Corners stick out more in high dimensions

Posted on by John

High-dimensional geometry is full of surprises. For example, nearly all the area of a high-dimensional sphere is near the equator, and by symmetry it doesn’t matter which equator you take.

Here’s another surprise: corners stick out more in high dimensions. Hypercubes, for example, become pointier as dimension increases.

How might we quantify this? Think of a pyramid and a flag pole. If you imagine a ball centered at the top of a pyramid, a fair proportion of the volume of the ball contains part of the pyramid. But if you do the same for a flag pole, only a small proportion of the ball contains pole; nearly all the volume of the ball is air.

So one way to quantify how pointy a corner is would be to look at a neighborhood of the corner and measure how much of the neighborhood intersects the solid that the corner is part of. The less volume, the pointier the corner.

Consider a unit square. Put a disk of radius r at a corner, with r < 1. One quarter of that disk will be inside the square. So the proportion of the square near a particular corner is πr²/4, and the proportion of the square near any corner is πr².

Now do the analogous exercise for a unit cube. Look at a ball of radius r < 1 centered at a corner. One eighth of the volume of that ball contains part of the cube. The proportion of cube’s volume located within a distance r of a particular corner is πr³/6, and the proportion located within a distance r of any corner is 4πr³/3.

The corner of a cube sticks out a little more than the corner of a square. 79% of a square is within a distance 0.5 of a corner, while the proportion is 52% for a cube. In that sense, the corners of a cube stick out a little more than the corners of a square.

Now let’s look at a hypercube of dimension n. Let V be the volume of an n-dimensional ball of radius r < 1. The proportion of the hypercube’s volume located within a distance r of a particular corner is V / 2n and the proportion located with a distance r of any corner is simply V.

The equation for the volume V is

V = \frac{\pi^{\frac{n}{2}} r^n}{\Gamma\left(\frac{n}{2} + 1\right)}

If we fix r and let n vary, this function decreases rapidly as n increases.

Saying that corners stick out more in high dimensions is a corollary of the more widely known fact that a ball in a box takes up less and less volume as the dimension of the ball and the box increase.

Let’s set r = 1/2 and plot how the volume of a ball varies with dimension n.

Volume of a ball of radius 1/2 in dimension n

You could think of this as the volume of a ball sitting inside a unit hypercube, or more relevant to the topic of this post, the proportion of the volume of the hypercube located with a distance 1/2 of a corner.

Discrete example of concentration of measure

Posted on by John

The previous post looked at a continuous example of concentration of measure. As you move away from a thin band around the equator, the remaining area in the rest of the sphere decreases as an exponential function of the dimension and the distance from the equator. This post will show a very similar result for discrete sequences.

Suppose you have a sequence X1, X2, …, Xn of n random variables that each take on the values {-1, 1} with equal probability. You could think of this as a random walk: you start at 0 and take a sequence of steps either to the left or the right.

Let Sn = X1 + X2 + … + Xn be the sum of the sequence. The expected value of Sn is 0 by symmetry, but it could be as large as n or as small as –n. We want to look at how large |Sn| is likely to be relative to the sequence length n.

Here’s the analytical bound:

P\left( \frac{|S_n|}{n} \geq r\right) \leq 2 \exp\left( - \frac{nr^2}{2}\right )

(If you’re reading this via email, you probably can’t see the equation. Here’s why and how to fix it.)

Here is a simulation in Python to illustrate the bound.

    from random import randint
    import numpy as np
    import matplotlib.pyplot as plt

    n = 400  # random walk length
    N = 10000 # number of repeated walks

    reps = np.empty(N)

    for i in range(N):
        random_walk = [2*randint(0, 1) - 1 for _ in range(n)]
        reps[i] = abs(sum(random_walk)) / n
    
    plt.hist(reps, bins=41)
    plt.xlabel("$|S_n|/n$")
    plt.show()

And here are the results.

Nearly all the area in a high-dimensional sphere is near the equator

Posted on by John

Nearly all the area of a high-dimensional sphere is near the equator.  And by symmetry, it doesn’t matter which equator you take. Draw any great circle and nearly all of the area will be near that circle.  This is the canonical example of “concentration of measure.”

What exactly do we mean by “nearly all the area” and “near the equator”? You get to decide. Pick your standard of “nearly all the area,” say 99%, and your definition of “near the equator,” say within 5 degrees. Then it’s always possible to take the dimension high enough that your standards are met. The more demanding your standard, the higher the dimension will need to be, but it’s always possible to pick the dimension high enough.

This result is hard to imagine. Maybe a simulation will help make it more believable.

In the simulation below, we take as our “north pole” the point (1, 0, 0, 0, …, 0). We could pick any unit vector, but this choice is convenient. Our equator is the set of points orthogonal to the pole, i.e. that have first coordinate equal to zero. We draw points randomly from the sphere, compute their latitude (i.e. angle from the equator), and make a histogram of the results.

The area of our planet isn’t particularly concentrated near the equator.

But as we increase the dimension, we see more and more of the simulation points are near the equator.

Here’s the code that produced the graphs.

from scipy.stats import norm
from math import sqrt, pi, acos, degrees
import matplotlib.pyplot as plt

def pt_on_sphere(n):
    # Return random point on unit sphere in R^n.
    # Generate n standard normals and normalize length.
    x = norm.rvs(0, 1, n)
    length = sqrt(sum(x**2))
    return x/length

def latitude(x):
    # Latitude relative to plane with first coordinate zero.
    angle_to_pole = acos(x[0]) # in radians
    latitude_from_equator = 0.5*pi - angle_to_pole
    return degrees( latitude_from_equator )

N = 1000 # number of samples

for n in [3, 30, 300, 3000]: # dimension of R^n
    
    latitudes = [latitude(pt_on_sphere(n)) for _ in range(N)]
    plt.hist(latitudes, bins=int(sqrt(N)))
    plt.xlabel("Latitude in degrees from equator")
    plt.title("Sphere in dimension {}".format(n))
    plt.xlim((-90, 90))
    plt.show()

Not only is most of the area near the equator, the amount of area outside a band around the equator decreases very rapidly as you move away from the band. You can see that from the histograms above. They look like a normal (Gaussian) distribution, and in fact we can make that more precise.

If A is a band around the equator containing at least half the area, then the proportion of the area a distance r or greater from A is bound by exp( -(n-1)r² ). And in fact, this holds for any set A containing at least half the area; it doesn’t have to be a band around the equator, just any set of large measure.

Related post: Willie Sutton and the multivariate normal distribution

DIEHARDER random number generator test results for PCG and MWC

Posted on by John

A few days ago I wrote about testing the PCG random number generator using the DIEHARDER test suite. In this post I’ll go into a little more background on this random number generator test suite. I’ll also show that like M. E. O’Neill’s PCG (“permuted congruential generator”), George Marsaglia’s MWC (“multiply with carry”) generator does quite well.

This is not to say that MWC is the best generator for every purpose, but any shortcomings of MWC are not apparent from DIEHARDER. The PCG family of generators, for example, is apparently superior to MWC, but you couldn’t necessarily conclude that from these tests.

Unless your application demands more of a random number generator than these tests do, MWC is probably adequate for your application. I wouldn’t recommend it for cryptography or for high-dimensional integration by darts, but it would be fine for many common applications.

DIEHARDER test suite

George Marsaglia developed the DIEHARD battery of tests in 1995. Physics professor Robert G. Brown later refined and extended Marsaglia’s original test suite to create the DIEHARDER suite. (The name of Marsaglia’s battery of tests was a pun on the Diehard car battery. Brown continued the pun tradition by naming his suite after Die Harder, the sequel to the movie Die Hard.) The DIEHARDER suite is open source. It is designed to be at least as rigorous as the original DIEHARD suite and in some cases more rigorous.

There are 31 distinct kinds of tests in the DIEHARDER suite, but some of these are run multiple times. In total, 114 tests are run.

Marsaglia’s MWC

The main strength of Marsaglia’s MWC algorithm is that it is very short. The heart of the code is only three lines:

    m_z = 36969 * (m_z & 65535) + (m_z >> 16);
    m_w = 18000 * (m_w & 65535) + (m_w >> 16);
    return (m_z << 16) + m_w;

You can find a full implementation of a random number generator class based in MWC here.

The heart of PCG is also very short, but a bit more mysterious.

    rng->state = oldstate * 6364136223846793005ULL + (rng->inc | 1);
    uint32_t xorshifted = ((oldstate >> 18u) ^ oldstate) >> 27u;
    uint32_t rot = oldstate >> 59u;
    return (xorshifted >> rot) | (xorshifted << ((-rot) & 31));

(These are the 64-bit state versions of MWC and PCG. Both have versions based on larger state.)

Because these generators require little code, they’d be relatively easy to step into with a debugger, compared to other RNGs such as Mersenne Twister that require more code and more state.

Test results

Out of the 114 DIEHARDER tests run on MWC, all but three returned a pass, and the rest returned a weak pass.

A few weak passes are to be expected. The difference between pass, weak pass, and failure is whether a p-value falls below a certain threshold. Theory says that ideally p-values would uniformly distributed, and so one would fall outside the region for a strong pass now and then.

Rather than counting strong and weak passes, let’s look at the p-values themselves. We’d expect these to be uniformly distributed. Let’s see if they are.

Here are the p-values reported by the DIEHARDER tests for MWC:

Histogram of p-values for MWC

Here are the corresponding values for PCG:

Histogram of p-values for PCG

Neither test has too many small p-values, no more than we’d expect. This is normally what we’re concerned about. Too many small p-values would indicate that the generated samples are showing behavior that would be rare for truly random input.

But both sets of test results have a surprising number of large p-values. Not sure what to make of that. I suspect it says more about the DIEHARDER test suite than the random number generators being tested.

Update: I went back to look at some results from Mersenne Twister to see if this pattern of large p-values persists there. It does, and in fact the p-values are even more biased toward the high end for Mersenne Twister.

Histogram of Mersenne Twister p-values

One thing I noticed is that the large p-values are disproportionately coming from some of the same tests each time. In particular, the repetitions of thests_serial test have an unexpectedly high number of large p-values for each generator.

The chaos game and the Sierpinski triangle

Posted on by John

The chaos game is played as follows. Pick a starting point at random. Then at each subsequent step, pick a triangle vertex at random and move half way from the current position to that vertex.

The result looks like a fractal called the Sierpinski triangle or Sierpinski gasket.

Here’s an example:

Unbiased chaos game results

If the random number generation is biased, the resulting triangle will show it. In the image below, the lower left corner was chosen with probability 1/2, the top with probability 1/3, and the right corner with probability 1/6.

Biased chaos game results

Update: Here’s an animated version that lets you watch the process in action.

animated gif

Here’s Python code to play the chaos game yourself.

from scipy import sqrt, zeros
import matplotlib.pyplot as plt
from random import random, randint

def midpoint(p, q):
    return (0.5*(p[0] + q[0]), 0.5*(p[1] + q[1]))

# Three corners of an equilateral triangle
corner = [(0, 0), (0.5, sqrt(3)/2), (1, 0)]

N = 1000
x = zeros(N)
y = zeros(N)

x[0] = random()
y[0] = random()
for i in range(1, N):
    k = randint(0, 2) # random triangle vertex
    x[i], y[i] = midpoint( corner[k], (x[i-1], y[i-1]) )
    
plt.scatter(x, y)
plt.show()

 

Update 2: Peter Norvig posted some Python code with variations on the game presented here, generalizing a triangle to other shapes. If you try the analogous procedure with a square, you simply get a square filled with random dots.

However, you can get what you might expect, the square analog of the Sierpinski triangle, the product of a Cantor set with itself, if you make a couple modifications. First, pick a side at random, not a corner. Second, move 1/3 of the way toward the chosen side, not 1/2 way.

Here’s what I got with these changes:

Chaos game for a square

Source: Chaos and Fractals

Testing the PCG random number generator

Posted on by John

M. E. O’Neill’s PCG family of random number generators looks very promising. It appears to have excellent statistical and cryptographic properties. And it takes remarkably little code to implement. (PCG stands for Permuted Congruential Generator.)

The journal article announcing PCG gives the results of testing it with the TestU01 test suite. I wanted to try it out by testing it with the DIEHARDER test suite (Robert G. Brown’s extension of George Marsaglia’s DIEHARD test suite) and the NIST Statistical Test Suite. I used what the generator’s website calls the “minimal C implementation.”

(The preprint of the journal article is dated 2015 but apparently hasn’t been published yet.)

For the NIST test suite, I generated 10,000,000 bits and divided them into 10 streams.

For the DIEHARDER test suite, I generated 800,000,000 unsigned 32-bit integers. (DIEHARDER requires a lot of random numbers as input.)

For both test suites I used the seed (state) 20170707105851 and sequence constant (inc) 42.

The PCG generator did well on all the NIST tests. For every test, at least least 9 out of 10 streams passed. The test authors say you should expect at least 8 out of 10 streams to pass.

Here’s an excerpt from the results. You can find the full results here.

--------------------------------------------------------
 C1  C2  C3 ...  C10  P-VALUE  PROPORTION  STATISTICAL TEST
--------------------------------------------------------
  2   0   2        0  0.213309     10/10   Frequency
  0   0   1        3  0.534146     10/10   BlockFrequency
  3   0   0        0  0.350485     10/10   CumulativeSums
  1   1   0        2  0.350485     10/10   CumulativeSums
  0   2   2        1  0.911413     10/10   Runs
  0   0   1        1  0.534146     10/10   LongestRun
  0   1   2        0  0.739918     10/10   Rank
  0   4   0        0  0.122325     10/10   FFT
  1   0   0        1  0.000439     10/10   NonOverlappingTemplate
  ...              
  2   1   0        0  0.350485      9/10   NonOverlappingTemplate
  0   2   1        0  0.739918     10/10   OverlappingTemplate
  1   1   0        2  0.911413     10/10   Universal
  1   1   0        0  0.017912     10/10   ApproximateEntropy
  1   0   1        1     ----       3/4    RandomExcursions
  ...              
  0   0   0        1     ----       4/4    RandomExcursions
  2   0   0        0     ----       4/4    RandomExcursionsVariant
  ...              
  0   0   3        0     ----       4/4    RandomExcursionsVariant
  1   2   3        0  0.350485      9/10   Serial
  1   1   1        0  0.739918     10/10   Serial
  1   2   0        0  0.911413     10/10   LinearComplexity

...

The DIEHARDER suite has 31 kinds tests, some of which are run many times, making a total of 114 tests. Out of the 114 tests, two returned a weak pass for the PCG input and all the rest passed. A few weak passes are to be expected from running so many tests and so this isn’t a strike against the generator. In fact, it might be suspicious if no tests returned a weak pass.

Here’s an edited version of the results. The full results are here.

#=============================================================================#
        test_name   |ntup| tsamples |psamples|  p-value |Assessment
#=============================================================================#
   diehard_birthdays|   0|       100|     100|0.46682782|  PASSED
      diehard_operm5|   0|   1000000|     100|0.83602120|  PASSED
  diehard_rank_32x32|   0|     40000|     100|0.11092547|  PASSED
    diehard_rank_6x8|   0|    100000|     100|0.78938803|  PASSED
   diehard_bitstream|   0|   2097152|     100|0.81624396|  PASSED
        diehard_opso|   0|   2097152|     100|0.95589325|  PASSED
        diehard_oqso|   0|   2097152|     100|0.86171368|  PASSED
         diehard_dna|   0|   2097152|     100|0.24812341|  PASSED
diehard_count_1s_str|   0|    256000|     100|0.75417270|  PASSED
diehard_count_1s_byt|   0|    256000|     100|0.25725000|  PASSED
 diehard_parking_lot|   0|     12000|     100|0.59288414|  PASSED
    diehard_2dsphere|   2|      8000|     100|0.79652706|  PASSED
    diehard_3dsphere|   3|      4000|     100|0.14978100|  PASSED
     diehard_squeeze|   0|    100000|     100|0.35356584|  PASSED
        diehard_sums|   0|       100|     100|0.04522121|  PASSED
        diehard_runs|   0|    100000|     100|0.39739835|  PASSED
        diehard_runs|   0|    100000|     100|0.99128296|  PASSED
       diehard_craps|   0|    200000|     100|0.64934221|  PASSED
       diehard_craps|   0|    200000|     100|0.27352733|  PASSED
 marsaglia_tsang_gcd|   0|  10000000|     100|0.10570816|  PASSED
 marsaglia_tsang_gcd|   0|  10000000|     100|0.00267789|   WEAK
         sts_monobit|   1|    100000|     100|0.98166534|  PASSED
            sts_runs|   2|    100000|     100|0.05017630|  PASSED
          sts_serial|   1|    100000|     100|0.95153782|  PASSED
...
          sts_serial|  16|    100000|     100|0.59342390|  PASSED
         rgb_bitdist|   1|    100000|     100|0.50763759|  PASSED
...
         rgb_bitdist|  12|    100000|     100|0.98576422|  PASSED
rgb_minimum_distance|   2|     10000|    1000|0.23378443|  PASSED
...
rgb_minimum_distance|   5|     10000|    1000|0.13215367|  PASSED
    rgb_permutations|   2|    100000|     100|0.54142546|  PASSED
...
    rgb_permutations|   5|    100000|     100|0.96040216|  PASSED
      rgb_lagged_sum|   0|   1000000|     100|0.66587166|  PASSED
...
      rgb_lagged_sum|  31|   1000000|     100|0.00183752|   WEAK
      rgb_lagged_sum|  32|   1000000|     100|0.13582393|  PASSED
     rgb_kstest_test|   0|     10000|    1000|0.74708548|  PASSED
     dab_bytedistrib|   0|  51200000|       1|0.30789191|  PASSED
             dab_dct| 256|     50000|       1|0.89665788|  PASSED
        dab_filltree|  32|  15000000|       1|0.67278231|  PASSED
        dab_filltree|  32|  15000000|       1|0.35348003|  PASSED
       dab_filltree2|   0|   5000000|       1|0.18749029|  PASSED
       dab_filltree2|   1|   5000000|       1|0.92600020|  PASSED

Simple random number generator does surprisingly well

Posted on by John

I was running the NIST statistical test suite recently. I wanted an example of a random number generator where the tests failed, and so I used a simple generator, a linear congruence generator. But to my surprise, the generator passed nearly all the tests, even though some more sophisticated generators failed some of the same tests.

This post will implement a couple of the simplest tests in Python and show that the generator does surprisingly well.

The linear congruential generator used here starts with an arbitrary seed, then at each step produces a new number by multiplying the previous number by a constant and taking the remainder by 231 – 1. The multiplier constant was chosen to be one of the multipliers recommended in [1].

We’ll need a couple math functions:

from math import sqrt, log

and we need to define the constants for our generator.

# Linear congruence generator (LCG) constants
z = 20170705   # seed
a = 742938285  # multiplier
e = 31         # will need this later
m = 2**e -1    # modulus

Next we form a long string of 0’s and 1’s using our generator

# Number of random numbers to generate
N = 100000     

# Format to print bits, padding with 0's on the left if needed
formatstr = "0" + str(e) + "b"

bit_string = ""
for _ in range(N):
    z = a*z % m # LCG
    bit_string += format(z, formatstr)

Next we run a couple tests. First, we count the number of 1’s in our string of bits. We expect about half the bits to be 1’s. We can quantify “about” as within two standard deviations.

def count_ones(string):
    ones = 0
    for i in range(len(string)):
        if string[i] == '1':
            ones += 1
    return ones

ones = count_ones(bit_string)
expected = e*N/2
sd = sqrt(0.25*N)
print( "Number of 1's: {}".format(ones) )
print( "Expected: {} to {}".format(expected - 2*sd, expected + 2*sd) )

The results are nothing unusual:

Number of 1's: 1550199
Expected: 1549683.8 to 1550316.2

Next we look at the length of the longest runs on 1’s. I’ve written before about the probability of long runs and the code below uses a couple results from that post.

def runs(string):
    max_run = 0
    current_run = 0
    for i in range(len(string)):
        if string[i] == '1':
            current_run += 1
        else:
            max_run = max(max_run, current_run)
            current_run = 0
    return max_run

runlength = runs(bit_string)
expected = -log(0.5*e*N)/log(0.5)
sd = 1/log(2)
print( "Run length: {}".format(runlength) )
print( "Expected: {} to {}".format(expected - 2*sd, expected + 2*sd) )

Again the results are nothing unusual:

Run length: 19
Expected: 17.7 to 23.4

Simple random number generators are adequate for many uses. Some applications, such as high dimensional integration and cryptography, require more sophisticated generators, but sometimes its convenient and sufficient to use something simple. For example, code using the LCG generator above would be easier to debug than code using the Mersenne Twister. The entire state of the LCG is a single number, whereas the Mersenne Twister maintains an internal state of 312 numbers.

One obvious limitation of the LCG used here is that it couldn’t possibly produce more than  231 – 1 values before repeating itself. Since the state only depends on the last value, every time it comes to a given output, the next output will be whatever the next output was the previous time. In fact, [1] shows that it does produce 231 – 1 values before cycling. If the multiplier were not chosen carefully it could have a shorter period.

So our LCG has a period of about two billion values. That’s a lot if you’re writing a little game, for example. But it’s not enough for many scientific applications.

* * *

[1] George S. Fishman and Louis R. Moore III, An exhaustive analysis of multiplicative congruential random number generators with modulus 231 – 1, SIAM Journal of Scientific and Statistical Computing, Vol. 7, no. 1, January 1986.

Least common multiple of the first n positive integers

Posted on by John

Here’s a surprising result: The least common multiple of the first n positive integers is approximately exp(n).

More precisely, let φ(n) equal the log of the least common multiple of the numbers 1, 2, …, n. There are theorems that give upper and lower bounds on how far φ(n) can be from n. We won’t prove or even state these bounds here. See [1] for that. Instead, we’ll show empirically that φ(n) is approximately n.

Here’s some Python code to plot φ(n) over n. The ratio jumps up sharply after the first few values of n. In the plot below, we chop off the first 20 values of n.

from scipy import arange, empty
from sympy.core.numbers import ilcm
from sympy import log
import matplotlib.pyplot as plt

N = 5000
x = arange(N)
phi = empty(N)

M = 1
for n in range(1, N):
    M = ilcm(n, M)
    phi[n] = log(M)

a = 20
plt.plot(x[a:], phi[a:]/x[a:])
plt.xlabel("$n$")
plt.ylabel("$\phi(n) / n$")
plt.show()

Here’s the graph this produces.

[1] J. Barkley Rosser and Lowell Schoenfeld. Approximate formulas for some functions of prime numbers. Illinois Journal of Mathematics, Volume 6, Issue 1 (1962), 64-94. (On Project Euclid)

Subscribing by email

Posted on by John

You can subscribe to my blog by email or RSS. I also have a brief newsletter you could sign up for. There are links to these in the sidebar of the blog:

subscription options

If you subscribe by email, you’ll get an email each morning containing the post(s) from the previous day.

I just noticed a problem with email subscription: it doesn’t show SVG images, at least when reading via Gmail; maybe other email clients display SVG correctly. Here’s what a portion of yesterday’s email looks like in Gmail:

screen shot of missing image

I’ve started using SVG for graphs, equations, and a few other images. The main advantage to SVG is that the images look sharper. Also, you can display the same image file at any resolution; no need to have different versions of the image for display at different sizes. And sometimes SVG files are smaller than their raster counterparts.

There may be a way to have web site visitors see SVG and email subscribers see PNG. If not, email subscribers can click on the link at the top of each post to open it in a browser and see all the images.

By the way, RSS readers handle SVG just fine. At least Digger Reader, the RSS reader I use, works well with SVG. The only problem I see is that centered content is always moved to the left.

* * *

The email newsletter is different from the email blog subscription. I only send out a newsletter once a month. It highlights the most popular posts and says a little about what I’ve been up to. I just sent out a newsletter this morning, so it’ll be another month before the next one comes out.

Effective sample size for MCMC

Posted on by John

In applications we’d like to draw independent random samples from complicated probability distributions, often the posterior distribution on parameters in a Bayesian analysis. Most of the time this is impractical.

MCMC (Markov Chain Monte Carlo) gives us a way around this impasse. It lets us draw samples from practically any probability distribution. But there’s a catch: the samples are not independent. This lack of independence means that all the familiar theory on convergence of sums of random variables goes out the window.

There’s not much theory to guide assessing the convergence of sums of MCMC samples, but there are heuristics. One of these is effective sample size (ESS). The idea is to have a sort of “exchange rate” between dependent and independent samples. You might want to say, for example, that 1,000 samples from a certain Markov chain are worth about as much as 80 independent samples because the MCMC samples are highly correlated. Or you might want to say that 1,000 samples from a different Markov chain are worth about as much as 300 independent samples because although the MCMC samples are dependent, they’re weakly correlated.

Here’s the definition of ESS:

\mbox{ESS} = \frac{n}{1 + 2\sum_{k=1}^\infty \rho(k)}

where n is the number of samples and ρ(k) is the correlation at lag k.

This behaves well in the extremes. If your samples are independent, your effective samples size equals the actual sample size. If the correlation at lag k decreases extremely slowly, so slowly that the sum in the denominator diverges, your effective sample size is zero.

Any reasonable Markov chain is between the extremes. Zero lag correlation is too much to hope for, but ideally the correlations die off fast enough that the sum in the denominator not only converges but also isn’t a terribly large value.

I’m not sure who first proposed this definition of ESS. There’s a reference to it in Handbook of Markov Chain Monte Carlo where the authors cite a paper [1] in which Radford Neal mentions it. Neal cites B. D. Ripley [2].

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[1] Markov Chain Monte Carlo in Practice: A Roundtable Discussion. Robert E. Kass, Bradley P. Carlin, Andrew Gelman and Radford M. Neal. The American Statistician. Vol. 52, No. 2 (May, 1998), pp. 93-100

[2] Stochlastic Simulation, B. D. Ripley, 1987.

Quicksort and prime numbers

Posted on by John

The average number of operations needed for quicksort to sort a list of n items is approximately 10 times the nth prime number.

Here’s some data to illustrate this.

|------+-----------------+---------|
|    n | avg. operations | 10*p(n) |
|------+-----------------+---------|
|  100 |          5200.2 |    5410 |
|  200 |         12018.3 |   12230 |
|  300 |         19446.9 |   19870 |
|  400 |         27272.2 |   27410 |
|  500 |         35392.2 |   35710 |
|  600 |         43747.3 |   44090 |
|  700 |         52297.8 |   52790 |
|  800 |         61015.5 |   61330 |
|  900 |         69879.6 |   69970 |
| 1000 |         78873.5 |   79190 |
| 1100 |         87984.4 |   88310 |
| 1200 |         97201.4 |   97330 |
| 1300 |        106515.9 |  106570 |
| 1400 |        115920.2 |  116570 |
| 1500 |        125407.9 |  125530 |
| 1600 |        134973.5 |  134990 |
| 1700 |        144612.1 |  145190 |
| 1800 |        154319.4 |  154010 |
| 1900 |        164091.5 |  163810 |
| 2000 |        173925.1 |  173890 |
|------+-----------------+---------|

The maximum difference between the quicksort and prime columns is about 4%. In the latter half of the table, the maximum error is about 0.4%.

What’s going on here? Why should quicksort be related to prime numbers?!

The real mystery is the prime number theorem, not quicksort. The prime number theorem tells us that the nth prime number is approximately n log n. And the number of operations in an efficient sort is proportional to n log n. The latter is easier to see than the former.

A lot of algorithms have run time proportional to n log n: mergesort, heapsort, FFT (Fast Fourier Transform), etc. All these have run time approximately proportional to the nth prime.

Now for the fine print. What exactly is the average run time for quicksort? It’s easy to say it’s O(n log n), but getting more specific requires making assumptions. I used as the average number of operations 11.67 n log n – 1.74 n based on Knuth’s TAOCP, Volume 3. And why 10 times the nth prime and not 11.67? I chose 10 to make the example work better. For very large values on n, a larger coefficient would work better.

 

Why do linear prediction confidence regions flare out?

Posted on by John

Suppose you’re tracking some object based on its initial position x0 and initial velocity v0. The initial position and initial velocity are estimated from normal distributions with standard deviations σx and σv. (To keep things simple, let’s assume our object is moving in only one dimension and that the distributions around initial position and velocity are independent.)

The confidence region for the object flares out over time, something like the bell of a trumpet.

Why does the region get larger? Because there’s uncertainty in the velocity, and the velocity gets multiplied by elapsed time.

Why isn’t the confidence region a cone? Because that would ignore the uncertainty in the initial position. The result would be too small.

Why isn’t the confidence region a truncated cone? That’s not a bad approximation, though it’s a bit too large. If we ignore probability for a moment and treat confidence intervals as deterministic limits, then we get a truncated cone. For example, suppose assume position and velocity are each within two standard deviations of their estimates. Then we’d estimate position to be between x0 – 2σx + (v0 – 2σv) t on the low end and x0 + 2σx + (v0 + 2σv) t on the high end. This is only an approximation because we’ve ignored probability, and it’s pessimistic because it assumes extreme error values for both estimates at the same time.

So what is the confidence region? It’s some where between the cone and the truncated cone.

The position x + t v is the sum of two random variables. The first has variance σx² and the second has variance t² σv². Variances of independent random variables add, so the standard deviation for the sum is

√(σx² + t² σv²) = t √(σx² / t² + σv²)

Note that as t increases, the latter approaches t σv from above. Ignoring the uncertainty in initial position underestimates standard deviation, but the relative error decreases as t increases.

For large t, a confidence interval for position at time t is approximately proportional to t, so the width of the confidence intervals over time look like a cone. But from small t, the dependence on t is less linear and more curved.

Polynomials evaluated at integers

Posted on by John

Let p(x) = a0 + a1x + a2x2 + … + anxn and suppose at least one of the coefficients ai is irrational for some i ≥ 1. Then a theorem by Weyl says that the fractional parts of p(n) are equidistributed as n varies over the integers. That is, the proportion of values that land in some interval is equal to the length of that interval.

Clearly it’s necessary that one of the coefficients be irrational. What may be surprising is that it is sufficient.

If the coefficients are all rational with common denominator N, then the sequence would only contain multiples of 1/N. The interval [1/3N, 2/3N], for example, would never get a sample. If a0 were irrational but the rest of the coefficients were rational, we’d have the same situation, simply shifted by a0.

This is a theorem about what happens in the limit, but we can look at what happens for some large but finite set of terms. And we can use a χ2 test to see how evenly our sequence is compared to what one would expect from a random sequence.

In the code below, we use the polynomial p(x) = √2 x² + πx + 1 and evaluate p at 0, 1, 2, …, 99,999. We then count how many fall in the bins [0, 0.01), [0.01, 0.02), … [0.99, 1] and compute a chi-square statistic on the counts.

from math import pi, sqrt, floor

def p(x):
    return 1 + pi*x + sqrt(2)*x*x

def chisq_stat(O, E):
    return sum( [(o - e)**2/e for (o, e) in zip(O, E)] )

def frac(x):
    return x - floor(x)

N = 100000
data = [frac(p(n)) for n in range(N)]

count = [0]*100
for d in data:
    count[ int(floor(100*d)) ] += 1

expected = [N/100]*100

print(chisq_stat(count, expected))

We get a chi-square statistic of 95.4. Since we have 100 bins, there are 99 degrees of freedom, and we should compare our statistic to a chi-square distribution with 99 degrees of freedom. Such a distribution has mean 99 and standard deviation √(99*2) = 14.07, so a value of 95.4 is completely unremarkable.

If we had gotten a very large chi-square statistic, say 200, we’d have reason to suspect something was wrong. Maybe a misunderstanding on our part or a bug in our software. Or maybe the sequence was not as uniformly distributed as we expected.

If we had gotten a very small chi-square statistic, say 10, then again maybe we misunderstood something, or maybe our sequence is remarkably evenly distributed, more evenly than one would expect from a random sequence.

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Leading digits of powers of 2

Posted on by John

The first digit of a power of 2 is a 1 more often than any other digit. Powers of 2 begin with 1 about 30% of the time. This is because powers of 2 follow Benford’s law. We’ll prove this below.

When is the first digit of 2n equal to k? When 2n is between k × 10p and (k+1) × 10p for some positive integer p. By taking logarithms base 10 we find that this is equivalent to the fractional part of n log102 being between log10 k and log10 (k + 1).

The map

x ↦ ( x + log10 2 ) mod 1

is ergodic. I wrote about irrational rotations a few weeks ago, and this is essentially the same thing. You could scale x by 2π and think of it as rotations on a circle instead of arithmetic mod 1 on an interval. The important thing is that log10 2 is irrational.

Repeatedly multiplying by 2 is corresponds to adding log10 2 on the log scale. So powers of two correspond to iterates of the map above, starting with x = 0. Birkhoff’s Ergodic Theorem tells us that the number of times iterates of this map fall in the interval [a, b] equals b – a. So for k = 1, 2, 3, … 9, the proportion of powers of 2 start with k is equal to  log10 (k + 1) –  log10 (k) =  log10 ((k + 1) / k).

This is Benford’s law. In particular, the proportion of powers of 2 that begin with 1 is equal to  log10 (2) = 0.301.

Note that the only thing special about 2 is that log10 2 is irrational. Powers of 3 follow Benford’s law as well because log10 3 is also irrational. For what values of b do powers of b not follow Benford’s law? Those with log10 b rational, i.e. powers of 10. Obviously powers of 10 don’t follow Benford’s law because their first digit is always 1!

[Interpret the “!” above as factorial or exclamation as you wish.]

Let’s look at powers of 2 empirically to see Benford’s law in practice. Here’s a simple program to look at first digits of powers of 2.

count = [0]*10
N = 10000

def first_digit(n):
    return int(str(n)[0])

for i in range(N):
    n = first_digit( 2**i )
    count[n] += 1

print(count)

Unfortunately this only works for moderate values of N. It ran in under a second with N set to 10,000 but for larger values of N it rapidly becomes impractical.

Here’s a much more efficient version that ran in about 2 seconds with N = 1,000,000.

from math import log10
N = 1000000
count = [0]*10

def first_digit_2_exp_e(e):
    r = (log10(2.0)*e) % 1
    for i in range(2, 11):
        if r < log10(i):
            return i-1
for i in range(N):
    n = first_digit_2_exp_e( i )
    count[n] += 1

print(count)

You could make it more efficient by caching the values of log10 rather than recomputing them. This brought the run time down to about 1.4 seconds. That’s a nice improvement, but nothing like the orders of magnitude improvement from changing algorithms.

Here are the results comparing the actual counts to the predictions of Benford’s law (rounded to the nearest integer).

|---------------+--------+-----------|
| Leading digit | Actual | Predicted |
|---------------+--------+-----------|
|             1 | 301030 |    301030 |
|             2 | 176093 |    176091 |
|             3 | 124937 |    124939 |
|             4 |  96911 |     96910 |
|             5 |  79182 |     79181 |
|             6 |  66947 |     66948 |
|             7 |  57990 |     57992 |
|             8 |  51154 |     51153 |
|             9 |  45756 |     45757 |
|---------------+--------+-----------|

The agreement is almost too good to believe, never off by more than 2.

Are the results correct? The inefficient version relied on integer arithmetic and so would be exact. The efficient version relies on floating point and so it’s conceivable that limits of precision caused a leading digit to be calculated incorrectly, but I doubt that happened. Floating point is precise to about 15 significant figures. We start with log10(2), multiply it by numbers up to 1,000,000 and take the fractional part. The result is good to around 9 significant figures, enough to correctly calculate which log digits the result falls between.

Update: See Andrew Dalke’s Python script in the comments. He shows a way to efficiently use integer arithmetic.