We let $\alpha = \alpha_1\alpha_2\alpha_3\ldots$ be an infinite random sequence (under the uniform measure) where $\alpha_i$ may be $1$ or $0$, and then define the boolean function $B_k$:

$$ B_k(\alpha_1\ldots\alpha_k) = \begin{cases} 1 \text{ if at least } \lceil k/2 \rceil \text{ of its inputs are } 1 \\ 0 \text{ otherwise} \end{cases} $$

Then we define two sequences:

$$B_3(\alpha_1\alpha_2\alpha_3)B_3(\alpha_4\alpha_5\alpha_6)B_3(\alpha_7\alpha_8\alpha_9)\ldots$$ $$B_4(\alpha_1\alpha_2\alpha_3\alpha_4)B_4(\alpha_5\alpha_6\alpha_7\alpha_8)B_4(\alpha_9\alpha_{10}\alpha_{11}\alpha_{12})\ldots$$

Which one of these two sequences is (algorithmically) random, and why? I should note that apparently there is an obvious measure-theoretic fact that gives away which one is not random.

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    $\begingroup$ If $k=2n+1$ then $P(B_k = 1) < P(B_k = 0)$. Isn't that sufficient, informally? $\endgroup$
    – OJFord
    Commented Jun 10, 2014 at 18:34
  • $\begingroup$ maybe a related question $\endgroup$
    – Nikos M.
    Commented Jun 10, 2014 at 19:05
  • $\begingroup$ Isnt each $B_k$ sequence the same as the partial sum sequence of $a_i$'s (or more correctly the difference of 2 partial sum sequences)? $\endgroup$
    – Nikos M.
    Commented Jun 10, 2014 at 19:08
  • $\begingroup$ i can also give another type of answer in sigal procesing terms. Each $B_k$ sequence acts as lowpass filter (filtering away high frequencies) as a result because random noise has especially high frequencies, each $B_k$ should be progresively less random (eventually equal a sequence of 1's). $\endgroup$
    – Nikos M.
    Commented Jun 10, 2014 at 19:11
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    $\begingroup$ @Newb According to the Wikipedia article you link, there are multiple possibilities and the default is a convention of the field. You should be careful using such conventions here without clarification -- not every reader is a domain expert. $\endgroup$
    – Raphael
    Commented Jun 11, 2014 at 9:04

1 Answer 1


The second sequence is not random. Let $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ be random, iid Bernoulli $1/2$ random variables. Let $\beta = B_4(\alpha_1 \alpha_2 \alpha_3 \alpha_4)$.

What is the distribution of the random variable $\beta$? Answer: $\beta=1$ if at least two of the $\alpha$'s are $1$, so $\Pr[\beta=1] = 11/16$.

In other words, $\beta$ is biased towards $1$. It follows that the second sequence is not algorithmically random: it is a set of independent Bernoulli random variables with $p=11/16$, i.e., the outcome of an infinite sequence of tosses of a biased coin.

  • $\begingroup$ All right! That makes perfect sense. Thanks. Do you have any reasoning as to why the first sequence is random? $\endgroup$
    – Newb
    Commented Jun 11, 2014 at 0:54
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    $\begingroup$ By contrast, it would be because $B_3$ is "fair": there are 4 3-bit sequences that $B_3$ maps to 1 and 4 3-bit sequences that it maps to 0, so assuming the original sequence is random, $Pr[B=1] = Pr[B=0] = \frac12$. $\endgroup$
    – gardenhead
    Commented Jun 11, 2014 at 1:18
  • $\begingroup$ i dont agree with this, it just states that the random variable $\beta$ does not have equal p1 and p2 probabilties. Why a random variable with P1 = 1/3 and P0 = 2/3, not be considered random? Why r.v's with a different distribution are not random (a gaussian rv is not random, it is biased towards the mean)? $\endgroup$
    – Nikos M.
    Commented Jun 12, 2014 at 8:15
  • $\begingroup$ In fact i am posting a question (soon?) addresing the relaton between probabilistic and algorithmic randomness (per the other question i linked in the comments) $\endgroup$
    – Nikos M.
    Commented Jun 12, 2014 at 8:21
  • $\begingroup$ I think you're right, while both are random in a sense, algorithmically random is not the same as probabilistically random. As I am not an expert on algorithmic randomness, that's really I can say. I would be interested for an elaboration of these points as well. $\endgroup$
    – gardenhead
    Commented Jun 17, 2014 at 21:03

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