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3 Proof

The key function is the spectral window:
\begin{displaymath}
G_{N}(\nu) = \frac{\mid \sum_{k=1}^{N} \exp(i 2\pi \nu t_{k})\mid^{2}}{N^{2}}\cdot\end{displaymath} (2)
We have $ G_{N}(\nu) \leq 1$ and $G_{N}(\nu)=1 \: \mbox{for} \:
\nu = m/p, m \in \mathbb{N}.$Replacing $t_{\rm k}$ with $n_{\rm k} p + t_1$, we notice that this function is periodic with period 1/p (each individual term in the sum has this same periodicity). Moreover as g(t) is real, the function $G_{N}(\nu)$ is symmetric around the origin. These properties of periodicity and symmetry imply that $G_{N}(\nu)$ is also symmetric around (2k+1)/2p. For k=0, we have 1/2p, which is just the Nyquist frequency. For the power spectrum, the proof is exactly the same. We can think of the set ti as a regular sampling with step p, where some (most) data have been dropped. The more random the data is, the smaller p will be and so the higher $\nu_{\mbox{\tiny Ny}}$.As in the regular case, the continuous signal f(t) should not contain any component above $\nu_{\mbox{\tiny Ny}}$. Otherwise, it will be mirrored into $2\nu_{\mbox{\tiny Ny}}-\nu$, and, without further assumptions, it will not be distinguishable from the mirror image. Of course, no frequency component filtered during the measurement process can be recovered by the random sampling. In particular, if $\Delta t_{\mathrm{exp}}$ is the exposure time of every measures, then in the frequency domain, the signal is filtered by $\sin^{2}(\pi \Delta
t_{\mathrm{exp}} \nu)/(\pi \Delta t_{\mathrm{exp}} \nu)^{2}$ and not much information will be recovered above $\nu_{\mathrm{max}}=1/(\kappa \Delta t_{\mathrm{exp}}), \; \kappa = \mbox{2 or 3}$.


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