Calculus7

Critical points of a cubic spline

The choice of piecewise polynomials of degree 3 for interpolation is justifiably popular: even-degree splines are algebraically awkward to construct, degree 1 is simply piecewise linear interpolation (not smooth), and degree 5, while feasible, entails juggling too many coefficients. Besides, a cubic polynomial minimizes the amount of wiggling (the integral of second derivative squared) for given values and slopes at the endpoints of an interval. (Recall Connecting dots naturally.)

But the derivative of a cubic spline is a quadratic spline. And one needs the derivative to find the critical points. This results in an awkward example in SciPy documentation, annotated with “(NB: sproot only works for order 3 splines, so we fit an order 4 spline)”.

Although not implemented in SciPy, the task of computing the roots of a quadratic spline is a simple one. Obtaining the roots from the internal representation of a quadratic spline in SciPy (as a linear combination of B-splines) would take some work and reading. But a quadratic polynomial is determined by three values, so sampling it at three points, such as two consecutive knots and their average, is enough.

Quadratic formula with values instead of coefficients

Suppose we know the values of a quadratic polynomial q at -1, 0, 1, and wish to find if it has roots between -1 and 1. Let’s normalize so that q(0)=1, and let x = q(-1), y = q(1). If either x or y is negative, there is definitely a root on the interval. If they are positive, there is still a chance: we need the parabola to be concave up, have a minimum within [-1, 1], and for the minimum to be negative. All of this is easily determined once we note that the coefficients of the polynomial are a = (x+y)/2 – 1, b = (y-x)/2, and c = 1.

The inequality ${(x-y)^2 \ge 8(x+y-2)}$ ensures the suitable sign of the discriminant. It describes a parabola with vertex (1, 1) and focus (2, 2), contained in the first quadrant and tangent to the axes at (4, 0) and (0, 4). Within the orange region there are no real roots.

The line x+y=2, tangent to the parabola at its vertex, separates convex and concave parabolas. While concavity in conjunction with x, y being positive definitely precludes having roots in [-1, 1], slight convexity is not much better: it results in real roots outside of the interval. Here is the complete picture: green means there is a root in [-1, 1], orange means no real roots, red covers the rest.

all_colors — Green = there is a root in the interval [-1, 1]

Back to splines

Since the derivative of a spline is implemented in SciPy (B-splines have a nice formula for derivatives), all we need is a root-finding routine for quadratic splines. Here it is, based on the above observations but using built-in NumPy polynomial solver np.roots to avoid dealing with various special cases for the coefficients.

def quadratic_spline_roots(spl):
    roots = []
    knots = spl.get_knots()
    for a, b in zip(knots[:-1], knots[1:]):
        u, v, w = spl(a), spl((a+b)/2), spl(b)
        t = np.roots([u+w-2*v, w-u, 2*v])
        t = t[np.isreal(t) & (np.abs(t) <= 1)]
        roots.extend(t*(b-a)/2 + (b+a)/2)
    return np.array(roots)

A demonstration, which plots the spline (blue), its critical points (red), and original data points (black) as follows:

import numpy as np
import matplotlib.pyplot as plt
from scipy.interpolate import InterpolatedUnivariateSpline

x = np.arange(7)
y = np.array([3, 1, 1, 2, 2, 4, 3])
f = InterpolatedUnivariateSpline(x, y, k=3)
crit_pts = quadratic_spline_roots(f.derivative())

t = np.linspace(x[0], x[-1], 500)
plt.plot(t, f(t))
plt.plot(x, y, 'kd')
plt.plot(crit_pts, f(crit_pts), 'ro')
plt.show()

Quasi-projections and isometries

A typical scenario: given a subset ${E}$ of a metric space ${X}$ and a point ${x\in X}$ , we look for a point ${y\in E}$ that is nearest to ${x}$ : that is, ${d(x, y) = \mathrm{dist}\,(x, E)}$ . Such a point is generally not unique: for example, if ${E}$ is the graph of cosine function and ${x = (\pi, \pi/2)}$ , then both ${(\pi/2, 0)}$ and ${(3\pi/2, 0)}$ qualify as nearest to ${x}$ . This makes the nearest-point projection onto ${E}$ discontinuous: moving ${x}$ slightly to the left or to the right will make its projection onto ${E}$ jump from one point to another. Not good.

proj1 — Discontinuous nearest-point projection

Even when the nearest point projection is well-defined and continuous, it may not be the kind of projection we want. For example, in a finite-dimensional normed space with strictly convex norm we have a continuous nearest-point projection onto any linear subspace, but it is in general a nonlinear map.

Let’s say that ${P\colon X\to E}$ is a quasi-projection if ${d(x, P(x)) \le C \mathrm{dist}\,(x, E)}$ for some constant ${C}$ independent of ${x}$ . Such maps are much easier to construct: indeed, every Lipschitz continuous map ${P\colon X\to E}$ such that ${P(x)=x}$ for ${x \in E}$ is a quasi-projection. For example, one quasi-projection onto the graph of cosine is the map ${(x, y)\mapsto (x, \cos x)}$ shown below.

If ${X}$ is a Banach space and ${E}$ is its subspace, then any idempotent operator with range ${E}$ is a quasi-projection onto ${E}$ . Not every subspace admits such an operator but many do (these are complemented subspaces; they include all subspaces of finite dimension or finite codimension). By replacing “nearest” with “close enough” we gain linearity. And even some subspaces that are not linearly complemented admit a continuous quasi-projection.

Here is a neat fact: if ${M}$ and ${N}$ are subspaces of a Euclidean space and ${\dim M = \dim N}$ , then there exists an isometric quasi-projection of ${M}$ onto ${N}$ with constant ${C=\sqrt{2}}$ . This constant is best possible: for example, an isometry from the ${y}$ -axis onto the ${x}$ -axis has to send ${(0, 1)}$ to one of ${(\pm 1, 0)}$ , thus moving it by distance ${\sqrt{2}}$ .

proj3 — An isometry must incur sqrt(2) distance cost

Proof. Let ${k}$ be the common dimension of ${M}$ and ${N}$ . Fix some orthonormal bases in ${M}$ and ${N}$ . In these bases, the orthogonal (nearest-point) projection from ${M}$ to ${N}$ is represented by some ${k\times k}$ matrix ${P}$ of norm at most ${1}$ . We need an orthogonal ${k\times k}$ matrix ${Q}$ such that the map ${M\to N}$ that it defines is a ${\sqrt{2}}$ -quasi-projection. What exactly does this condition mean for ${Q}$ ?

Let’s say ${x\in M}$ , ${y\in N}$ is the orthogonal projection of ${x}$ onto ${N}$ , and ${z\in N}$ is where we want to send ${x}$ by an isometry. Our goal is ${\|x-z\|\le \sqrt{2}\|x-y\|}$ , in addition to ${\|z\|=\|x\|}$ . Squaring and expanding inner products yields ${2\langle x, y\rangle - \langle x, z \rangle \le \|y\|^2}$ . Since both ${y}$ and ${z}$ are in ${N}$ , we can replace ${x}$ on the left by its projection ${y}$ . So, the goal simplifies to ${\|y\|^2 \le \langle y, z\rangle}$ . Geometrically, this means placing ${z}$ so that its projection onto the line through ${y}$ lies on the continuation of this line beyond ${y}$ .

So far so good, but the disappearance of ${x}$ from the inequality is disturbing. Let’s bring it back by observing that ${\|y\|^2 \le \langle y, z\rangle}$ is equivalent to ${4(\|y\|^2 - \langle y, z\rangle) + \|z\|^2 \le \|x\|^2}$ , which is simply ${\|2y-z\| \le \|x\|}$ . So that’s what we want to do: map ${x}$ so that the distance from its image to ${2y}$ does not exceed ${\|x\|}$ . In terms of matrices and their operator norm, this means ${\|2P-Q\|\le 1}$ .

It remains to show that every square matrix of norm at most ${2}$ (such as ${2P}$ here) is within distance ${1}$ of some orthogonal matrix. Let ${2P = U\Sigma V^T}$ be the singular value decomposition, with ${U, V}$ orthogonal and ${\Sigma}$ a diagonal matrix with the singular values of ${2P}$ on the diagonal. Since the singular values of ${2P}$ are between ${0}$ and ${2}$ , it follows that ${\|\Sigma-I\|\le 1}$ . Hence ${\|2P - UV^T\|\le 1}$ , and taking ${Q=UV^T}$ concludes the proof.

(The proof is based on a Stack Exchange post by user hypernova.)

The sum of pairwise distances and the square of CDF

Suppose we have ${n}$ real numbers ${x_0,\dots, x_{n-1}}$ and want to find the sum of all distances ${|x_j-x_k|}$ over ${j < k}$ . Why? Maybe because over five years ago, the gradient flow of this quantity was used for "clustering by collision" (part 1, part 2, part 3).

If I have a Python console open, the problem appears to be solved with one line:

>>> 0.5 * np.abs(np.subtract.outer(x, x)).sum()

where the outer difference of x with x creates a matrix of all differences ${x_i-x_j}$ , then absolute values are taken, and then they are all added up. Double-counted, hence the factor of 0.5.

But trying this with, say, one million numbers is not likely to work. If each number takes 8 bytes of memory (64 bits, double precision), then the array x is still pretty small (under 8 MB) but a million-by-million matrix will require over 7 terabytes, and I won’t have that kind of RAM anytime soon.

In principle, one could run a loop adding these values, or store the matrix on a hard drive. Both are going to take forever.

There is a much better way, though. First, sort the numbers in nondecreasing order; this does not require much time or memory (compared to quadratic memory cost of forming a matrix). Then consider the partial sums ${s_k = x_0+\dots+x_k}$ ; the cost of computing them is linear in time and memory. For each fixed ${k}$ , the sum of distances to ${x_j}$ with ${j<k}$ is simply ${kx_k - s_{k-1}}$ , or, equivalently, ${(k+1)x_k - s_k}$ . So, all we have to do is add these up. Still one line of code (after sorting), but a much faster one:

>>> x.sort()
>>> (np.arange(1, n+1)*x - np.cumsum(x)).sum()

For example, x could be a sample from some continuous distribution. Assuming the distribution has a mean (i.e., is not too heavy tailed), the sum of all pairwise distances grows quadratically with n, and its average approaches a finite limit. For the uniform distribution on [0, 1] the computation shows this limit is 1/3. For the standard normal distribution it is 1.128… which is not as recognizable a number.

As ${n\to \infty}$ , the average distance of a sample taken from a distribution converges to the expected value of |X-Y| where X, Y are two independent variables with that distribution. Let’s express this in terms of the probability density function ${p}$ and the cumulative distribution function ${\Phi}$ . By symmetry, we can integrate over ${x> y}$ and double the result:

$\frac12 E|X-Y| = \int_{-\infty}^\infty p(x)\,dx \int_{-\infty}^x (x-y) p(y)\,dy$

Integrate by parts in the second integral: ${p(y) = \Phi'(y)}$ , and the boundary terms are zero.

$\frac12 E|X-Y| = \int_{-\infty}^\infty p(x)\,dx \int_{-\infty}^x \Phi(y)\,dy$

Integrate by parts in the other integral, throwing the derivative onto the indefinite integral and thus eliminating it. There is a boundary term this time.

$\frac12 E|X-Y| = \Phi(\infty) \int_{-\infty}^\infty \Phi(y)\,dy - \int_{-\infty}^\infty \Phi(x)^2\,dx$

Since ${\Phi(\infty) = 1}$ , this simplifies nicely:

$\frac12 E|X-Y| = \int_{-\infty}^\infty \Phi(x) (1-\Phi(x))\,dx$

This is a lot neater than I expected: ${E|X-Y|}$ is simply the integral of ${2\Phi(1-\Phi)}$ . I don’t often see CDF squared, like here. Some examples: for the uniform distribution on [0,1] we get

$E|X-Y| = \int_0^1 2x(1-x)\,dx = \frac13$

and for the standard normal, with ${\Phi(x) = (1+\mathrm{erf}\,(x/\sqrt{2}))/2}$ , it is

$\int_{-\infty}^\infty \frac12 \left(1-\mathrm{erf}\,(x/\sqrt{2}) ^2 \right)\,dx = \frac{2}{\sqrt{\pi}}\approx 1.12838\ldots$

The trick with sorting and cumulative sums can also be used to find, for every point ${x_k}$ , the sum (or average) of distances to all other points. To do this, we don’t sum over ${k}$ but must also add ${|x_j-x_k|}$ for ${j>k}$ . The latter sum is simply ${S - s_k - (n-k-1)x_k}$ where ${S}$ is the total sum. So, all we need is

>>> (2*np.arange(1,n+1)-n)*x - 2*np.cumsum(x) + x.sum()

Unfortunately, the analogous problems for vector-valued sequences are not as easy. If the Manhattan metric is used, we can do the computations for each coordinate separately, and add the results. For the Euclidean metric…

Moderatorship connections between Stack Exchange sites

Consider the graph in which the vertices are Stack Exchange sites, connected if two sites share a moderator. What does this graph look like? As being a moderator correlates with some level of expertise or at least interest in the topic, we can expect the structure of the graph to highlight topical connections between the sites.

Unsurprisingly, there is a dominant component (55 sites) centered at Stack Overflow. (Click the image for the larger version.)

Its radius is 6: every site can be joined to Stack Overflow by a path of length at most 6. The five sites at distance 6 from SO are Lifehacks, Martial Arts, Physical Fitness, Space Exploration, and Travel.

By the triangle inequality, the diameter of the SO component is at most 12. In fact it is exactly 12: it takes 12 steps to go from Space to either Fitness, Lifehacks, or Martial Arts.

The center of the SO component is not unique: Web Applications is another site from which every other one can be reached in 6 steps. And it has an advantage over SO in that only four sites in the component are at distance 6 from Web Apps. Naturally, these are the four sites that realize the diameter 12 of the component.

That said, Stack Overflow is the vertex of highest degree (8), meaning that Stack Overflow moderators also take part in moderating eight other sites.

What about other components? Excluding isolated vertices, we have the following picture of small components.

The largest of these is the “language component”: English Language & Usage, English Language Learners, Japanese Language, Portuguese Language, and less logically, Sustainable Living. Other notable components are Unix (with bioinformatics thrown in) and Mathematics.

Source of the information: on the page listing moderators grouped by users I executed a JavaScript one-liner

JSON.stringify(Array.from(document.querySelectorAll('.mods-summary-list')).filter(e=>e.children.length>1).map(e=>Array.from(e.children).map(x=>x.hostname.split('.')[0])))

which gave a list of connections. The rest was done in Python:

import networkx as nx
from itertools import chain, combinations
connections = # copied JS output
edges = list(chain.from_iterable(combinations(c, 2) for c in connections))
G = nx.Graph()
G.add_edges_from(edges)
components = sorted(list(nx.connected_components(G)), key=len)
main_component = G.subgraph(components[-1])
pos = nx.kamada_kawai_layout(main_component)
nx.draw_networkx(main_component, pos=pos, node_size=100)

I used different layout algorithms for the dominant component and for the rest. The default “spring” layout makes the SO component too crowded, but is okay for small components:

other_components = G.subgraph(chain.from_iterable(components[:-1]))
pos = nx.spring_layout(other_components, k=0.25) 
nx.draw_networkx(other_components, pos=pos, node_size=100)

The rest was done by various functions of the NetworkX library such as

nx.degree(main_component)
nx.center(main_component)
nx.diameter(main_component)
nx.periphery(main_component)
nx.radius(main_component)
nx.shortest_path_length(main_component, source="stackoverflow")
nx.shortest_path_length(main_component, source="webapps")

The infinitely big picture: tanh-tanh scale

When plotting the familiar elementary functions like x² or exp(x), we only see whatever part of the infinitely long curve fits in the plot window. What if we could see the entire curve at once?

The double-tanh scale can help with that. The function u = tanh(x) is a diffeomorphism of the real line onto the interval (-1, 1). Its inverse, arctanh or artanh or arth or ${\tanh^{-1}x}$ or ${\frac12 \log((1+x)/(1-x))}$ , whatever you prefer to call it, does the opposite. So, conjugating any function ${f\colon \mathbb R\to \mathbb R}$ by the hyperbolic tangent produces a function ${g\colon (-1, 1)\to (-1,1)}$ which we can plot in its entirety. Let’s try this.

Out of linear functions y = kx, only y=x and y=-x remain lines.

The powers of x, from 1 to 4, look mostly familiar:

Sine, cosine, and tangent functions are not periodic anymore:

The exponential function looks concave instead of convex, although I don’t recommend trying to prove this by taking the second derivative of its tanh-conjugate.

The Gaussian loses its bell-shaped appearance and becomes suspiciously similar to a semicircle.

This raises the question: which function does appear as a perfect semi-circle of radius 1 on the tanh-tanh scale? Turns out, it is ${f(x) = \log|\coth(x/2)|}$ . Here it is shown in the normal coordinate system.

Discrete Cosine Transforms: Trapezoidal vs Midpoint

The existence of multiple versions of Discrete Cosine Transform (DCT) can be confusing. Wikipedia explains that its 8 types are determined by how one reflects across the boundaries. E.g., one can reflect 1, 2, 3 across the left boundary as 3, 2, 1, 2, 3 or as 3, 2, 1, 1, 2, 3, and there are such choices for the other boundary too (also, the other reflection can be odd or even). Makes sense enough.

But there is another aspect to the two most used forms, DCT-I and DCT-II (types I and II): they can be expressed in terms of the Trapezoidal and Midpoint rules for integration. Here is how.

The cosines ${\cos kx}$ , ${k=0,1,\dots}$ are orthogonal on the interval ${[0,\pi]}$ with respect to the Lebesgue measure. The basis of discrete transforms is that these cosines are also orthogonal with respect to some discrete measures, until certain frequency is reached. Indeed, if cosines are orthogonal with respect to measure ${\mu}$ whose support consists of ${n}$ points, then we can efficiently use them to represent any function defined on the support of ${\mu}$ , and those are naturally identified with sequences of length n.

How to find such measures? It helps that some simple rules of numerical integration are exact for trigonometric polynomials up to some degree.

For example, the trapezoidal rule with n sample points exactly integrates the functions ${\cos kx}$ for ${k=0,\dots, 2n-3}$ . This could be checked by converting to exponential form and summing geometric progressions, but here is a visual explanation with n=4, where the interval ${[0,\pi]}$ is represented as upper semi-circle. The radius of each red circle indicates the weight placed at that point; the endpoints get 1/2 of the weight of other sample points. To integrate ${\cos x}$ correctly, we must have the x-coordinate of the center of mass equal to zero, which is obviously the case.

Replacing ${\cos x}$ by ${\cos kx}$ means multiplying the polar angle of each sample point by ${k}$ . This is what we get:

Argument of cosine multiplied by k=2 or k=4

In all cases the x-coordinate of the center of mass is zero. With k=6 this breaks down, as all the weight gets placed in one point. And this is how it goes in general with integration of sines and cosines: equally spaced points work perfectly until they don’t work at all, which happens when the step size is equal to the period of the function. When ${k=2n-2}$ , the period ${2\pi/k}$ is equal to ${\pi/(n-1)}$ , the spacing of points in the trapezoidal rule.

The orthogonality of cosines has to do with the formula ${\cos kx \cos j x = \frac12(\cos (k-j)x) + \frac12(\cos (k+j)x)}$ . Let ${\tau_n}$ be the measure expressing the trapezoidal rule on ${[0,\pi]}$ with ${n}$ sample points; so it’s the sum of point masses at ${0, \pi/(n-1), \dots , \pi}$ . Then ${\{\cos k x \colon k=0,\dots, n-1\}}$ are orthogonal with respect to ${\tau_n}$ because any product ${\cos k x\cos j x }$ with ${k >j}$ taken from this range will have ${k-j, k+j \le 2n-3}$ . Consequently, we can compute the coefficients of any function ${f}$ in the cosine basis as

$c_k = \int f(x)\cos kx\,d\tau_n(x) \bigg/ \int \cos^2 kx\,d\tau_n(x)$

The above is what DCT-I (discrete cosine transform of type 1) does, up to normalization.

The DCT-II transform uses the Midpoint rule instead of the Trapezoidal rule. Let ${\mu_n}$ be the measure expressing the Midpoint rule on ${[0,\pi]}$ with ${n}$ sample points; it gives equal mass to the points ${(2j-1)\pi/(2n)}$ for ${j=1,\dots, n}$ . These are spaced at ${\pi/n}$ and therefore the midpoint rule is exact for ${\cos kx}$ with ${k=0,\dots, 2n-1}$ which is better than what the trapezoidal rule does. Perhaps more significantly, by identifying the given data points with function values at the midpoints of subintervals we stay away from the endpoints ${0,\pi}$ where the cosines are somewhat restricted by having to have zero slope.

Let’s compare DCT-I and DCT-II on the same data set, ${y=(0, \sqrt{1}, \sqrt{2}, \dots, \sqrt{8})}$ . There are 9 numbers here. Following DCT-I we place them at the sample points of the trapezoidal rule, and expand into cosines using the inner product with respect to ${\tau_n}$ . Here is the plot of the resulting trigonometric polynomial: of course it interpolates the data.

But DCT-II does it better, despite having exactly the same cosine functions. The only change is that we use ${\mu_n}$ and so place the ${y}$ -values along its support.

Less oscillation means the high-degree coefficients are smaller, and therefore easier to discard in order to compress information. For example, drop the last two coefficients in each expansion, keeping 6 numbers instead of 8. DCT-II clearly wins in accuracy then.

Okay, so the Midpoint rule is better, no surprise. After all, it’s in general about twice as accurate as the Trapezoidal rule. What about Simpson’s rule, would it lead to some super-efficient form of DCT? That is, why don’t we let ${\sigma_n}$ be the discrete measure that expresses Simpson’s rule and use the inner product ${\int fg\,d\sigma_n}$ for cosine expansion? Alas, Simpson’s rule on ${n}$ points is exact only for ${\cos kx}$ with ${k=0,\dots, n-2}$ , which is substantially worse than either Trapezoidal or Midpoint rules. As a result, we don’t get enough orthogonal cosines with respect to ${\sigma_n}$ to have an orthogonal basis. Simpson’s rule has an advantage when dealing with algebraic polynomials, not with trigonometric ones.

Finally, the Python code used for the graphics; I did not use SciPy’s DCT method (which is of course more efficient) to keep the relation to numerical integration explicit in the code. The method trapz implements the trapezoidal rule, and the midpoint rule is just the summation of sampled values. In both cases there is no need to worry about factor dx, since it cancels out when we divide one numerical integral by the other.

import numpy as np
import matplotlib.pyplot as plt
#   Setup 
y = np.sqrt(np.arange(9))
c = np.zeros_like(y)
n = y.size
#   DCT-I, trapezoidal
x = np.arange(n)*np.pi/(n-1)
for k in range(n):
  c[k] = np.trapz(y*np.cos(k*x))/np.trapz(np.cos(k*x)**2)
t = np.linspace(0, np.pi, 500)
yy = np.sum(c*np.cos(np.arange(9)*t.reshape(-1, 1)), axis=1)
plt.plot(x, y, 'ro')
plt.plot(t, yy)
plt.show()
#   DCT-II, midpoint
x = np.arange(n)*np.pi/n + np.pi/(2*n)
for k in range(n):
  c[k] = np.sum(y*np.cos(k*x))/np.sum(np.cos(k*x)**2)
t = np.linspace(0, np.pi, 500)
yy = np.sum(c*np.cos(np.arange(9)*t.reshape(-1, 1)), axis=1)
plt.plot(x, y, 'ro')
plt.plot(t, yy)
plt.show()

Experiments with the significance of autocorrelation

Given a sequence of numbers ${x_j}$ of length ${L}$ one may want to look for evidence of its periodic behavior. One way to do this is by computing autocorrelation, the correlation of the sequence with a shift of itself. Here is one reasonable way to do so: for lag values ${\ell=1,\dots, \lfloor L/2 \rfloor}$ compute the correlation coefficient of ${(x_1,\dots, x_{L-\ell}}$ with ${(x_{\ell+1},\dots, x_L)}$ . That the lag does not exceed ${L/2}$ ensures the entire sequence participates in the computation, so we are not making a conclusion about its periodicity after comparing a handful of terms at the beginning and the end. In other words, we are not going to detect periodicity if the period is more than half of the observed time period.

Having obtained the correlation coefficients, pick one with the largest absolute value; call it R. How large does R have to be in order for us to conclude the correlation is not a fluke? The answer depends on the distribution of our data, but an experiment can be used to get some idea of likelihood of large R.

I picked ${x_j}$ independently from the standard normal distribution, and computed ${r}$ as above. After 5 million trials with a sequence of length 100, the distribution of R was as follows:

ac100-5M — Extremal correlation coefficient in a sequence of length 100

Based on this experiment, the probability of obtaining |R| greater than 0.5 is less than 0.0016. So, 0.5 is pretty solid evidence. The probability of ${|R| > 0.6}$ is two orders of magnitude less, etc. Also, |R| is unlikely to be very close to zero unless the data is structured in some strange way. Some kind of correlation ought to be present in the white noise.

Aside: it’s not easy to construct perfectly non-autocorrelated sequences for the above test. For length 5 an example is 1,2,3,2,3. Indeed, (1,2,3,2) is uncorrelated with (2,3,2,3) and (1,2,3) is uncorrelated with (3,2,3). For length 6 and more I can’t construct these without filling them with a bunch of zeros.

Repeating the experiment with sequences of length 1000 shows a tighter distribution of R: now |R| is unlikely to be above 0.2. So, if a universal threshold is to be used here, we need to adjust R based on sequence length.

ac1000-5M — Extremal correlation coefficient in a sequence of length 1000

I did not look hard for statistical studies of this subject, resorting to an experiment. Experimentally obtained p-values are pretty consistent for the criterion ${L^{0.45}|R| > 4}$ . The number of trials was not very large (10000) so there is some fluctuation, but the pattern is clear.

Length, L	P(L^0.45\|R\| > 4)
100	0.002
300	0.0028
500	0.0022
700	0.0028
900	0.0034
1100	0.0036
1300	0.0039
1500	0.003
1700	0.003
1900	0.0042
2100	0.003
2300	0.0036
2500	0.0042
2700	0.0032
2900	0.0043
3100	0.0042
3300	0.0025
3500	0.0031
3700	0.0027
3900	0.0042

Naturally, all this depends on the assumption of independent normal variables.

And this is the approach I took to computing r in Python:

import numpy as np
n = 1000
x = np.random.normal(size=(n,))
acorr = np.correlate(x, x, mode='same')
acorr = acorr[n//2+1:]/(x.var()*np.arange(n-1, n//2, -1))
r = acorr[np.abs(acorr).argmax()]