2

Let’s call a sequence of functions $\{f_n\}_{n = 1}^\infty$ where $f_n: {0; 1}^n \to \mathbb{R}$ a Bernoulli estimator, if for any sequence of $\{X_i\}_{i = 1}^\infty$ of i.i.d. Bernoulli random variables the sequence of random variables $\{f_n(X_1, … , X_n)\}_{n = 1}^\infty$ converges in probability to $P(X_1 = 1)$.

Some notable Bernoulli estimators include:

$f_n(x_1, … , x_n) = \frac{\sum_{i = 1}^n x_i}{n}$, which is based on the Law of Large numbers

$f_n(x_1, … , x_n) = n^{\frac{1}{L_n(x_1, … , x_n)}}$, where $L_n(x_1, … , x_n)$ stands for the longest streak of $1$-s in $x_1, … , x_n$.

Suppose $BE$ is the set of all Bernoulli estimators. Let’s call a sequence $\{a_i\}_{i = 1}^\infty$ $p$-quasirandom iff $\forall \{f_n\}_{n = 1}^\infty \in BE$ $\lim_{n \to \infty} f_n(a_1, … , a_n) = p$.

Does there exist a $\mu$-recursive $p$-quasirandom sequence for some $p \in (0; 1)$?

I managed only to prove the following two facts:

A periodic sequence is never quasirandom.

Proof:

If $x_1, x_2, … $ is periodic, then $L_n(x_1, … , x_n)$ is bounded and thus $n^{\frac{1}{L_n(x_1, … , x_n)}}$ (which is a Bernoulli estimator) is unbounded.

If a $\mu$-recursive $p$-quasirandom sequence exists for some $p \in (0; 1)$, then it exists for $p = \frac{1}{2}$.

Proof:

It can be constructed through Von Neumann procedure

Chain Markov
  • 15,564
  • 6
  • 36
  • 116
  • 1
    In the definition of $p$-quasirandom, are you putting any restrictions on the ${f_n}_{n = 1}^\infty$'s considered besides requiring them to be Bernoulli estimators? For instance, are you requiring them to be computable or anything like that? If you don't put some additional restrictions, then there are no quasirandom sequences at all, let alone computable ones. – Patrick Lutz Dec 15 '19 at 18:00
  • @PatrickLutz, the proof that there are no quasirandom sequences at all will be the valid answer to my question. Feel free to post it as an answer. – Chain Markov Dec 15 '19 at 22:09

1 Answers1

1

Claim: There are no $p$-quasirandom sequences for any $p \in (0,1)$.

Proof: For the sake of contradiction, suppose that the sequence $\{a_i\}_{i=1}^\infty$ is a $p$-quasirandom sequence, where $0 < p < 1$. Obviously this sequence must contain infinitely many $1$'s. Let $i_1 < i_2 < i_3 < \ldots$ be a sequence of indices of $1$'s in the original sequence, i.e. $a_{i_j} = 1$ for each $j$. Now define a Bernoulli estimator $f_n$ as follows: $$ f_n(x_1, \ldots, x_n) = \frac{x_{i_1} + \ldots + x_{i_k}}{k} $$ where $k$ is the largest number such that $i_k \leq n$.

Basically $\{f_n\}$ looks at a fixed subsequence of whatever sequence it is given and tries to estimate the Bernoulli parameter based on that. Hopefully it is clear that $\{f_n\}$ is a Bernoulli estimator and that $\lim_{n\to\infty}f_n(a_1,\ldots,a_n) = 1$ so $\{a_i\}$ is not $p$-quasirandom.

If you change the definition of quasirandom to only allow computable Bernoulli estimators then there are quasirandom sequences, however there are no computable ones. The study of sequences with properties like this is known as "algorithmic randomness." I don't know if your definition of "quasirandom" is equivalent to any definition studied in algorithmic randomness, but it is at least close. For vastly more detail on this topic see the book Algorithmic Randomness and Complexity by Rod Downey and Denis Hirschfeldt, especially chapters 6 and 7.

Patrick Lutz
  • 875