If we have a equally spaced sample, we can use the conventionally used periodogram:

$$P_x(\omega) = \frac{1}{N_0}\left[\left(\sum_j X_j\cos \omega t_j\right)^2+\left(\sum_j X_j\sin \omega t_j\right)^2\right]$$

But in a unevenly spaced sample, it's also well known the Lomb-Scargle periodogram:

$$P_x(\omega) = \frac{1}{2}\left(\frac{(\sum_j X_j\cos\omega(t_j-\tau))^2}{\sum_j \cos^2\omega(t_j-\tau)}+\frac{(\sum_j X_j\sin\omega(t_j-\tau))^2}{\sum_j \sin^2\omega(t_j-\tau)}\right),$$

with $\tan(2\omega\tau)=(\sum_j\sin 2\omega t_j)/(\sum_j\cos 2\omega t_j).$

I want to see that, if we apply the L-S periodogram to an equally spaced sample, it becomes the classical periodogram, e.g, the denominators are equal to $N_0/2$ and it has time-translation invariance, e.g., $\tau$ doesn't affect anything (Scargle, 1982):

$$\sum_{j=1}^{N_0} \sin^2\omega(t_j-\tau)= N_0/2 = \sum_{j=1}^{N_0} \cos^2\omega(t_j-\tau)$$

Some books I've seen says it's purely trigonometrical reasons, but I don't see it. Also, it's empirically shown that, even for an unevenly spaced sample, this denominators are never far from $N_0/2:$

$$\sum_{j=1}^{N_0} \sin^2\omega(t_j-\tau)\approx N_0/2 \approx \sum_{j=1}^{N_0} \cos^2\omega(t_j-\tau),$$

is there some analytical explanation of why this happens?


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