Subject: Duplicate strings within parallel passages

From: Jim Deardorff <[email-masked]>
Date: Wed, 3 Sep 1997 14:45:02 -0700 (PDT)

I worked up some standard deviations for the exponential distributions
that the computer simulation can give; it was Vinton Dearing, I think, who 
urged that they be presented.

This was from the computer runs that simulated the frequency distribution
(FRQ in Tim's notation) of duplicate strings of identical words within the
many parallels of 1 Esdras and Ezra.  What this latter test case yielded,
by my tallies from the Septuagint, is:

  n=   1    2    3   4   5   6   7   8  9  10  11  12  13  14  15  16
FRQ=       232  114 44  36  24  12   6  1  1   1   1    1   0   0   0

The computer runs, for a probability of .577, on the average, that each 
successive word choice would be the same as in the parallel passage, gave:

  n =  1    2    3   4   5   6   7   8  9  10  11  12  13  14  15  16
 FRQ  339  162  97  67  39  27  10   5  5   5   0   2   1   1   0   0 
 FRQ  319  184  96  68  35  23  11  12  3   2   2   1   0   0   0   0  
 FRQ  325  200 107  63  39  21  13   8  4   0   0   1   0   0   0   0
 FRQ  336  177 119  65  36  21  10   7  3   4   0   1   1   0   0   0
       etc.
MEAN  322  187 108  62  37  21  12 7.2 4.1 2.3     .80 .39 .25 .13 .07
                                               1.4
FIT   322  186 107  62  36  21  12 6.9 4.0 2.3     .76 .44 .25 .15 .08
                                               1.3

STD   28   17 11.5 6.4     4.2 2.4 2.0 2.0 2.0 1.3     .52 .36
                       4.6                         .79

The mean was taken from the average of 320 runs, four of which are shown
for the four FRQ rows of data shown.  The standard deviation
(STD) was taken from only 16 runs (since I had to hand-calculate them).

For n=5, for example, STD is 4.6, so one expects that the FRQ you may get
from an independently translated data set of the length involved here for
n=5 has a 95% chance of lying between 27 and 45 (+- 2 std's from the
mean).  This indicates the difficulty of estimating the best fitting mean
curve to represent FRQ from any one real data set.  

The FIT is given by FRQ = 559 exp(-0.55n), where n is the number of words
in the duplicate-word string.  (The 0.55 value for b is related to the
.577 conditional-probability value by .577 = exp(-.55).)

The STD decreases with increasing n, along with the FRQ, but not quite as
fast. In fact, for n>12 STD becomes larger than the mean.  This is why I
couldn't give any firm estimate of the STDs from the Biblical cases, which
each had to be considered separately, except for what one can estimate by
fitting the data with a mean curve and noting the scatter of the
individual data points from the mean curve of FRQ(n), which I previously
called Y(I).  But if I were to simulate the Gospel cases, for which the
curve with b=-.45 fit their initial portions, I think I could end up with
pretty good estimates of their STDs also.

I think the STDs from the computer runs above are a little larger than for
the runs in which the probability of the translator's choosing the same
word as did the other translator (or drawing a black ball from the mixed
bag) was fixed at 0.577 rather than randomly fluctuating about 0.577. 

The point of it all is in noting if the data exhibits departures,
especially for large n, that are significantly in excess of random
variations, in which case it is due to purposeful variations by the
writer/scribe.  For the synoptic problem, where the assumption of
independence is often assumed between Matthew and Luke, this has clear
implications.  The one writer is assumed not to have used the other's
work, but the analysis may preclude this.  

For TC, it seems much less clear to me that this kind of analysis would
have any point, since there the goal had been complete dependence, not
independence -- the head scribe or lector usually didn't want independent
readings to crop up.  But when they did, it was sometimes due to
accidental mistakes and other times due to purposeful changes or attempted
corrections, all by different persons.  The question of the existence of
variants is not at issue, one knows they exist.  With the synoptic
problem, the question of whether Mt & Lk were independently written is
still at issue.  

I do need to assume that the TC of the past half century has been good
enough to allow analyses of duplicate-string FRQ distributions from the
Gospel parallels within N-S to represent the Gospels as they stood shortly
after their writing/translation.  The representation doesn't need to be
perfect, though, since the FRQ distributions will tend to be little
changed if both unsuspected variants and unsuspected harmonistic
corruptions exist and tend to counterbalance each other.

>From reading W. R. Farmer's work, the long strings of duplicate words
between Mt and Lk (not in Mark) were known to Herbert Marsh back in 1798,
causing him even then to postulate a verbal dependence between those two
gospels.  I wonder if Marsh used the Textus Receptus to deduce this, does
anyone know?  And would the TR be similar to N-S 27 in this respect? 

Jim Deardorff

---

From: Jim Deardorff <[email-masked]>
Date: Thu, 4 Sep 1997 15:55:54 -0700 (PDT)

I heard back from the statistician I had asked about this problem.  The
procedure for subjectively fitting data that looks like it falls off
exponentially with increasing n is pretty simple.  One first converts the
data (frequencies of occurrence in this case) into their (natural) 
logarithms, causing the converted data (y) to look roughly linear with
"n", which is called x; that is, y = a0 + a1 x.  Then one does the
least-squares analysis on that data, using the standard technique, as in
CRC's Standard Mathematical Tables, e.g., to determine a0 and a1.

The coefficient (a1) for the slope of the best fitting line is then the
exponential decline coefficient b.  The constant (a0), when exp is raised
to that power, gives the constant C, in FRQ(n) = C exp(-bn).

Using the data for n=4,5,....10 or 12 or so, my eyeball fit for the
1Esdras-Ezra duplicated-strings data had given: 

       FRQ(n) = 510 exp(-.55n).

Using the data for n=4,5...12, least squares gives FRQ = 528 exp(-.578n).

Using the data for n=4,5...13, least squares gives FRQ = 335 exp(-.510n).

It seems that one can't escape some subjectivity because you need to
decide just how far out in n to go in your analysis.  But in the
least-squares fitting, you can't include n so large that FRQ = 0, since
its logarithm is -infinity.

My paper describing the use of the duplicate-strings frequency
distributions in the Gospel priority problem is now in the electronic
Journal of Biblical Studies, http://web.infoave.net/~jwest/Articles.htm .

Jim Deardorff

---

From: "Robert B. Waltz" <[email-masked]>
Date: Thu, 4 Sep 1997 18:37:39 -0500 (CDT)

>CRC's Standard Mathematical Tables, e.g., to determine a0 and a1.

[ ... ]

>its logarithm is -infinity.

This isn't exactly "subjectivity." There is a difference between saying
"assume a solution of the form" and "I guess the answer is."

What you are doing is *assuming* an exponential distribution and
a particular value for n. Having achieved the results, you should
calculate a correlation coefficient. If this is close enough, you
can consider your fit adequate.(I should note that "close enough"
is a rather uncertain term; in biology, a CC of 50% is good, whereas
in Freshman Physics even the worst klutzes among us -- I being one
of them -- could achieve at least 95% CC. From where I sit, unfortunately,
I can't give you a fixed number -- but I won't be too impressed if you're
much short of 90%.)

If you do not get a good enough Correlation Coefficient, it's back
to the curve-fitting board. In this case you might try another n.
If that doesn't work, you'll need a different equation of fit.

Ultimately this is a sort of scientific method: Hypothesise, test,
hypothesise again. There is "guesswork" in forming the hypothesis,
there is none in testing it.

-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-*-

                        Robert B. Waltz
                     [email-masked]

Want more loudmouthed opinions about textual criticism?
Try my web page: http://www.skypoint.com/~waltzmn
(A site inspired by the Encyclopedia of NT Textual Criticism)

---

From: Jim Deardorff <[email-masked]>
Date: Thu, 4 Sep 1997 18:29:47 -0700 (PDT)

> "assume a solution of the form" and "I guess the answer is."

Robert, 

You may have misunderstood what n is here.  It's not anything subjective,
but is the independent coordinate which the frequency of occurrence is a
function of.  Large n defines the exponential tail of the frequency
distribution where the occurrence is either 1 (or sometimes 2) or 0.  The
curve fit should be in the region where n is smaller than that and the
number of occurrences for a given n is several or many.  The decision on
where n becomes too large is subjective, and depends partly upon the
scatter of the data. 

> a particular value for n.

So one doesn't assume anything for n.  But one does need to examine the
data and make your assessment on whether it looks linear, quadratic,
Gaussian, exponential or whatever before you try to curve-fit it.  I think
Tim Finney pointed this out a little while back.  One's guess on this
should make use of any other available information, such as an aymptotic
value that FRQ may approach as n becomes very large. 

> Having achieved the results, you should

I suspect that for the relatively small data sets available in the
Biblical-parallels duplicate-strings study, the results will behave more
like in biology than in physics.  

Jim Deardorff

---

From: "Robert B. Waltz" <[email-masked]>
Date: Thu, 4 Sep 1997 21:16:19 -0500 (CDT)

On Thu, 4 Sep 1997, Jim Deardorff <[email-masked]> wrote,
in part:

>function of.

Understood. But as you yourself point out, you can choose which
values for n you examine. But that's minor.

[ ... ]

Accepting that, what pre
>Gaussian, exponential or whatever before you try to curve-fit it.

Yes and (more to the point) NO!

That is, you look at that data, assume a solution, and try to make
it work. But the fact that you've assumed it doesn't make it right
or wrong. Try an exponential; fine. But if you're being thorough,
you should also try polynomials of various orders, and perhaps other
functions. The data may look exponential (as Tim's and mine both
did). So what? At values close to zero or multiples of pi over two,
trig functions mimic linear functions. Doesn't make them the same.

Look at your subject line: "Fitting the Exponential distribution
for duplicate strings." You can fit an exponential. You could,
for that matter, fit a Bessel Function. But unless you give us
a measure of goodness-of-fit, and a comparison with other fits,
we don't know whether that's really the correct form or not.

>I think

True, of course. But you must be willing to try other forms. That's
all I'm saying.

>like in biology than in physics.  

Probably true. But if you only get a 50% correlation coefficient, you
should at least try some other functions, e.g. polynomials (Tim's and
my data also look like they might fit a 1/x type curve after all).

BTW (this addressed to any innocent bystanders still reading this :-) --
let's be careful not to compare apples and oranges. Jim Deardorff's data
on repetitions and Tim's and mine on number of readings per variant
are completely different problems. The only thing we have in common
is that we're curve fitting. There is no reason to assume, e.g., that
the same sort of functions are used.

                        Robert B. Waltz
                     [email-masked]

---

From: Jim Deardorff <[email-masked]>
Date: Thu, 4 Sep 1997 22:11:57 -0700 (PDT)

Robert,

Polynomials are obviously no good, as they don't approach zero as n grows
large.  And I'm still interested in fitting them with a 2-parameter curve,
not 3 or 4 or more.  Trig functions also don't asymptotically
approach zero.

I did briefly look at FRQ = a/(b+n), fitting it at two points, but then an
intermediate point was a factor of 5 too small, and the approach to zero
way too slow.  Also tried FRQ = a/n^p, but it's too steep a curve at small
n.  If a 3rd parameter were added here, one might get a fit about as good
as with the exponential, but 2 is to be preferred over 3.

It's also important to me that the function be derivable through a
feasible underlying mechanism.  Such exists for the 2-parameter
exponential curve: the independent translator or editor, at each step in
his work after each duplicate string has by chance been initiated, has a
certain conditional probability of selecting as his next word either the
same next word as exists in the parallel text, or a different word.  This
leads to the exponential form, even when the conditional probability jumps
around in value from one word choice to another.  

> we don't know whether that's really the correct form or not.

Why don't you suggest a two-parameter function that has the proper
asymptotic limit for large n -- yet something other than an exponential,
and one that has a plausible, explainable mechanism underlying it.  Then I
would have something reasonable against which to compare the exponential. 

In the meantime, I could see what I can learn about goodness-of-fit.  When
speaking of correlations, I'm only used to correlating one variable
against another variable: obtaining the average product of their
fluctuations divided by the product of the two standard deviations. 
Something different is needed here, perhaps just comparing the rms of the
deviations of the FRQ data from the fitted curve #1 against the same
relative to fitted curve #2. 

Jim Deardorff

---

From: "Robert B. Waltz" <[email-masked]>
Date: Fri, 5 Sep 1997 08:30:22 -0500 (CDT)

On Thu, 4 Sep 1997, Jim Deardorff <[email-masked]> wrote:

>large.

Not quite true... a full-fledged polynomial is of the form

                 -3       -2       -1                 2      3
  f(x) = ... a  x   + a  x   + a  x   + a  + a x + a x  + a x....
              -3       -2       -1       0    1     2      3

And even *this* leaves out fractional or irrational exponents.

Of course, we prefer positive integral powers. They're so much
*easier*! But they set of all polynomials with positive integral
powers is only a subset of the set of all polynomials.

>And I'm still interested in fitting them with a 2-parameter curve,
>not 3 or 4 or more.

This is understandable, but it's not good math. If you are fitting
a function to data, the assumption is that there is an underlying
formula for the data that you are trying to fit. What if it isn't
something that can be fitted with two parameters. Assume, for the
sake of the argument, that the actual formula were

           -x     .5
  f(x) = ae   - bx   + c

How do you fit *that* with two parameters?

Note that I am not speaking of your particular distribution. With
only a few data points, it's hard to determine more than two (or
even one) parameter. But there is a principle involved here: You
cannot simply say "I believe the solution to be of this form"
and expect anyone to accept the results as having any validity.
(At least not without a really impressive correlation coefficient --
say 98%.)

>Trig functions also don't asymptotically
>approach zero.

True in the specific case, but I was speaking of the general case.

>as with the exponential, but 2 is to be preferred over 3.

I assume you were trying this for p an integer? How about trying a
square or cube root. (I don't know if this is better, but it sounds
like it might help.)

>around in value from one word choice to another.  

This sounds like hand-waving. Can you give a rigorous argument for
that?

BTW -- Be it noted that it think you're probably right and the
distribution *is* exponential. I am merely demanding an adequate
standard of proof.

>would have something reasonable against which to compare the exponential. 

BZZT! Why does it have to be explainable? The explanation for a function
is that it works! The data should be telling you how it behaves; you
shouldn't be telling it. The best fit is the best fit -- even if it's
a fourth-order Bessel function.

Now I'll admit that I've been throwing out the past letters in this
conversation; remember, my interest is not in the answers but in the
rigour with which they are computed. But if you want to send me (off-list)
the data and a graph, I will see if I can offer alternate suggestions.
I already offered one, which would generalize to
              a
  f(x) =  ----------
                p/q
           (b+x)

For this form, I would just try values for p/q (e.g. 1/2, 1/3, 2/3).
Then you only have two parameters to find.

>relative to fitted curve #2. 

That's how you calculate a correlation coefficient: Comparing the
actual data against the calculated value. It should be in your
CRC handbook. (I hope, because my books on the subject are too deeply
buried for me to find easily. :-)

                        Robert B. Waltz
                     [email-masked]

---

From: Jim Deardorff <[email-masked]>
Date: Fri, 5 Sep 1997 17:21:42 -0700 (PDT)

> square or cube root. (I don't know if this is better, but it sounds

Hello Robert,

No, I fit FRQ = a/n^p to two points on a smooth curve passing through the
data points, and so naturally obtained non-integer p, as in
FRQ=23091/n^4.19 . 

> This sounds like hand-waving. Can you give a rigorous argument for

My apologies to others who may be tiring of the repeated explanation!

It should be clear that each word the independent translator chooses as
his next word following a string of duplicate words will either be the
same or will be different from what exists in the parallel text unknown to
him.  Is that where the difficulty lies? 

This point is what brings in the analogy of drawing a red or a
black ball from a bag of mixed balls, or of tossing a coin, say. The toss 
will turn out either heads or tails.  

Whether a mathematician or not, you know that the odds of getting three
heads in a row is .5x.5x.5.  And so the odds of getting four heads in a
row is .5x.5x.5x.5, and so on.  So this gives a geometric progression, or
exponential decline, for the frequency distribution of length of strings
of heads.

Now if the coin is lopsidedly weighted and the odds are 0.6, say, for
getting "heads," one replaces .5 above with .6.  Obviously you still have
a geometric progression for the frequency distribution of successive
heads, the odds then being .6x.6x.6 for three heads in a row, .6x.6x.6x.6
for four heads (n=4), etc.  The exponential form for it is FRQ(n) = C
exp(-.51n), since FRQ(n+1)/FRQ(n) = exp(-.51) = 0.6. 

Next we generalize the thought experiment to have the increased order of
complexity of the case of the translator.  Instead of always tossing the
same weighted coin, we have dozens of coins of different weightings for
heads versus tails.  For each toss of the coin, the coin tosser chooses
one of these coins at random and keeps track of how many strings of how
many heads in a row he gets.  Whatever the average weighting of all the
coins, he will still obtain an exponential distribution for the frequency
of these strings of successive heads.  And if the average weighting
turns out to be 0.6 as before, then FRQ(n) = C exp(-.51n) as before, upon
averaging together the results of a very large number of tosses.

With the weighting randomized, I don't know how to mathematically prove
the resulting frequency distribution will still be exponential. Perhaps a
mathematician wouldn't have too much trouble proving this.  But I had to
resort to the computer, where I had 320 runs calculated in which the
weighting was allowed to vary randomly with each toss of the coin (or each
draw of a red or black ball from a mixed bag of balls).  Each run
contained several thousand tosses of the coin, to correspond to the amount
of data in the test case I had analyzed (1Esdras-Ezra parallels).  The
resulting distribution was indeed exponential, with only a tiny scatter
from the mean, as you may have noted in a posting of mine a day ago.  The
computer program kept track of the average weighting that occurred in each
of the 320 sets of runs, which was designed to be 0.577 on the average,
but ranged from .570 to .584.  In each individual coin toss, however, this
weighting randomly varied between .254 and 0.900. 

Is it difficult to see the analogy here with the independent translator of
Aramaic text into Greek, and comparing his word choices with that of
another translator? I don't see why.  His each successive word is either
identical or not with what the other translator chose.  The conditional
probabilities involved in each word choice range widely, depending upon
where you are in the phrase/clause/sentence/word-string.  How then could
this lead to any distribution of duplicate word strings other than the
exponential, unless the one translator made use of the other's work in
some systematic fashion?  

> a fourth-order Bessel function.

Science and scholastics just don't work that way, since we want to answer
how & why as well as what.  Recall during the 1700s when meteorites were
noticed by a few astronomers as being unearthly. They couldn't convince
their peers of it for about a century, since it wasn't considered
desirable for dirty rocks to fall from the heavens.  It wasn't until after
some astronomer could postulate a mechanism whereby such chunks of rock
could be out there and occasionally fall to earth that meteorites could be
accepted as real.  Other examples abound. 

> Then you only have two parameters to find.

Sorry, but you haven't set forth an explainable mechanism how that form
could come into being, and since it is 3-parameter instead of 2, it can't
compete with the 2-parameter exponential.  

But if you're interested in spending more time on this, it would really be
valuable to me if you could mathematically prove that the exponential
distribution arises for the distribution of successive identical yes-or-no
outcomes when the probability of each outcome is allowed to vary randomly
rather than be fixed.

In this problem you have duplicate strings involved where the individual
progression of terms in each string is A, A*ri, A*ri*rj, A*ri*rj*rk,
A*ri*rj*rk*rl, etc., where the r's lie between 0 and 1 but themselves have
some mean intermediate value, and (subscripts) i,j,k.. denote the randomly
different values of r.  After averaging or summing over a huge number of
such strings, you examine the ratio Avg(ri*rj*rk*rl)/Avg(ri*rj*rk), and
see if you can show that in the limit it becomes just Avg(ri) = Avg (rj)..
If so, it produces an exponential distribution.

Jim Deardorff

---

From: "Robert B. Waltz" <[email-masked]>
Date: Sat, 6 Sep 1997 08:56:58 -0500 (CDT)

>data points, and so naturally obtained non-integer p, as in

OK, you were using p as a parameter. Sorry. I would be interested in
knowing whether this was a better or worse fit than the other.

To me, though, the result above suggests (and note that this is
NOT PROOF, merely a suggestion for something you should fit) that
you actually have an inverse fourth power result. In other words,
you should try fitting

            a
  f(x) = -------
         (b+x)^4

with a and b being parameters.

***

I was hoping to keep the next part brief. It didn't work too well.
I'm tempted to say that we should drop this whole subject. It
seems I can't convince Jim Deardorff, since he dislikes rigour,
and he can't convince me, because I refuse to accept non-rigorous
arguments.

>him.  Is that where the difficulty lies? 

Yes, I suppose. You're saying that things "ought" to work this way.
I say, so what? If you want to use math, and scientific technique,
the word "ought" must not be in your vocabulary. You must proof
rigorously, or else you must give statistical evidence. (Usually,
in the sciences, you give statistical evidence.) Your argument
may be strong, but it is not rigorous. And your statistical evidence
is insufficient.

I am sorry if I am not saying this gently. But you must either offer
rigour, or disclaim scientific accuracy for your results.

I'm going to make comments below on the rigour of your arguments.
I don't know how you're going to feel about this. But I hope they'll
show why I don't approve of your approach.

>will turn out either heads or tails.  

You are beginning your argument with an analogy. That is very
dangerous. Analogies serve to help us understand what we see.
They do not explain it.

>exponential decline, for the frequency distribution of length of strings

No problems so far.

>exp(-.51n), since FRQ(n+1)/FRQ(n) = exp(-.51) = 0.6.

OK, I'll freely concede you lost me, though this may just be a
difference in terminology. Assume the frequency for an event
(for convenience, call it a "head") is .6. Then, as you say, the
table of probabilities is

event     probability
-----     -----------
1 head    .6^1 = .6
2 heads   .6^2 = .36
3 heads   .6^3 = .216
4 heads   .6^3 = .1296
etc.

This is the *binomial* distribution; it has nothing to do with
exponentials.

(BTW -- you're having a minor problem with significant digits here.
How many are you carrying? It doesn't matter in an analogy, but it
will matter in an actual proof.) 

[ ... ]

>mathematician wouldn't have too much trouble proving this.

On the contrary, I don't have any idea how you can prove it. In fact,
I was inclined to believe it until I read this argument. I *never*
trust an argument by analogy.

You are not called upon to prove your argument rigorously. You
can validate it statistically with enough data. *If* you analyse
the data sufficiently.

>But I had to

If the result was indeed exponential (and *where* is your measure of
goodness-of-fit?) then your analogy to a binomial distribution goes
out the window.

[ ... ]

>how & why as well as what.

Are you kidding me? Scholastics may work that way (I'm tempted to say
that's their problem, not mine). But in sceince, the question is
*what* is the behavior. Explanations come afterward, if at all.
Often there is no explanation (go ahead, ask someone to explain the
"why" of quantum physics. The only answer you'll get is "because
that's the way it is").

Let me give you an analogy. The Lorenz-Fitzgerald contraction and
and the constant velocity of light are observed phenomena. We accept
those based on collected data.

To explain these, we have Einstein's theories of relativity. Note the
word *theories*. Relativity is a theory -- a very solid one; it has
passed experimental test. But it is *only a theory.*

>Recall during the 1700s when meteorites were

Agreed in part. Certainly it was a long time before meteors were
accepted. But the "mechanism," as you describe it, was worked out
by Newton, centuries before meteors were accepted. The fact that
even scientists are prejudiced against certain things doesn't mean
that they *should be* prejudiced. Don't turn the faults of the few
into an excuse for being non-rigorous!

>compete with the 2-parameter exponential.  

I don't need to offer an explanation; if it works, it WORKS. If your
way works (as demonstrated by a measure of goodness-of-fit) then it
works, no explanations needed. But I repeat, for the I don't know
how many-eth time, that you CANNOT START WITH THE EXPLANATION. You
start with the data.

Also, you have no basis whatsoever for rejecting three-parameter
fits. (Though I didn't ask you to fit three; I asked you to fit
two given certain assumptions.) You may not need a third parameter
(if your two-parameter fit is good enough). But first you must
demonstrate the goodness of the fit. (Yes, I'm repeating myself.
You can't repeat the truth too often.)

>outcomes when the probability of each outcome is allowed to vary randomly

Since I don't necessarily believe it, how can I prove it? Sorry.

[ ... ]

My apologies to the list for dragging this on so long. I believe I've
said all I can possibly say on the subject. Unless someone has a
purely mathematical query, I will drop this thread.

                        Robert B. Waltz
                     [email-masked]

---

From: Jim Deardorff <[email-masked]>
Date: Sat, 6 Sep 1997 12:35:33 -0700 (PDT)

Robert,

The relative deviations of my 1Esdras-Ezra (Greek) test data from
FRQ=23091/n^4.19 are:

 n   4     5     6     7     8     9
   -37%  +32%  +89%  +81%  +58%  -57%

The relative deviations of the same data from FRQ=510 exp(-.55n) are:
 n   4     5     6     7     8     9  
   -22%  +10%  +28%  +11%  -4%   -72%

The exponential is clearly better, though I had previously overstated the
degree to which a/n^b fails.  The degree of scatter from the exponential
is of the magnitude expected from examination of the std's in the
synthetic computer runs that simulated the test data.  If deviations of
this general magnitude had *not* been observed, one would have to be
especially suspicious of the data analysis.

If I had utilized the n=4 data point, rather than a larger value, in
fitting FRQ=a/n^b, the large positive deviations for n=5 to 8 would have
been larger still for the a/n^b trial.

These two different, but parallel, Greek texts were translated into
English by two different translators, the latter of whom apparently did
not make use of, or plagiarize, the former's work.  (Details in my paper.) 
So I regard it also as an independent translation against which to test
the exponential hypothesis. It contains over twice as much
duplicate-string data, since there are fewer degrees of freedom for
different word arrangements, choices and spellings in English than in
Greek. So the scatter in those test results for the English, which I'll
show below, is considerably less than for the Greek (above). Consequently,
the non-zero data below extend to larger values of n:

  n       4    5    6    7    8    9   10   11   12   13   14   15   16
Obs FRQ  122  63   47   34   13    14   3   6    2    1    4    2    0 

Here, Obs FRQ is the freq of observed duplicate strings of length n words
each appearing in my analysis of the English 1Esdras-Ezra data.  Below,
Exp is the (eyeball) fit given by FRQ=900 exp(-0.5n), and 
a/n^b is the fit to the data points at n=4 and n=12 (FRQ=21846/n^3.74).
(These two data points appear to lie close to any mean curve faired
through them all.)

  n       4    5    6    7    8    9   10   11   12   13   14   15   16
Exp     122  73.9 44.8 27.2 16.5 10.0  6.1  3.7 2.23 1.35 0.82 0.50 0.30
% Dev    0%  -15% +5%  +25% -21% +40% -51% +62% -10% -26% +388 +300 -100%

a/b^n   122  53.1 26.8 15.1  9.2  5.9  4.0  2.8 2.0  1.49 1.13 0.87 0.69
% Dev    0%  +19% +75% +125 +41% +137 -25% +114  0%  -33% +254 +130 -100%

Again the exponential is much better, until n is large enough that the
scatter of the data exceeds the mean, around n=13.  Since a/b^n approaches
infinity as n --> 0, this causes its curve to be too steep at small n,
whereas the exponential is often pretty good at n = 3 and 2, also, since
its n=0 value is well behaved.

> with a and b being parameters.

This should indeed do better.  However, you can't really say it's a
2-parameter family, since if the power of 3 gave better results than 4,
then you'd use 3, etc. So the power parameter was a third parameter.

 

 
> table of probabilities is

In still simpler terms it's also a geometric progression, and thus has
everything to do with exponentials.  Let "r" generalize the "0.6" above. 
The geometric progression is the 2nd line below:

n  =  0    1      2      3      4    
      a    ar    ar^2   ar^3   ar^4   etc.
      1   .6     .36    .216  .1296

If a = 1 and r = 0.6, this last line (above) is what you've listed above.
The exponential that gives these values is 

            Exp =   a*exp[n(ln r)].

(For those whose college math is rusty, "ln" stands for natural
logarithm.  We use ^3 to represent the cube power, etc.)

For r = 0.6 and a=1, the above exponential becomes exp(-.5108n).  If one
raises e=2.718282 to the power -.5108n, one gets the 2nd line below:

n  =  0    1      2      3      4    
Exp:  1   .6     .36    .216  .1296  

which of course is the same geometric progression.  If this somehow
slipped by you, then I can understand some of the confusion.  But this is
why we've been calling the estimated fit to the frequency distribution a
geometric progression or exponential, interchangeably. 

In ending this little exchange, I'd like to remind other readers that I'm
still interested in hearing opinions on the extent to which the lengthy
strings of duplicate words between parallels of Luke and Matthew (Q
verses) could reflect the way the Gospels appeared within a decade or so
of their appearance, as opposed to the possibility that the long strings
were generated through scribes' harmonistic corruptions and/or the manner
in which the majority text was deduced.  Whether the duplicate strings
obey an exponential or some other frequency distribution was just a
digression.

Jim Deardorff

---

From: "Robert B. Waltz" <[email-masked]>
Date: Sat, 6 Sep 1997 15:09:11 -0500 (CDT)

Only one, very brief, mathematical comment which should illustrate
why Deardorff and I are not getting anywhere.

>            Exp =   a*exp[n(ln r)].

Did it really never occur to you to *simplify* that expression
and observe that it's

            n
   Exp = a*r     ?????

Try it, you'll like it. And there is no use of logarithms or the
exponential function.

                        Robert B. Waltz
                     [email-masked]

---

From: "Ronald L. Minton" <[email-masked]>
Date: Mon, 8 Sep 1997 16:28:13 -0500 (CDT)

I am trying to determine the number of extant NT mss.
Aland TEXT sometimes (as I do) uses 5400 as a round number.  I understand
why his p. 83 totals only 4102; there are some mss. that are Gospels/Paul,
etc.
Aland TEXT, p. 81 indicates 5197
Aland TEXT, p. 82 indicates 5269, but he may be omitting some combinations
which are really only one mss. or some others I am unaware of.
I do not have it readily available, but I have read that in 1994, Aland
_Kurzgefasste Liste der grieschen Handschriften des Neuen
Testaments_, gives a total of 5656.  Is this correct and is it accurate?  
Thanks for any help anyone can offer.

--
Prof. Ron Minton: [email-masked]   W (417)268-6053  H 833-9581
Baptist Bible Graduate School 628 E. Kearney St. Springfield, MO 65803

---

From: "Robert B. Waltz" <[email-masked]>
Date: Mon, 8 Sep 1997 17:38:18 -0500 (CDT)

>Thanks for any help anyone can offer.

I'm not the one to answer this, but I must ask for a point of
clarification.

Are you asking for the *actual* number of manuscripts, or the
*nominal* number? The two aren't the same.

For example, the uncials 070, 0110, 0124, 0178, 0179, 0180, 0190,
0191, 0193, 0202 are all (regarded as) one manuscript, but they
have separate numbers. This works both ways: For a while 0121
was applied to two different manuscripts (one of them now merged
with 0243). And a series of ostraca are merged under one number.

And does one count the harmony 0212 as a manuscript?

Before one can say how many manuscripts there are, one must
determine the meaning of the phrase "how many manuscripts." :-)

And, of course, the number is constantly changing.

If it were me, though, I'd go with the number in the _Kurzgefasste
Liste_. It is, if nothing else, the most current.

                        Robert B. Waltz
                     [email-masked]