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Imagine a memoryless source that outputs 0's and 1's with probabilities $P_X(0)$ and $P_X(1)$. For example, $P_{X^2}(00)=P_X(0)P_X(0)$.

How would you calculate the probability that the sequence $010$ is present in an $n$-length binary sequence?

What I have thought so far is that,

$$P[010 \text{ is in an } n\text{-length sequence}]=(n-3)P_X(0)^{\#0}P_X(1)^{\#1}$$

I am sure that I have to multiply the probabilities $P_X(0)$ and $P_X(1)$ by $n-3$, because I need to take into consideration all the possible combinations in which $010$ can appear (e.g. $\{010...x\}$,$\{x010...x\}$,etc.). But I am not sure about the number of zeros $\#0$ and the number of ones $\#1$.

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  • $\begingroup$ This is not clear. Are you considering the universe of binary sequences generated by, say, the repeated tosses of a (possibly unfair) coin? Something else? $\endgroup$
    – lulu
    Jun 2, 2021 at 17:15
  • $\begingroup$ You seem to have a lot of double counting problems. If the sequence starts $01010$ aren't you counting it twice? $\endgroup$
    – saulspatz
    Jun 2, 2021 at 17:19
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    $\begingroup$ To your question: You'll need to work recursively to avoid double counting. I suggest starting from the fact that every sufficiently long "good" sequence must end in one of $1,00, 110$. $\endgroup$
    – lulu
    Jun 2, 2021 at 17:22
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    $\begingroup$ @BrianMoehring My comment was pretty vague. In the given problem, I'd divide the good sequences into $4$ types, according to whether they end in $11,01,00$ or $10$. I haven't written it out, but I think that's good enough...after all, the only banned action is to add a $0$ to the type ending in $01$. For longer 'forbidden' blocks, you'd need to consider more endings. $\endgroup$
    – lulu
    Jun 2, 2021 at 17:46
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    $\begingroup$ Should note that the posted solution, from @Onir, follows a more or less similar methodolgy to the one I am sketching. $\endgroup$
    – lulu
    Jun 2, 2021 at 17:52

1 Answer 1

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Let $a(00,n), a(01,n), a(10,n), a(11,n)$ be the probability we get a sequence that doesn't contain $010$ and end in each of the finishes.

We get:

$a(00,n+1)=p(a(00,n) + a(10,0))$

$a(01,n+1) = (1-p)(a(00,n)+a(10,n))$

$a(10,n+1) = pa(11,n)$

$a(11,n+1) = (1-p)(a(01,n) + a(11,n))$

We can write this as:

$\begin{pmatrix} p & 0 & p & 0 \\ 1-p & 0 & 1-p & 0 \\ 0 & 0 & 0 & p \\ 0 & 1-p & 0 & 1-p \\ \end{pmatrix} \begin{pmatrix} a(00,n) \\ a(01,n) \\ a(10,n)\\ a(11,n) \end{pmatrix} = \begin{pmatrix} a(00,n+1) \\ a(01,n+1) \\ a(10,n+1)\\ a(11,n+1) \end{pmatrix} $

When $n=2$ the values are $(p^2,p(1-p),(1-p)p,(1-p)^2)$. Hence we have:

$\begin{pmatrix} p & 0 & p & 0 \\ 1-p & 0 & 1-p & 0 \\ 0 & 0 & 0 & p \\ 0 & 1-p & 0 & 1-p \\ \end{pmatrix}^{n-2} \begin{pmatrix} p^2 \\ p(1-p)\\ (1-p)p\\ (1-p)^2 \end{pmatrix} = \begin{pmatrix} a(00,n) \\ a(01,n) \\ a(10,n)\\ a(11,n) \end{pmatrix} $

If the matrix happens to be diagonalizable you can get explicit formulas, even if it isn't you can expect to put it in a good form. You can also use exponentiation by squaring for rapid computations.

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  • $\begingroup$ Oh, there is $p$ and $1-p$ ? I need to change the transition matrix then. $\endgroup$
    – Asinomás
    Jun 2, 2021 at 17:50
  • $\begingroup$ "When $n=2$ all the values are $1$"? $\endgroup$
    – saulspatz
    Jun 2, 2021 at 19:27
  • $\begingroup$ Oh I forgot to remove that part. $\endgroup$
    – Asinomás
    Jun 2, 2021 at 19:28
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    $\begingroup$ When $n=2$, don't we have $a_2(0,0)=p^2,\ a_2(0,1)=p(1-p),$ etc? $\endgroup$
    – saulspatz
    Jun 2, 2021 at 19:32
  • $\begingroup$ Yes, thank you. $\endgroup$
    – Asinomás
    Jun 2, 2021 at 19:36

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