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Say I have a database with 10000 protein sequences of the same species in it. I was to study if mutations at positions i and j are correlated. I can decompose each sequence into strings of binary variables

11 - i is mutated, j is mutated
10 - i is mutated, j is not mutated
01 - i is not mutated, j is mutated
00 - i is not mutated, j is not mutated

I can sum these up and place them into a contingency table

        j
      1 | 0
    ---------
  1 | a | b |
i------------
  0 | c | d |
    ---------

If I normalize this contingency table by N=a+b+c+d, I get joint frequencies

             j
         1   |    0
    -------------------
  1 | f(1,1) | f(1,0) |
i----------------------
  0 | f(0,1) | f(0,0) |
    -------------------

Are these joint proportions (e.g. f(1,1)) probabilities? If not, under what assumptions can I say that these joint proportions approximate the probabilities of two mutations occuring together in the population? If I was given a database of 50,000 sequences and I made two contingency tables from random sample of 10,000, I assume the two tables, and thus the proportions calculated from them, would be different. So I am asking, under what circumstances are the joint proportions from the contingency table an appropriate approximation of the probabilities of seeing these events in the true population?

I know that I can use similarity coefficients like Jaccard, Dice, or Ochiai to determine whether the positions are correlated, but I am confused as to the difference between probability distributions and the normalized contingency table cells.

To me, the normalized contingency table cells look like bivariate marginal distributions (if my protein is 100 amino acids long, then I have marginalized the total joint probability distribution over the other 98 amino acids). Then the normalized row/column sums look like univariate marginal distributions. Is this wrong? Is it that these normalized contingency table cells are maximum likelihood estimates to the joint probability distributions?

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  • $\begingroup$ Sorry, to me your question and terminology (e.g. covary in a correlated way is unclear. Those joint proportions ("frequencies") can of cause be called probabilities. f(1,1) = a/N is called Russell and Rao similarity coefficient. $\endgroup$ – ttnphns Jul 1 '13 at 21:44
  • $\begingroup$ Hi, I edited my question to make it clearer, hopefully. $\endgroup$ – wflynny Jul 1 '13 at 21:58
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This depends on your definition of probability. This has been a very heated debate in statistics. My take on it now is that the debate has become more polite but is still unresolved and is probably unresolvable. You can find an introduction on this theme in: Ian Hacking (2001) An Introduction to Probability and Inductive Logic. Cambridge: Cambridge University Press.

On a less philosophical and more pragmatic level I have no problem with designating those proportions as (frequentist) estimates of probabilities.

If I were to describe the association in that cross-tabulation I would probably compute an odds ratio $\frac{a d}{b c}$ or equivalently $\frac{f(1,1)f(0,0)}{f(1,0)f(0,1)}$, but other measures of association in a cross-tabulation exist.

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  • 1
    $\begingroup$ This is not really a question about frequency or non-frequency definitions of probability - it is a question about finite population inference from a selected sample. $\endgroup$ – probabilityislogic Jul 2 '13 at 13:14
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There are two considerations for approximating the probability using these counts - bias and variance.

  • The first is that your sample of 10,000 sequences should be representative of the general population. This ensures that any particular characteristic/feature of these sequences, in this case the combination of mutations of i and j, will occur in the same proportion in your sample in expectation, as in the population, i.e., there is no bias in your estimate.
  • The second is that your sample should be large enough for you to obtain certain confidence intervals for your estimate of probabilities. The sample size required will depend on the size of the confidence interval. For example, a 10% confidence interval will require around 100 samples, while a 1% confidence interval will require around 10,000 samples. Larger sample sizes will reduce the variance of your estimate.

See: http://en.wikipedia.org/wiki/Sample_size_determination, for the second point.

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