Kramer defines the sample cross correlation (Eq. 1) as:
$$C_{ij}[\tau] = \frac{1}{\hat{\sigma_i}\hat{\sigma_j}(n-|l|)} \sum_{t=1}^{n-\tau} (x_i[t]-\bar{x}_i)(x_j[t+\tau]-\bar{x}_j)$$
Actually, Kramer's normalization of $(n-2l)$ doesn't make sense to me for two reasons: (1) there's a spurious factor of $2$ which does not capture the number of overlap terms in the case of finite signals, and (2) $l$ should have the absolute value applied. I find that my corrected formula is consistent with Eq. 2 in Kramer2009, whereas the original definition is not.
Kramer also cites the following (Bartlett's) estimate for the variance of $C_{ij}[l]$ when signals $x_i$ and $x_j$ are uncoupled:
$$var(C_{ij}[l]) = \frac{1}{n-|l|}\sum_{\tau=-n}^n C_{ii}[\tau]C_{jj}[\tau]$$.
(Again, I added the absolute value.)
The corresponding mean is clearly $E\left[C_{ij}[l]\right]=0$.
For this inaugural post, I would like to verify the Bartlett estimate. I begin with two uncorrelated white Gaussian noise $x_1[t]$ and $x_2[t]$.
Here is the simulated distribution of $C_{ij}[l]$ at a few values of $l$, and a normal distribution whose variance is given by the Bartlett estimate. Not bad at all!:
Next, Kramer asks us to consider the Fisher transformation of the $C_{ij}$, as follows:
$$C_{ij}^F[\tau] = \frac{1}{2}\log\frac{1+C_{ij}[\tau]}{1-C_{ij}[\tau]}$$.
Oh, this bit is trivial. The Fisher transform maps $[-1, 1] \to [-\infty, \infty]$, so $C_{ij}^F$ is better described by the normal distribution than $C_{ij}$. I checked the correspondence of the above experiment, when the $C_{ij}$ values underwent a Fisher transform. The agreement with the Bartlett estimated distribution is still good (the transform does little to change the values of $C_{ij}$ above).
Next, let me consider the distribution in $C_{ij}$ in the case of an actual correlation between $x_1$ and $x_2$. I will follow Kramer's example and define the two signals as follows: $x_1 = w_1$ and $x_1 = w_2 + \alpha w_1$ ($\alpha = 0.4$) where $w_i$ are independent WGN instances. As I understand it, it is not required to estimate the distribution of $C_{ij}$ under the alternate hypothesis ("H1: Coupling") for Kramer's framework, but I want to see the distribution since I have the scripts already written:
Interestingly, it does not appear to be the case that the distribution is simply the normal distribution with a mean at the coupling value $\alpha$ and the Bartlett estimate of the variance. The distribution of $C_{ij}$ (left panel in above figure) is definitely not well described by this candidate distribution; the distribution of $C_{ij}^F$ fares better (right panel) but there is still a systematic offset.
[2013 10 16]: Actually, I wonder if the mean of the Bartlett estimate normal needs to be inverse Fisher-transformed...
Note that the correlated $C_{ij}$ at $\alpha=0.4$ would all be extremely improvable (very low $p$-value) under the null distribution. So, correlation at $\alpha=0.4$ is highly detectable whereas $\alpha \approx 0.1$ would be harder to distinguish, as judging from the null hypothesis distribution.
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