Miscellaneous Observations on Randomization

Peter Occil

Contents

On a Binomial Sampler

Take the following sampler of a binomial(n, 1/2) distribution (where n is even), which is equivalent to the one that appeared in (Bringmann et al. 2014)(1), and adapted to be more programmer-friendly.

  1. If n is less than 4, generate n unbiased random bits (zeros or ones) and return their sum. Otherwise, if n is odd, set ret to the result of this algorithm with n = n − 1, then add an unbiased random bit's value to ret, then return ret.
  2. Set m to floor(sqrt(n)) + 1.
  3. (First, sample from an envelope of the binomial curve.) Generate unbiased random bits until a zero is generated this way. Set k to the number of ones generated this way.
  4. Set s to an integer in [0, m) chosen uniformly at random, then set i to k*m + s.
  5. Generate an unbiased random bit. If that bit is 0, set ret to (n/2)+i. Otherwise, set ret to (n/2)−i−1.
  6. (Second, accept or reject ret.) If ret < 0 or ret > n, go to step 3.
  7. With probability choose(n, ret)*m*2(kn)+2, return ret. Otherwise, go to step 3. (Here, choose(n, k) is a binomial coefficient, or the number of ways to choose k out of n labeled items.(2))

This algorithm has an acceptance rate of 1/16 regardless of the value of n. However, step 7 will generally require a growing amount of storage and time to exactly calculate the given probability as n gets larger, notably due to the inherent factorial in the binomial coefficient. The Bringmann paper suggests approximating this factorial via Spouge's approximation; however, it seems hard to do so without using floating-point arithmetic, which the paper ultimately resorts to. Alternatively, the logarithm of that probability can be calculated that is much more economical in terms of storage than the full exact probability. Then, an exponential random variate can be generated, negated, and compared with that logarithm to determine whether the step succeeds.

More specifically, step 7 can be changed as follows:

My implementation of loggamma and the natural logarithm (interval.py) relies on rational interval arithmetic (Daumas et al. 2007)(4) and a fast converging version of Stirling's formula for the factorial's natural logarithm (Schumacher 2016)(5).

Also, according to the Bringmann paper, m can be set such that m is in the interval [sqrt(n), sqrt(n)+3], so I implement step 1 by starting with u = 2floor((1+β(n))/2), then calculating v = floor((u+floor(n/u))/2), w = u, u = v until v >= w, then setting m to w + 1. Here, β(n) = ceil(ln(n+1)/ln(2)), or alternatively the minimum number of bits needed to store n (with β(0) = 0).

Notes:

On a Geometric Sampler

The following algorithm is equivalent to the geometric(px/py) sampler that appeared in (Bringmann and Friedrich 2013)(7), but adapted to be more programmer-friendly. As used in that paper, a geometric(p) random variate expresses the number of failing trials before the first success, where each trial is independent and has success probability p. (Note that the terminology "geometric random variate" has conflicting meanings in academic works. Note also that the algorithm uses the rational number px/py, not an arbitrary real number p; some of the notes in this section indicate how to adapt the algorithm to an arbitrary p.)

  1. Set pn to px, k to 0, and d to 0.
  2. While pn*2 <= py, add 1 to k and multiply pn by 2. (Equivalent to finding the largest k >= 0 such that p*2k <= 1. For the case when p need not be rational, enough of its binary expansion can be calculated to carry out this step accurately, but in this case any k such that p is greater than 1/(2k+2) and less than or equal to 1/(2k) will suffice, as the Bringmann paper points out.)
  3. With probability (1−px/py)2k, add 1 to d and repeat this step. (To simulate this probability, the first sub-algorithm below can be used.)
  4. Generate a uniform random integer in [0, 2k), call it m, then with probability (1−px/py)m, return d*2k+m. Otherwise, repeat this step. (The Bringmann paper, though, suggests to simulate this probability by sampling only as many bits of m as needed to do so, rather than just generating m in one go, then using the first sub-algorithm on m. However, the implementation, given as the second sub-algorithm below, is much more complicated and is not crucial for correctness.)

The first sub-algorithm returns 1 with probability (1−px/py)n, assuming that n*px/py <= 1. It implements the approach from the Bringmann paper by rewriting the probability using the binomial theorem. (For the case when p need not be rational, the probability (1−p)n can be simulated using Bernoulli factory algorithms, or by calculating its digit expansion or series expansion and using the appropriate algorithm for simulating irrational constants. Run that algorithm n times or until it outputs 1, whichever comes first. This sub-algorithm returns 1 if all the runs return 0, or 1 otherwise.)

  1. Set pnum, pden, and j to 1, then set r to 0, then set qnum to px, and qden to py, then set i to 2.
  2. If j is greater than n, go to step 5.
  3. If j is even, set pnum to pnum*qden + pden*qnum*choose(n,j). Otherwise, set pnum to pnum*qdenpden*qnum*choose(n,j).
  4. Multiply pden by qden, then multiply qnum by px, then multiply qden by py, then add 1 to j.
  5. If j is less than or equal to 2 and less than or equal to n, go to step 2.
  6. Multiply r by 2, then add an unbiased random bit's value (either 0 or 1 with equal probability) to r.
  7. If r <= floor((pnum*i)/pden) − 2, return 1. If r >= floor((pnum*i)/pden) + 1, return 0. If neither is the case, multiply i by 2 and go to step 2.

The second sub-algorithm returns an integer m in [0, 2k) with probability (1−px/py)m, or −1 with the opposite probability. It assumes that 2k*px/py <= 1.

  1. Set r and m to 0.
  2. Set b to 0, then while b is less than k:
    1. (Sum b+2 summands of the binomial equivalent of the desired probability. First, append an additional bit to m, from most to least significant.) Generate an unbiased random bit (either 0 or 1 with equal probability). If that bit is 1, add 2kb to m.
    2. (Now build up the binomial probability.) Set pnum, pden, and j to 1, then set qnum to px, and qden to py.
    3. If j is greater than m or greater than b + 2, go to the sixth substep.
    4. If j is even, set pnum to pnum*qden + pden*qnum*choose(m,j). Otherwise, set pnum to pnum*qdenpden*qnum*choose(m,j).
    5. Multiply pden by qden, then multiply qnum by px, then multiply qden by py, then add 1 to j, then go to the third substep.
    6. (Now check the probability.) Multiply r by 2, then add an unbiased random bit's value (either 0 or 1 with equal probability) to r.
    7. If r <= floor((pnum*2b)/pden) − 2, add a uniform random integer in [0, 2k*b) to m and return m (and, if requested, the number kb−1). If r >= floor((pnum*2b)/pden) + 1, return −1 (and, if requested, an arbitrary value). If neither is the case, add 1 to b.
  3. Add an unbiased random bit to m. (At this point, m is fully sampled.)
  4. Run the first sub-algorithm with n = m, except in step 1 of that sub-algorithm, set r to the value of r built up by this algorithm, rather than 0, and set i to 2k, rather than 2. If that sub-algorithm returns 1, return m (and, if requested, the number −1). Otherwise, return −1 (and, if requested, an arbitrary value).

As used in the Bringmann paper, a bounded geometric(p, n) random variate is a geometric(p) random variate or n (an integer greater than 0), whichever is less. The following algorithm is equivalent to the algorithm given in that paper, but adapted to be more programmer-friendly.

  1. Set pn to px, k to 0, d to 0, and m2 to the smallest power of 2 that is greater than n (or equivalently, 2bits where bits is the minimum number of bits needed to store n).
  2. While pn*2 <= py, add 1 to k and multiply pn by 2.
  3. With probability (1−px/py)2k, add 1 to d and then either return n if d*2k is greater than or equal to m2, or repeat this step if less. (To simulate this probability, the first sub-algorithm above can be used.)
  4. Generate a uniform random integer in [0, 2k), call it m, then with probability (1−px/py)m, return min(n, d*2k+m). In the Bringmann paper, this step is implemented in a manner equivalent to the following (this alternative implementation, though, is not crucial for correctness):
    1. Run the second sub-algorithm above, except return two values, rather than one, in the situations given in the sub-algorithm. Call these two values m and mbit.
    2. If m < 0, go to the first substep.
    3. If mbit >= 0, add 2mbit times an unbiased random bit to m and subtract 1 from mbit. If that bit is 1 or mbit < 0, go to the next substep; otherwise, repeat this substep.
    4. Return n if d*2k is greater than or equal to m2.
    5. Add a uniform random integer in [0, 2mbit+1) to m, then return min(n, d*2k+m).

Sampling Unbounded Monotone Density Functions

This section shows a preprocessing algorithm to generate a random variate in [0, 1] from a distribution whose probability density function (PDF)—

The trick here is to sample the peak in such a way that the result is either forced to be 0 or forced to belong to the bounded part of the PDF. This algorithm does not require the area under the curve of the PDF in [0, 1] to be 1; in other words, this algorithm works even if the PDF is known up to a normalizing constant. The algorithm is as follows.

  1. Set i to 1.
  2. Calculate the cumulative probability of the interval [0, 2i] and that of [0, 2−(i − 1)], call them p and t, respectively.
  3. With probability p/t, add 1 to i and go to step 2. (Alternatively, if i is equal to or higher than the desired number of fractional bits in the result, return 0 instead of adding 1 and going to step 2.)
  4. At this point, the PDF at [2i, 2−(i − 1)) is bounded from above, so sample a random variate in this interval using any appropriate algorithm, including rejection sampling. Because the PDF is monotonically decreasing, the peak of the PDF at this interval is located at 2i, so that rejection sampling becomes trivial.

It is relatively straightforward to adapt this algorithm for monotonically increasing PDFs with the unbounded peak at 1, or to PDFs with a different domain than [0, 1].

This algorithm is similar to the "inversion–rejection" algorithm mentioned in section 4.4 of chapter 7 of Devroye's Non-Uniform Random Variate Generation (1986)(3). I was unaware of that algorithm at the time I started writing the text that became this section (Jul. 25, 2020). The difference here is that it assumes the whole distribution (including its PDF and cumulative distribution function) can take on any value in the interval [0, 1] and only those values (that is, its support is [0, 1]), while the algorithm presented in this article doesn't make that assumption (e.g., the interval [0, 1] can cover only part of the PDF's support).

By the way, this algorithm arose while trying to devise an algorithm that can generate an integer power of a uniform random variate, with arbitrary precision, without actually calculating that power (a naïve calculation that is merely an approximation and usually introduces bias); for more information, see my other article on partially-sampled random numbers. Even so, the algorithm I have come up with in this note may be of independent interest.

In the case of powers of a uniform [0, 1] random variate X, namely Xn, the ratio p/t in this algorithm has a very simple form, namely (1/2)1/n, which is possible to simulate using a so-called Bernoulli factory algorithm without actually having to calculate this ratio. Note that this formula is the same regardless of i. This is found by taking the PDF f(x) = x1/n/(x * n) and finding the appropriate p/t ratios by integrating f over the two intervals mentioned in step 2 of the algorithm.

Certain Families of Distributions

This section is a note on certain families of univariate (one-variable) probability distributions, with emphasis on sampling random variates from them. Some of these families are described in Ahmad et al. (2019)(8).

The following definitions are used:

In general, families of the form "X-G" (such as "beta-G" (Eugene et al., 2002)(9)) use two distributions, X and G, where X is a continuous distribution whose support is the interval [0, 1] and G is a distribution with an easy-to-compute quantile function. The following algorithm samples a random variate following a distribution from this kind of family:

  1. Generate a random variate that follows the distribution X. (Or generate a uniform partially-sampled random number (PSRN) that follows the distribution X.) Call the number x.
  2. Calculate the quantile for G of x, and return that quantile. (If x is a uniform PSRN, see "Random Variate Generation via Quantiles", later.)

Certain special cases of the "X-G" families, such as the following, use a specially designed distribution for X:

In fact, the "X-G" families are a special case of the so-called "transformed–transformer" family of distributions introduced by Alzaatreh et al. (2013)(14) that uses two distributions, X and G, where X (the "transformed") is an arbitrary continuous distribution, G (the "transformer") is a distribution with an easy-to-compute quantile function, and W is a nondecreasing function that maps a number in [0, 1] to a number that has the same support as X and meets certain other conditions. The following algorithm samples a random variate from this kind of family:

  1. Generate a random variate that follows the distribution X. (Or generate a uniform PSRN that follows X.) Call the number x.
  2. Calculate the quantile for G of W−1(x) (where W−1(.) is the inverse of W), and return that quantile. (If x is a uniform PSRN, see "Random Variate Generation via Quantiles", later.)

The following are special cases of the "transformed–transformer" family:

Many special cases of the "transformed–transformer" family have been proposed in many papers, and usually their names suggest the distributions that make up this family. Some members of the "odd X G" family have names that begin with the word "generalized", and in most such cases this corresponds to W−1(x) = (x/(1+x))1/a, where a > 0 is a shape parameter; examples include the "generalized odd gamma-G" family (Hosseini et al. 2018)(17).

A family very similar to the "transformed–transformer" family uses a decreasing W. When distribution X's support is [0, ∞), one such W that has been proposed is W(x) = −ln(x) (W−1(x) = exp(−x); examples include the "Rayleigh-G" family or "Rayleigh–Rayleigh" distribution (Al Noor and Assi 2020)(18), as well as the "generalized gamma-G" family, where "generalized gamma" refers to the Stacy distribution (Boshi et al. 2020)(19)).

A compound distribution is simply the minimum of N random variates distributed as X, where N >= 1 is an integer distributed as the discrete distribution Y (Tahir and Cordeiro 2016)(20). For example, the "beta-G-geometric" family represents the minimum of N beta-G random variates, where N is a random variate expressing 1 plus the number of failures before the first success, with each success having the same probability.

A complementary compound distribution is the maximum of N random variates distributed as X, where N >= 1 is an integer distributed as the discrete distribution Y. An example is the "geometric zero-truncated Poisson distribution", where X is the distribution of 1 plus the number of failures before the first success, with each success having the same probability, and Y is the zero-truncated Poisson distribution (Akdoğan et al., 2020)(21).

An inverse X distribution (or inverted X distribution) is generally the distribution of the reciprocal of a random variate distributed as X. For example, an inverse exponential random variate (Keller and Kamath 1982)(22) is the reciprocal of an exponential random variate with rate 1 (and so is distributed as −1/ln(U) where U is a uniform(0, 1) random variate) and may be multiplied by a parameter θ > 0.

A weighted X distribution uses a distribution X and a weight function w(x) whose values lie in [0, 1] everywhere in X's support. The following algorithm samples from a weighted distribution (see also (Devroye 1986, p. 47)(3)):

  1. Generate a random variate that follows the distribution X. (Or generate a uniform PSRN that follows X.) Call the number x.
  2. With probability w(x), return x. Otherwise, go to step 1.

Some weighted distributions allow any weight function w(x) whose values are non-negative everywhere in X's support (Rao 1985)(23). (If w(x) = x, the distribution is often called a length-biased or size-biased distribution; if w(x) = x2, area-biased.) Their probability density functions are proportional to the original density functions multiplied by w(x).

To generate an inflated X (also called c-inflated X) random variate with parameters c and α, generate—

For example, a zero-inflated beta random variate is 0 with probability α and a beta random variate otherwise (the parameter c is 0) (Ospina and Ferrari 2010)(24) A zero-and-one inflated X distribution is 0 or 1 with probability α and distributed as X otherwise. For example, to generate a zero-and-one-inflated unit Lindley random variate (with parameters α, θ, and p) (Chakraborty and Bhattacharjee 2021)(25):

  1. With probability α, return a number that is 0 with probability p and 1 otherwise.
  2. Generate a unit Lindley(θ) random variate, that is, generate x/(1+x) where x is a Lindley(θ) random variate.

Certain Distributions

In the table below, U is a uniform(0, 1) random variate.

This distribution: Is distributed as: And uses these parameters:
Power function(a, c). c*U1/a. a > 0, c > 0.
Right-truncated Weibull(a, b, c) (Jodrá 2020)(26). Minimum of N power function(b, c) random variates, where N is zero-truncated Poisson(a*cb). a, b, c > 0.
Lehmann Weibull(a1, a2, β) (Elgohari and Yousof 2020)(27). (ln(1/U)/β)1/a1/a2 or E1/a1/a2 a1, a2, β > 0. E is exponential with rate β.
Marshall–Olkin(α). (1−U)/(U*(α−1) + 1). α in [0, 1].
Lomax(α). (−1/(U−1))1/α−1. α > 0.
Power Lomax(α, β) (Rady et al. 2016)(28). L1/β β > 0; L is Lomax(α).

Batching Random Samples via Randomness Extraction

Devroye and Gravel (2020)(29) suggest the following randomness extractor to reduce the number of random bits needed to produce a batch of samples by a sampling algorithm. The extractor works based on the probability that the algorithm consumes X random bits to produce a specific output Y (or P(X | Y) for short):

  1. Start with the interval [0, 1].
  2. For each pair (X, Y) in the batch, the interval shrinks from below by P(X−1 | Y) and from above by P(X | Y). (For example, if [0.2, 0.8] (range 0.6) shrinks from below by 0.1 and from above by 0.8, the new interval is [0.2+0.1*0.6, 0.2+0.8*0.6] = [0.26, 0.68]. For correctness, though, the interval is not allowed to shrink to a single point, since otherwise step 3 would run forever.)
  3. Extract the bits, starting from the binary point, that the final interval's lower and upper bound have in common (or 0 bits if the upper bound is 1). (For example, if the final interval is [0.101010..., 0.101110...] in binary, the bits 1, 0, 1 are extracted, since the common bits starting from the point are 101.)

After a sampling method produces an output Y, both X (the number of random bits the sampler consumed) and Y (the output) are added to the batch and fed to the extractor, and new bits extracted this way are added to a queue for the sampling method to use to produce future outputs. (Notice that the number of bits extracted by the algorithm above grows as the batch grows, so only the new bits extracted this way are added to the queue this way.)

The issue of finding P(X | Y) is now discussed. Generally, if the sampling method implements a random walk on a binary tree that is driven by unbiased random bits and has leaves labeled with one outcome each (Knuth and Yao 1976)(30), P(X | Y) is found as follows (and Claude Gravel clarified to me that this is the intention of the extractor algorithm): Take a weighted count of all leaves labeled Y up to depth X (where the weight for depth z is 1/2z), then divide it by a weighted count of all leaves labeled Y at all depths (for instance, if the tree has two leaves labeled Y at z=2, three at z=3, and three at z=4, and X is 3, then P(X | Y) is (2/22+3/23) / (2/22+3/23+3/24)). In the special case where the tree has at most 1 leaf labeled Y at every depth, this is implemented by finding P(Y), or the probability to output Y, then chopping P(Y) up to the Xth binary digit after the point and dividing by the original P(Y) (for instance, if X is 4 and P(Y) is 0.101011..., then P(X | Y) is 0.1010 / 0.101011...).

Unfortunately, P(X | Y) is not easy to calculate when the number of values Y can take on is large or even unbounded. In this case, I can suggest the following ad hoc algorithm, which uses a randomness extractor that takes bits as input, such as the von Neumann, Peres, or Zhou–Bruck extractor (see "Notes on Randomness Extraction"). The algorithm counts the number of bits it consumes (X) to produce an output, then feeds X to the extractor as follows.

  1. Let z be abs(XlastX), where lastX is either the last value of X fed to this extractor for this batch or 0 if there is no such value.
  2. If z is greater than 0, feed the bits of z from most significant to least significant to a queue of extractor inputs.
  3. Now, when the sampler consumes a random bit, it checks the input queue. As long as 64 bits or more are in the input queue, the sampler dequeues 64 bits from it, runs the extractor on those bits, and adds the extracted bits to an output queue. (The number 64 can instead be any even number greater than 2.) Then, if the output queue is not empty, the sampler dequeues a bit from that queue and uses that bit; otherwise it generates an unbiased random bit as usual.

Random Variate Generation via Quantiles

This note is about generating random variates from a continuous distribution via inverse transform sampling (or via quantiles), using uniform partially-sampled random numbers (PSRNs). See "Certain Families of Distributions" for a definition of quantile functions. A uniform PSRN is ultimately a number that lies in an interval; it contains a sign, an integer part, and a fractional part made up of digits sampled on demand.

Take the following situation:

Then the following algorithm transforms that number to a random variate for the desired distribution, which comes within the desired error tolerance of ε with probability 1 (see (Devroye and Gravel 2020)(29)):

  1. Generate additional digits of x uniformly at random—thus shortening the interval [a, b]—until a lower bound of Q(f(a)) and an upper bound of Q(f(b)) differ by no more than 2*ε. Call the two bounds low and high, respectively.
  2. Return low+(highlow)/2.

In some cases, it may be possible to calculate the needed digit size in advance.

As one example, if f(t) = t (the identity function) and the quantile function is Lipschitz continuous on the interval [a, b], which roughly means that it's a continuous function with no vertical slope on that interval, then the following algorithm generates a quantile with error tolerance ε:

  1. Let d be ceil((ln(max(1,L)) − ln(ε)) / ln(β)), where L is an upper bound of the quantile function's maximum slope (also known as the Lipschitz constant). For each digit among the first d digits in x's fractional part, if that digit is unsampled, set it to a digit chosen uniformly at random.
  2. The PSRN x now lies in the interval [a, b]. Calculate lower and upper bounds of Q(a) and Q(b), respectively, that are within ε/2 of the true quantiles, call the bounds low and high, respectively.
  3. Return low+(highlow)/2.

This algorithm chooses a random interval of size equal to βd, and because the quantile function is Lipschitz continuous, the values at the interval's bounds are guaranteed to vary by no more than 2*ε (actually ε, but the calculation in step 2 adds an additional error of at most ε), which is needed to meet the tolerance ε (see also Devroye and Gravel 2020(29)).

A similar algorithm can exist even if the quantile function Q is not Lipschitz continuous on the interval [a, b].

Specifically, if—

then d in step 1 above can be calculated as—
  max(0, ceil(−ln(ω−1(ε))/ln(β))),
where ω−1(ε) is the inverse of the modulus of continuity. (Loosely speaking, a modulus of continuity ω(δ) gives the quantile function's maximum range in a window of size δ, and the inverse modulus ω−1(ε) finds a window small enough that the quantile function differs by no more than ε in the window.(31)).(32)

For example—

The algorithms given earlier in this section have a disadvantage: the desired error tolerance has to be made known to the algorithm in advance. To generate a quantile to any error tolerance (even if the tolerance is not known in advance), a rejection sampling approach is needed. For this to work:

Here is a sketch of how this rejection sampler might work:

  1. After using one of the algorithms given earlier in this section to sample digits of x as needed, let a and b be x's upper and lower bounds. Calculate lower and upper bounds of the quantiles of f(a) and f(b) (the bounds are [alow, ahigh] and [blow, bhigh] respectively).
  2. Given the density function, sample a uniform PSRN, y, in the interval [alow, bhigh] using an arbitrary-precision rejection sampler such as Oberhoff's method (described in an appendix to the PSRN article).
  3. Accept y (and return it) if it clearly lies in [ahigh, blow]. Reject y (and go to the previous step) if it clearly lies outside [alow, bhigh]. If y clearly lies in [alow, ahigh] or in [blow, bhigh], generate more digits of x, uniformly at random, and go to the first step.
  4. If y doesn't clearly fall in any of the cases in the previous step, generate more digits of y, uniformly at random, and go to the previous step.

ExpoExact

This algorithm ExpoExact, samples an exponential random variate given the rate rx/ry with an error tolerance of 2-precision; for more information, see "Partially-Sampled Random Numbers"; see also Morina et al. (2019)(33); Canonne et al. (2020)(34). In this section, RNDINT(1) generates an independent unbiased random bit.

METHOD ZeroOrOneExpMinus(x, y)
  if y <= 0 or x<0: return error
  if x==0: return 1 // exp(0) = 1
  if x > y
    x = rem(x, y)
    if x>0 and ZeroOrOneExpMinus(x, y) == 0: return 0
    for i in 0...floor(x/y): if ZeroOrOneExpMinus(1,1) == 0: return 0
    return 1
  end
  r = 1
  oy = y
  while true
    if ZeroOrOne(x, y) == 0: return r
    r=1-r; y = y + oy
  end
END METHOD

METHOD ExpoExact(rx, ry, precision)
   ret=0
   for i in 1..precision
    // This loop adds to ret with probability 1/(exp(2^-prec)+1).
    // References: Alg. 6 of Morina et al. 2019; Canonne et al. 2020.
    denom=pow(2,i)*ry
    while true
       if RNDINT(1)==0: break
       if ZeroOrOneExpMinus(rx, denom) == 1:
         ret=ret+MakeRatio(1,pow(2,i))
    end
   end
   while ZeroOrOneExpMinus(rx,ry)==1: ret=ret+1
   return ret
END METHOD

Note: After ExpoExact is used to generate a random variate, an application can append additional binary digits (such as RNDINT(1)) to the end of that number while remaining accurate to the given precision (Karney 2014)(35).

A sampler for distributions with nonincreasing or nondecreasing weights

An algorithm for sampling an integer in the interval [a, b) with probability proportional to weights listed in nonincreasing order (example: [10, 3, 2, 1, 1] when a = 0 and b = 5) can be implemented as follows (Chewi et al. 2021)(36). It has a logarithmic time complexity in terms of setup and sampling.

For nondecreasing rather than nonincreasing weights, the algorithm is as follows instead:

Note: The weights can be base-β logarithms, especially since logarithms preserve order, but in this case the algorithm requires changes. In the setup step 2, replace "q[m]*min((ba)" with "q[m]+ln(min((ba))/ln(β)" (which is generally inexact unless β is 2); in sampling step 1, use an algorithm that takes base-β logarithms as weights; and replace sampling step 3 with "Generate an exponential random variate with rate ln(β). If that variate is greater than q[k] minus w[x], return x. Otherwise, go to step 1." These modifications can introduce numerical errors unless care is taken, such as by using partially-sampled random numbers (PSRNs).

A sampler for unimodal distributions of weights

The following is an algorithm for sampling an integer in the interval [a, b) with probability proportional to a unimodal distribution of weights (that is, nondecreasing on the left and nonincreasing on the right) (Chewi et al. 2021)(36). It assumes the mode (the point with the highest weight) is known. An example is [1, 3, 9, 4, 4] when a = 0 and b = 5, and the mode is 2, which corresponds to the weight 9. It has a logarithmic time complexity in terms of setup and sampling.

Log-Uniform Distribution

Samples from the so-called "log uniform distribution" as used by the Abseil programming library. This algorithm takes a maximum mx and a logarithmic base b, and chooses an integer in [0, mx] such that two values are equally likely to be chosen if their base-b logarithms are equal in their integer parts (which roughly means that lower numbers are exponentially more likely to occur). Although this algorithm works, in principle, for any b > 0, Abseil supports only integer bases b.

  1. Let L be ceil(ln(mx+1)/ln(b)). Choose a uniform random integer in the closed interval [0, l], call it u.
  2. If u is 0, return 0.
  3. Set st to min(mx, ceil(bu−1)).
  4. Set en to min(mx, ceil(bu) − 1).
  5. Choose a uniform random integer in the closed interval [st, en], and return it.

Notes

License

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