Working through an exercise in Probability (the question can be found in Lamperti).

Let $X_1,\dots$ be independent Bernoulli random variables with $\mathbb{P}(X_i=1) = p$ and $\mathbb{P}(X_i=0)=1-p$. Let $$Z = \sum_{k=1}^\infty X_k 2^{-k}$$ and $F_p(x) = \mathbb{P}(Z \le x)$.

Show that if $p = 1/2$ then $F_{1/2}(x) = x$, $0<x<1$. That is, $Z \sim U(0,1)$ and if $p\in(0,1)\setminus\{1/2\}$, then $F_p$ is continuous, strictly increasing and singular. That is, $F'_p(x) = 0$ almost-everywhere with respect to the Lebesgue measure.

We did the first part as an assignment question, where you approach it from a partial sum up to $n$ Bernoullis and take the characteristic function of that and then let $n\to\infty$. I'm not sure how to approach the singular part, because it was given as an exercise before we covered characteristic functions. Lampert hints that the distribution is singular if and only if the corresponding measure puts probability 1 on a set of Lebesgue measure 0.

Any hints or help would be appreciated! Thanks


1 Answer 1


This might be complicated to describe rigorously, using elementary notions, but the underlying concept is simple: almost all real numbers between $0$ and $1$, when written in binary, have equally many zeros and ones. Because $Z$ assigns zero probability to such numbers--it favors an imbalance of zeros and ones when $p\ne 1-p$--it must have a singular distribution.

The statement in bold needs a clear definition. I am going to sneak up on the definition by developing a series of simpler definitions and notations. But before we proceed, let's look at some of these distributions.

Figure 1: CDF for p=1/6, 1/2, 3/4, and 199/200

The colored curves plot the CDFs. They are all self-similar in the sense suggested by the underlying rectangles: The curve in the lower left rectangle is obtained by taking the entire curve and multiplying all x-values by $1/2$ and multiplying all y=values by $1-p$. That is, the curve is shrunk towards the origin $(0,0)$ by factors of $1/2$ and $1-p$. Likewise, the curve in the upper right rectangle is obtained by shrinking the entire curve towards the coordinate $(1,1)$ by factors of $1$ and $p$. (This description completely determines each curve!)

Here are the major steps in the analysis.

  1. Let $a_1, a_2, \ldots, a_n$ be any sequence of zeros and ones. It represents a binary number in the interval $[0,1)$ by "putting a binary point in front of it." As a matter of notation, and to define precisely what this means, let $$[a_1\,a_2\,\cdots\,a_n] = a_1 2^{-1} + a_2 2^{-2} + \cdots + a_n 2^{-n}.$$

  2. Since we will be studying the proportions of ones within binary representations, let's invent a notation for that. Define $$\rho([a_1\,a_2\,\cdots\,a_n]) = \frac{\#\{i \in \{1,2,\ldots, n\}\mid a_i=1\}}{n}$$ to be that proportion.

  3. We cannot analyze all the infinitely many binary digits of any real number $x$ all at once: we have to look at finitely many of them and take limits. To this end, extend the function $\rho$ to entire intervals of real numbers determined by their initial sequences. That is, if $$[a_1\,a_2\,\cdots\,a_n] \le x \lt [a_1\,a_2\,\cdots\,a_n] + 2^{-n},$$ then define $$\rho_n(x) = \rho([a_1\,a_2\,\cdots\,a_n]).$$ The functions $\rho_n$ are defined on the entire interval $[0,1)$.

    If you like, you could also define the $\rho_n(x)$ numerically in terms of the binary expansion of the integer part of $2^n x$ (which is how the next figure was produced), but I thought the preceding description might give a clearer idea of the underlying concept.

    ![Figure 1: graphs of rho_n, n=2,5,8,11,14

  4. The key idea is to collect all numbers that look, based on the initial (finite!) part of their binary representation, like they have a given proportion of ones. We should allow for that proportion to be in a narrow range, say between $\alpha$ and $\beta$ where $0 \le \alpha \le \beta \le 1$. To this end, define the (Borel measurable) set $R$ by $$R(n,\alpha,\beta) = \{x\in [0,1)\mid \alpha \le \rho_n(x) \le \beta\}.$$ That is, the proportion of ones within the first $n$ binary digits of $x$ lies between $\alpha$ and $\beta$.

    In the preceding figure, you may visualize $R(n,\alpha,\beta)$ by referring to the graph of $\rho_n$, drawing the interval $[\alpha,\beta]$ on the vertical axis, identifying all parts of the graph whose heights are within that interval, and looking down at the horizontal ($x$) axis to see which numbers correspond to those heights: that (highly disconnected) set is $R(n,\alpha,\beta)$.

  5. At this point we may leave the world of binary representations and focus on the digits themselves. When those first $n$ binary digits are generated by $X_1, X_2, \ldots, X_n$, each having a chance $p$ of being $1$, then the proportion of ones has a Binomial$(n,p)$ distribution. Thus, unless $X_{n+1}=X_{n+2}=\cdots = 1$, $$\rho_n(Z) = \rho[X_1\,X_2\,\cdots\,X_n] \sim \frac{1}{n}\text{Binomial}(n,p).$$ We may exclude all values of $Z$ that end in all ones because (as is easy to see and show) their chance of occurring is zero.

  6. Consequently, writing $F_{n,p}$ for the Binomial$(n,p)$ distribution function, $$\mathbb{P}(Z \in R(n,\alpha,\beta)) = F_{n,p}(n\beta) - F_{n,p}(n\alpha).$$

  7. Now suppose $p\ne 1/2$. Pick $\alpha,\beta$ such that $p\in[\alpha,\beta]$ but $1/2\notin[\alpha,\beta]$. It is easy to prove (in many ways, ranging from the Weak Law of Large Numbers to Chebyshev's Inequality) that as $n$ grows large, the value of $F_{n,p}(n\beta)-F_{n,p}(n\alpha)$ approaches $1$: that is, the Binomial distribution becomes concentrated around its expected value $p$. This means as $n$ grows large, all the probability of $Z$ becomes concentrated within the (very complicated union of intervals) $R(n,\alpha,\beta)$.

  8. However, the result for the case $p=1/2$ (for which the distribution of $Z$ is Lebesgue measure) shows that the Lebesgue measure of $R(n,\alpha,\beta)$ approaches zero. Together with (7) we conclude that for $p\ne 1/2$, the distribution of $Z$ is singular (with respect to Lebesgue measure).

  9. In fact, the same argument shows that the various distributions of $Z$ for any two different $p$ are all mutually singular: as $n$ grows large, the probability of one concentrates within a region that has vanishingly small probability for the other.

Example R code to plot one of these distribution functions follows. Function f generates the sequence of integers $n\rho_n(0), n\rho_n(2^{-n}), n\rho_n((2)2^{-n}), \ldots, n\rho_n((2^n-1)2^{-n}).$ These count the number of $X_i$ for $i=1,2,\ldots, n$ that are equal to $1$. It then computes the associated probability and accumulates it to approximate the CDF of $Z$ across an equally-spaced set of $2^n$ points in the interval $[0,1]$. This process emulates the self-similarity transformation described at the beginning of this answer.

# The distribution function, approximated to an accuracy of 2^(-n) in Z.
f <- function(n, p=1/2) {
  y <- rep(0, 2^n)
  if (n > 0) 
    for (i in 1:n) {
      j <- 1:(2^(i-1)); y[j + 2^(i-1)] <- y[j] + 1
  cumsum(p^y * (1-p)^(n-y))
# Example: plot a distribution for given `p`.
n <- 11
p <- 1/6
z <- c(0:(2^n) / 2^n)
Probability <- c(0, f(n, p))
plot(z, Probability, type="S")

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