Re: Pitman CSI Formula

On Aug 7, 12:20 pm, hersheyh <hershe...@xxxxxxxxx> wrote:

< snip >

My formula, unlike yours *is* measuring a probability. And it is not
any more convoluted.

Again, the probability that a randomly chosen sequence from total
sequence space will be within hd sequence differences from a reference
sequence is, mathematically,

Sum (i=0,1,2,...hd) n!/(n-hd)! hd!)/2^n

Basically it is the sum of the number of sequences between 0
mismatches and hd mismatches (the n!/(n-hd)! hd!) term) divided by the
total universe of all possible sequences of length n. That gives me
the fraction of the sequence universe that would be that close or
closer to the reference sequence. *That* and only *that* equation
will tell me the probability that my test sequence could have been
this close to the reference sequence by chance alone. I would have
had to decide what level of probability I would consider to be
unlikely to be due to chance beforehand. That is pretty basic
probability.

By your formula, a Hamming Distance of n/2 would have only a 50%
chance of occurring "by chance". Yet, in reality, a randomly produced
string would be expected to be at a HD of n/2, plus or minus a few,
the majority of the time. Also, by your formula, the maximum possible
HD would have 100% odds of being produced by random generation. This
is clearly wrong since the maximum possible HD has the same odds of
being produced by random generation as a sequence with the minimum
possible HD of 1.

You see, the actual statistical distribution for a randomly generated
string is maximum around HD of n/2 and drops off from there on either
side of n/2 exponentially - right in line with the increase in the CSI
value as per my formula.

Sean Pitman
www.DetectingDesign.com

.

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