The Central Limit and Pairing Theorems

We now come to the statistics of $E(t)$. For example, we already know that $\langle E^2(t)\rangle = \sum r_n^2/2$. How about $\langle E^3(t)\rangle$? Quite easy to check that it is zero because

$$\langle r_lr_mr_n\cos(\omega_mt+\varphi_m)\cos(\omega_nt + \varphi_n) \cos(\omega_l t+ \varphi_l)\rangle=0$$

when we let the $\varphi$'s each vary independently over the full circle $0$ to $2\pi$. This is true whether $l, m, n$ are distinct or not. But coming to even powers like $\langle E^4(t)\rangle$, something interesting happens. When we integrate a product like $r_lr_mr_nr_p\cos(\omega_mt+\varphi_m)\cos(\omega_nt+\varphi_n)\cos(\omega_\ell t+\varphi_l)\cos(\omega_pt+\varphi_p)$ over all the four $\varphi$'s we can get non-zero answers, provided the $\varphi$'s occur in pairs, i.e., if $l=m$ and $n=p$, then we encounter $\cos^2\varphi_l \times \cos^2\varphi_n$ which has a non-zero average. (We saw a particular case of this when we calculated $\langle E(t) E(t+\tau)\rangle$ and only $r^2_m$ type terms survived).

Because of the random and independent phases of the large number of different frequencies, we can now state the ``pairing theorem''.

$$\langle E(t_1)E(T_2) \ldots E(t_{2k})\rangle = \sum_{\rm pair... ...E(t_1)E(t_2)\rangle \ldots \langle E(t_{2k-1)}E(t_{2k}) \rangle$$

As discussed in Section 1.8, this pairing theorem proves that the statistics is gaussian. (A careful treatment shows that only the $r^2_m r^2_n$ terms are equal on the two sides- we have not quite got the $r_m^4$ terms right, but there are many more (of the order of $N$ times more) of the former type and they dominate as $T\rightarrow \infty$ and the numbers of sines and cosines we are adding is very large). This result -- that the sum of a large number of small, finite variance, independent terms has a gaussian distribution -- is a particular case of the ``central limit theorem''. We only need the particular case where these terms are cosines with random phases.