ) \; (B - \langle B \rangle) \rangle = \langle A B - \langle A \rangle B - A \langle B \rangle + \langle A \rangle \langle B \rangle \rangle$$

and $\langle \langle X \rangle Y \rangle = \langle X \rangle \langle Y \rangle$ (you can pull a number out of mean) so two terms cancel and we're left with

$$\langle A B \rangle - \langle A \rangle \langle B \rangle$$