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Suppose I am working with sequences like this (monotonic but not strictly monotonic where each member of the alphabet repeats an identical number of times):

$x = \{1,1,1,1,2,2,2,2,3,3,3,3\}$, then $|x| = N = 12$ and the alphabet is $\Sigma = \{1,2,3\}$

Ultimately I am interested in the contiguous identical sub-sequences in $x$ and a corrupted version of $x$ let's call it $x'$, and let's say it has a value:

$x' = \{1,1,2,2,2,3,3,3,3\}$.

With this I am hoping to measure the goodness-of-fit between the corrupted version and the reference version $x$ by looking at the distribution over the identical sub-sequences, instead of raw counts of the alphabet, since the sub-sequences carry information on the monotonic nature of both sequences. That is my intution at least (looking at the distribution of the alphabet would simply yield a uniform distribution, and so would loose the monotonic repeating trend).

As an example, suppose we simply just count the identical $n-$tuples (let's call these $a_n$) when $n=2$, then we simply count all the identical tuples which:

  1. are exactly of length 2 i.e. $|a_2| = 2$.
  2. have all symbols in $a_2$ identical i.e. $a[0]=a[1]$.

There are three 2-tuples (i.e. three 2-tuples which look like so: $[1,1]$) in the first sub-sequence of ones, three 2-tuples in the second sub-sequence of twos and the same for the 2-tuples in the last sub-sequence of threes. In total there are 9 2-tuples in $x$.

Thus the probability of finding an identical 2-tuple in $x$ is:

$$P[a_2] = \frac{1}{9}$$

(not entirely sure this is right, I am primarily interested in counting the number of identical $n-$tuples).

Hence, now I would like to extend this so that I get a PMF for all the available tuples i.e. for $n \in \{2,3,4\}$ as $4$ is the maximum identical tuple we can consider given what $x$ looks like.

The histogram for this problem:

$n=2 : \#9$ $n=3 : \#6$ $n=4 : \#3$

Possible solution:

$f$ : identical sub-sequence count

$f(n) \triangleq N - (n-1)|\Sigma|$

Is it as simple as this? Though of course, if you construct the PMF from this, it does not sum to 1.

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  • $\begingroup$ Where does probability come into the picture? What is being randomly chosen? $\endgroup$ – Mike Earnest Jan 18 at 21:07
  • $\begingroup$ I may have formulated this the wrong way; I am trying to find the distribution over all possible identical $n-$tuples in $x$ for $n\geq2$. $\endgroup$ – Astrid Jan 18 at 21:11
  • $\begingroup$ Well, your formula for $f(n)$ is correct. If you want a probability distribution, you can set $p(n) = f(n)/(\sum_{i=2}^{N/|\Sigma|}f(i))$. The interpretation is that you put all of the possible identical tuples in a bag, pull one out, and observe its length. $\endgroup$ – Mike Earnest Jan 18 at 21:23
  • $\begingroup$ Hmm... Yeah I don't think that I am formulating this very well, since that does not sound very intelligent. In your bag example, it does not make sense since a 2-tuple is much more common than a 4-tuple so should have more weight. But letting only the normalising constant take care of that does not seem like the right way to go about it. $\endgroup$ – Astrid Jan 18 at 21:24
  • $\begingroup$ Maybe give some more context? Why do you need this PMF? $\endgroup$ – Mike Earnest Jan 18 at 21:25

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