> For any $f(t)$, denote the moving average over the past year by:

$$\langle f(t) \rangle =\frac{1}{365} \sum_{d=0}^{364} f(t-d)$$

Despite this definition it seems though that they don't define

$$\langle \langle T_i(t) \rangle \rangle := \frac{1}{365} \sum_{d=0}^{364} \langle T_i(t-d) \rangle = \frac{1}{365^2} \sum_{d=0}^{364} \sum_{D=0}^{364} T_i(t-d-D) $$

but probably (?)

$$\langle \langle T_i(t) \rangle \rangle := \langle T_i(t) \rangle$$

That is I don't have the time to check this but on a first glance it looks as if one would use the first definition then the cross covariance would be different from the usual (?) covariance, that is it seems on a first glance that for this case:

$$ C_{i,j}^{t}(\tau) \stackrel{?}{\neq}\langle (T_i(t - \tau) - \langle T_i(t - \tau) \rangle )(T_j(t) - \langle T_j(t) \rangle )\rangle$$

This first glance might be wrong but if this is a different definition on purpose, this might be mentioned in the text.

$$\langle f(t) \rangle =\frac{1}{365} \sum_{d=0}^{364} f(t-d)$$

Despite this definition it seems though that they don't define

$$\langle \langle T_i(t) \rangle \rangle := \frac{1}{365} \sum_{d=0}^{364} \langle T_i(t-d) \rangle = \frac{1}{365^2} \sum_{d=0}^{364} \sum_{D=0}^{364} T_i(t-d-D) $$

but probably (?)

$$\langle \langle T_i(t) \rangle \rangle := \langle T_i(t) \rangle$$

That is I don't have the time to check this but on a first glance it looks as if one would use the first definition then the cross covariance would be different from the usual (?) covariance, that is it seems on a first glance that for this case:

$$ C_{i,j}^{t}(\tau) \stackrel{?}{\neq}\langle (T_i(t - \tau) - \langle T_i(t - \tau) \rangle )(T_j(t) - \langle T_j(t) \rangle )\rangle$$

This first glance might be wrong but if this is a different definition on purpose, this might be mentioned in the text.