## Recursive randomness of reals: summing a random decreasing sequence

In a comment to Recursive randomness of integers Rahul pointed out a continuous version of the same problem: pick ${x_0}$ uniformly in ${[0,1]}$, then ${x_1}$ uniformly in ${[0,x_0]}$, then ${x_2}$ uniformly in ${[0, x_1]}$, etc. What is the distribution of the sum ${S=\sum_{n=0}^\infty x_n}$?

The continuous version turns out to be easier to analyse. To begin with, it’s equivalent to picking uniformly distributed, independent ${y_n\in [0,1]}$ and letting ${x_n = y_0y_1\cdots y_n}$. Then the sum is

${\displaystyle y_0+y_0y_1 + y_0y_1y_2 + \cdots }$

which can be written as

${\displaystyle y_0(1+y_1(1+y_2(1+\cdots )))}$

So, ${S}$ is a stationary point of the random process ${X\mapsto (X+1)U}$ where ${U}$ is uniformly distributed in ${[0,1]}$. Simply put, ${S}$ and ${(S+1)U}$ have the same distribution. This yields the value of ${E[S]}$ in a much simpler way than in the previous post:

${E[S]=E[(S+1)U] = (E[S] + 1) E[U] = (E[S] + 1)/2}$

hence ${E[S]=1}$.

We also get an equation for the cumulative distribution function ${C(t) = P[S\le t]}$. Indeed,

${\displaystyle P[S\le t] = P[(S+1)U \le t] = P[S \le t/U-1]}$

The latter probability is ${\int_0^1 P[S\le t/u-1]\,du = \int_0^1 C(t/u-1)\,du}$. Conclusion: ${C(t) = \int_0^1 C(t/u-1)\,du}$. Differentiate to get an equation for the probability density function ${p(t)}$, namely ${p(t) = \int_0^1 p(t/u-1)\,du/u}$. It’s convenient to change the variable of integration to ${v = t/u-1}$, which leads to

${\displaystyle p(t) = \int_{t-1}^\infty p(v)\,\frac{dv}{v+1}}$

Another differentiation turns the integral equation into a delay differential equation,

${\displaystyle p'(t) = - \frac{p(t-1)}{t}}$

Looks pretty simple, doesn’t it? Since the density is zero for negative arguments, it is constant on ${[0,1]}$. This constant, which I’ll denote ${\gamma}$, is ${\int_{0}^\infty p(v)\,\frac{dv}{v+1}}$, or simply ${E[1/(S+1)]}$. I couldn’t get an analytic formula for ${\gamma}$. My attempt was ${E[1/(S+1)] = \sum_{n=0}^\infty (-1)^n M_n}$ where ${M_n=E[S^n]}$ are the moments of ${S}$. The moments can be computed recursively using ${E[S^n] = E[(S+1)^n]E[U^n]}$, which yields

${\displaystyle M_n=\frac{1}{n} \sum_{k=0}^{n-1} \binom{n}{k}M_k}$

The first few moments, starting with ${M_0}$, are 1, 1, 3/2, 17/6, 19/3, 81/5, 8351/80… Unfortunately the series ${\sum_{n=0}^\infty (-1)^n M_n}$ diverges, so this approach seems doomed. Numerically ${\gamma \approx 0.5614}$ which is not far from the Euler-Mascheroni constant, hence the choice of notation.

On the interval (1,2) we have ${p'(t) = -\gamma/t}$, hence

${p(t) = \gamma(1-\log t)}$ for ${1 \le t \le 2}$.

The DDE gets harder to integrate after that… on the interval ${[2,3]}$ the solution already involves the dilogarithm (Spence’s function):

${\displaystyle p(t) = \gamma(1+\pi^2/12 - \log t + \log(t-1)\log t + \mathrm{Spence}\,(t))}$

following SciPy’s convention for Spence. This is as far as I went… but here is an experimental confirmation of the formulas obtained so far (link to full size image).

To generate a sample from distribution S, I begin with a bunch of zeros and repeat “add 1, multiply by U[0,1]” many times. That’s it.

import numpy as np
import matplotlib.pyplot as plt
trials = 10000000
terms = 10000
x = np.zeros(shape=(trials,))
for _ in range(terms):
np.multiply(x+1, np.random.uniform(size=(trials,)), out=x)
_ = plt.hist(x, bins=1000, normed=True)
plt.show()

I still want to know the exact value of ${\gamma}$… after all, it’s also the probability that the sum of our random decreasing sequence is less than 1.

### Update

The constant I called “${\gamma}$” is in fact ${\exp(-\gamma)}$ where ${\gamma}$ is indeed Euler’s constant… This is what I learned from the Inverse Symbolic Calculator after solving the DDE (with initial value 1) numerically, and calculating the integral of the solution. From there, it did not take long to find that

Oh well. At least I practiced solving delay differential equations in Python. There is no built-in method in SciPy for that, and although there are some modules for DDE out there, I decided to roll my own. The logic is straightforward: solve the ODE on an interval of length 1, then build an interpolating spline out of the numeric solution and use it as the right hand side in the ODE, repeat. I used Romberg’s method for integrating the solution; the integration is done separately on each interval [k, k+1] because of the lack of smoothness at the integers.

import numpy as np
from scipy.integrate import odeint, romb
from scipy.interpolate import interp1d
numpoints = 2**12 + 1
solution = [lambda x: 1]
integrals = [1]
for k in range(1, 15):
y0 = solution[k-1](k)
t = np.linspace(k, k+1, numpoints)
rhs = lambda y, x: -solution[k-1](np.clip(x-1, k-1, k))/x
y = odeint(rhs, y0, t, atol=1e-15, rtol=1e-13).squeeze()
solution.append(interp1d(t, y, kind='cubic', assume_sorted=True))
integrals.append(romb(y, dx=1/(numpoints-1)))
total_integral = sum(integrals)
print("{:.15f}".format(1/total_integral))


As a byproduct, the program found the probabilities of the random sum being in each integer interval:

• 56.15% in [0,1]
• 34.46% in [1,2]
• 8.19% in [2,3]
• 1.1% in [3,4]
• 0.1% in [4,5]
• less than 0.01% chance of being greater than 5

## Recursive randomness of integers

Entering a string such as “random number 0 to 7” into Google search brings up a neat random number generator. For now, it supports only uniform probability distributions over integers. That’s still enough to play a little game.

Pick a positive number, such as 7. Then pick a number at random between 0 and 7 (integers, with equal probability); for example, 5. Then pick a number between 0 and 5, perhaps 2… repeat indefinitely. When we reach 0, the game becomes really boring, so that is a good place to stop. Ignoring the initial non-random number, we got a random non-increasing sequence such as 5, 2, 1, 1, 0. The sum of this one is 9… how are these sums distributed?

Let’s call the initial number A and the sum S. The simplest case is A=1, when S is the number of returns to 1 until the process hits 0. Since each return to 1 has probability 1/2, we get the following geometric distribution

 Sum Probability 0 1/2 1 1/4 2 1/8 3 1/16 k 1/2k+1

When starting with A=2, things are already more complicated: for one thing, the probability mass function is no longer decreasing, with P[S=2] being greater than P[S=1]. The histogram shows the counts obtained after 2,000,000 trials with A=2.

The probability mass function is still not too hard to compute: let’s say b is the number of times the process arrives at 2, then the sum is 2b + the result with A=1. So we end up convolving two geometric distributions, one of which is supported on even integers: hence the bias toward even sums.

 Sum Probability 0 1/3 1 1/6 2 5/36 3 7/72 k ((4/3)[k/2]+1-1)/2k

For large k, the ratio P[S=k+2]/P[s=k] is asymptotic to (4/3)/4 = 1/3, which means that the tail of the distribution is approximately geometric with the ratio of ${1/\sqrt{3}}$.

I did not feel like computing exact distribution for larger A, resorting to simulations. Here is A=10 (ignore the little bump at the end, an artifact of truncation):

There are three distinct features: P[S=0] is much higher than the rest; the distribution is flat (with a bias toward even, which is diminishing) until about S=n, and after that it looks geometric. Let’s see what we can say for a general starting value A.

Perhaps surprisingly, the expected value E[S] is exactly A. To see this, consider that we are dealing with a Markov chain with states 0,1,…,A. The transition probabilities from n to any number 0,…,n are 1/(n+1). Ignoring the terminal state 0, which does not contribute to the sum, we get the following kind of transition matrix (the case A=4 shown):

${\displaystyle M = \begin{pmatrix}1/2 & 0 & 0 & 0 \\ 1/3 & 1/3 & 0 & 0 \\ 1/4 & 1/4 & 1/4 & 0 \\ 1/5 & 1/5 & 1/5 & 1/5\end{pmatrix} }$

The initial state is a vector such as ${v = (0,0,0,1)}$. So ${vM^j}$ is the state after j steps. The expected value contributed by the j-th step is ${vM^jw}$ where ${w = (1,2,3,4)^T}$ is the weight vector. So, the expected value of the sum is

${\displaystyle \sum_{j=1}^\infty vM^jw = v\left(\sum_{j=1}^\infty M^j\right)w = vM(I-M)^{-1}w}$

It turns out that the matrix ${M(I-M)^{-1}}$ has a simple form, strongly resembling M itself.

${\displaystyle M(I-M)^{-1} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 1 & 1/2 & 0 & 0 \\ 1 & 1/2 & 1/3 & 0 \\ 1 & 1/2 & 1/3 & 1/4 \end{pmatrix} }$

Left multiplication by v extracts the bottom row of this matrix, and we are left with a dot product of the form ${(1,1/2,1/3,1/4)\cdot (1,2,3,4) = 1 + 1 + 1 + 1 = 4 }$. Neat.

What else can we say? The median is less than A, which is no surprise given the long tail on the right. Also, P[S=0] = 1/(A+1) since the only way to have zero sum is to hit 0 at once. A more interesting question is: what is the limit of the distribution of T = S/A as A tends to infinity? Here is the histogram of 2,000,000 trials with A=50.

It looks like the distribution of T tends to a limit, which has constant density until 1 (so, until A before rescaling) and decays exponentially after that. Writing the supposed probability density function as ${f(t) = c}$ for ${0\le t\le 1}$, ${f(t) = c\exp(k(1-t))}$ for ${t > 1}$, and using the fact that the expected value of T is 1, we arrive at ${c = 2-\sqrt{2} \approx 0.586}$ and ${k=\sqrt{2}}$. This is a pretty good approximation in some aspects: the median of this distribution is ${1/(2c)}$, suggesting that the median of S is around ${n/(4-2\sqrt{2})}$ which is in reasonable agreement with experiment. But the histogram for A=1000 still has a significant deviation from the exponential curve, indicating that the supposedly geometric part of T isn’t really geometric:

One can express S as a sum of several independent geometric random variables, but the number of summands grows quadratically in A, and I didn’t get any useful asymptotics from this. What is the true limiting distribution of S/A, if it’s not the red curve above?

## Normalizing to zero mean or median

Pick two random numbers ${x_1,x_2}$ from the interval ${[0,1]}$; independent, uniformly distributed. Normalize them to have mean zero, which simply means subtracting their mean ${(x_1+x_2)/2}$ from each. Repeat many times. Plot the histogram of all numbers obtained in the process.

No surprise here. In effect this is the distribution of ${Y=(X_1-X_2)/2}$ with ${X_1,X_2}$ independent and uniformly distributed over ${[0,1]}$. The probability density function of ${Y}$ is found via convolution, and the convolution of ${\chi_{[0,1]}}$ with itself is a triangular function.

Repeat the same with four numbers ${x_1,\dots,x_4}$, again subtracting the mean. Now the distribution looks vaguely bell-shaped.

With ten numbers or more, the distribution is not so bell-shaped anymore: the top is too flat.

The mean now follows an approximately normal distribution, but the fact that it’s subtracted from uniformly distributed ${x_1,\dots,x_{10}}$ amounts to convolving the Gaussian with ${\chi_{[0,1]}}$. Hence the flattened top.

What if we use the median instead of the mean? With two numbers there is no difference: the median is the same as the mean. With four there is.

That’s an odd-looking distribution, with convex curves on both sides of a pointy maximum. And with ${10}$ points it becomes even more strange.

Scilab code:

k = 10
A = rand(200000,k)
A = A - median(A,'c').*.ones(1,k)
histplot(100,A(:))

## B-splines and probability

If one picks two real numbers ${X_1,X_2}$ from the interval ${[0,1]}$ (independent, uniformly distributed), their sum ${S_2=X_1+X_2}$ has the triangular distribution.

The sum ${S_3}$ of three such numbers has a differentiable probability density function:

And the density of ${S_4=X_1+X_2+X_3+X_4}$ is smoother still: the p.d.f. has two
continuous derivatives.

As the number of summands increases, these distributions converge to normal if they are translated and scaled properly. But I am not going to do that. Let’s keep the number of summands to four at most.

The p.d.f. of ${S_n}$ is a piecewise polynomial of degree ${n-1}$. Indeed, for ${S_1=X_1}$ the density is piecewise constant, and the formula

$\displaystyle S_n(x) = \int_{x-1}^x S_{n-1}(t)\,dt$

provides the inductive step.

For each ${n}$, the translated copies of function ${S_n}$ form a partition of unity:

$\displaystyle \sum_{k\in\mathbb Z}S_n(x-k)\equiv 1$

The integral recurrence relation gives an easy proof of this:

$\displaystyle \sum_{k\in\mathbb Z}\int_{x-k-1}^{x-k} S_{n-1}(t)\,dt = \int_{\mathbb R} S_{n-1}(t)\,dt = 1$

And here is the picture for the quadratic case:

A partition of unity can be used to approximate functions by piecewise polynomials: just multiply each partition element by the value of the function at the center of the corresponding interval, and add the results.

Doing this with ${S_2}$ amounts to piecewise linear interpolation: the original function ${f(x)=e^{-x/2}}$ is in blue, the weighted sum of hat functions in red.

With ${S_4}$ we get a smooth curve.

Unlike interpolating splines, this curve does not attempt to pass through the given points exactly. However, it has several advantages over interpolating splines:

• Is easier to calculate; no linear system to solve;
• Yields positive function for positive data;
• Yields monotone function for monotone data

## Tossing a continuous coin

To generate a random number ${X}$ uniformly distributed on the interval ${[0,1]}$, one can keep tossing a fair coin, record the outcomes as an infinite sequence ${(d_k)}$ of 0s and 1s, and let ${X=\sum_{k=1}^\infty 2^{-k} d_k}$. Here is a histogram of ${10^6}$ samples from the uniform distribution… nothing to see here, except maybe an incidental interference pattern.

Let’s note that ${X=\frac12 (d_1+x_1)}$ where ${X_1=\sum_{k=1}^\infty 2^{-k} d_{k+1}}$ has the same distribution as ${X}$ itself, and ${d_1}$ is independent of ${X_1}$. This has an implication for the (constant) probability density function of ${X}$:

$\displaystyle p(x) = p(2x) + p(2x-1)$

because ${2 p(2x)}$ is the p.d.f. of ${\frac12 X_1}$ and ${2p(2x-1)}$ is the p.d.f. of ${\frac12(1+X_1)}$. Simply put, ${p}$ is equal to the convolution of the rescaled function ${2p(2x)}$ with the discrete measure ${\frac12(\delta_0+\delta_1)}$.

Let’s iterate the above construction by letting each ${d_k}$ be uniformly distributed on ${[0,1]}$ instead of being constrained to the endpoints. This is like tossing a “continuous fair coin”. Here is a histogram of ${10^6}$ samples of ${X=\sum_{k=1}^\infty 2^{-k} d_k}$; predictably, with more averaging the numbers gravitate toward the middle.

This is not a normal distribution; the top is too flat. The plot was made with this Scilab code, putting n samples put into b buckets:

n = 1e6
b = 200
z = zeros(1,n)
for i = 1:10
z = z + rand(1,n)/2^i
end
c = histplot(b,z)


If this plot too jagged, look at the cumulative distribution function instead:

It took just more line of code: plot(linspace(0,1,b),cumsum(c)/sum(c))

Compare the two plots: the c.d.f. looks very similar to the left half of the p.d.f. It turns out, they are identical up to scaling.

Let’s see what is going on here. As before, ${X=\frac12 (d_1+X_1)}$ where ${X_1=\sum_{k=1}^\infty 2^{-k} d_{k+1}}$ has the same distribution as ${X}$ itself, and the summands ${d_1,X_1}$ are independent. But now that ${d_1}$ is uniform, the implication for the p.d.f of ${X}$ is different:

$\displaystyle p(x) = \int_0^{1} 2p(2x-t)\,dt$

This is a direct relation between ${p}$ and its antiderivative. Incidentally, if shows that ${p}$ is infinitely differentiable because the right hand side always has one more derivative than the left hand side.

To state the self-similarity property of ${X}$ in the cleanest way possible, one introduces the cumulative distribution function ${F}$ (the Fabius function) and extends it beyond ${[0,1]}$ by alternating even and odd reflections across the right endpoint. The resulting function satisfies the delay-differential equation ${F\,'(x)=2F(2x)}$: the derivative is a rescaled copy of the function itself.

Since ${F}$ vanishes at the even integers, it follows that at every dyadic rational, all but finitely many derivatives of ${F}$ are zero. The Taylor expansion at such points is a polynomial, while ${F}$ itself is not. Thus, ${F}$ is nowhere analytic despite being everywhere ${C^\infty}$.

This was, in fact, the motivation for J. Fabius to introduce this construction in 1966 paper Probabilistic Example of a Nowhere Analytic ${C^\infty}$-Function.