Discuss number of sampling at each lag

docs/movie/sampling.md

@@ -20,3 +20,61 @@ Scientifically, we are interested in the statistical properties of an M87* movie
 In other word, we are interested in measuring the structure function of M87*.
 For simplicity, we assume EHT is capable of obtaining an uniform resolution movie.
 We focus on estimating the time structure function for each of the pixel in such movie.
+
+## Definitions
+
+Suppose we work with a discrete time series $f_i = f(t_i)$.
+Let $\tau_k = t_j - t_i > 0$ be a lag, the (2nd-order) discrete structure function is defined by
+$$
+SF(\tau_k) = \sum_{i,j}\frac{[f_i - f_j]^2}{n_k},
+$$
+where $n_k$ is the number of different $i,j$ pairs.
+Note that this definition does not assume uniform time sampling $t_i = i\Delta t$.
+
+Nevertheless, if the function is periodic, uniform time sampling provide the same number of sampling at each lag.
+
++++
+
+## Non-Periodic Function
+
+However, this is not true for non-periodic functions.
+In fact, the number of sampling at each lag decreases as the lag increase.
+To see this, we can simply plot the histogram of the lags.
+
+```{code-cell} ipython3
+import numpy as np
+
+def count(t):
+    tau   = np.arange(len(t)+1)
+    n_tau = np.zeros(len(t)+1)
+
+    for ti in t:
+        for tj in t:
+            if tj > ti:
+                tauk = tj - ti
+                n_tau[round(tauk)] += 1
+    return tau, n_tau
+```
+
+```{code-cell} ipython3
+from matplotlib import pyplot as plt
+
+N = 64
+t = np.arange(N)
+
+tau, n_tau = count(t)
+plt.step(tau, n_tau, where='mid')
+```
+
+Similar trend appears for non-uniform sampling.
+
+```{code-cell} ipython3
+t = np.sort(np.random.uniform(size=N)*N)
+
+tau, n_tau = count(t)
+plt.step(tau, n_tau, where='mid')
+```
+
+```{code-cell} ipython3
+
+```