Carrier's numbers and math in OHJ

[Bernard Muller]

to Peter,

Does he arrive at the correct result, given the inputs, or does he not? It's a simple question.

He doesn't

Cordially, Bernard

[Peter Kirby]

to Peter,

Does he arrive at the correct result, given the inputs, or does he not? It's a simple question.

He doesn't

Using the same inputs, what do you believe is the mathematically correct result?

[Bernard Muller]

to Peter,

Using the same inputs, what do you believe is the mathematically correct result?

For sure, I would use the addition law of probability, or the sum rule for the consequent odds:

Another important property is:
P(A \cup B) = P(A) + P(B) - P(A \cap B).
This is called the addition law of probability, or the sum rule. That is, the probability that A or B will happen is the sum of the probabilities that A will happen and that B will happen, minus the probability that both A and B will happen.
Note: the math sign for \cup looks like a "U" and the math sign for \cap looks like a "U" reversed.

I am not going to make calculations (but you can, if you want). That would be a waste of time considering I would be channelled into having to deal with garbage input data and dubious categories.
I have some details here:
viewtopic.php?f=3&t=1317&p=30320&hilit=hebrews#p30320

Cordially, Bernard

[Peter Kirby]

to Peter,

Using the same inputs, what do you believe is the mathematically correct result?

For sure, I would use the addition law of probability, or the sum rule for the consequent odds:

Another important property is:
P(A \cup B) = P(A) + P(B) - P(A \cap B).
This is called the addition law of probability, or the sum rule. That is, the probability that A or B will happen is the sum of the probabilities that A will happen and that B will happen, minus the probability that both A and B will happen.
Note: the math sign for \cup looks like a "U" and the math sign for \cap looks like a "U" reversed.

I am not going to make calculations (but you can, if you want). That would be a waste of time considering I would be channelled into having to deal with garbage input data and dubious categories.
I have some details here:
viewtopic.php?f=3&t=1317&p=30320&hilit=hebrews#p30320

Cordially, Bernard

So this stuff...

But even if we have 4 cases in favor of h, each one rated at only 30 %, the total will come to a probability of 76 %.

Yikes... This is just absurd. We've gone round in a circle, and your understanding is none improved for the trip.

Proverbs 26:11
Like a dog that returns to its vomit Is a fool who repeats his folly.

I was far too optimistic, and I must have overestimated your potential for comprehension.

By all means, keep up the good work though. Publish it somewhere. Whatever.

[Peter Kirby]

For the record, I guess:

Carrier's Inputs - a fortiori ("upper bound")

P( h ) = 0.33333
P( ~h ) = 0.66667

P( e₁ | h ) = 0.4608
P( e₂ | h ) = 0.72
P( e₃ | h ) = 1
P( e₄ | h ) = 1

P( e₁ | ~h ) = 1
P( e₂ | ~h ) = 1
P( e₃ | ~h ) = 1
P( e4 | ~h ) = .34722

e = e₁, e₂, e₃, e₄ [a definition of "e" with "," for intersection (AND)]

Unstated assumption - conditional independence

Concept explained here:

http://pages.cs.wisc.edu/~dyer/cs540/no ... ainty.html
http://www.inf.ed.ac.uk/teaching/course ... -bayes.pdf
http://cs.wellesley.edu/~anderson/writi ... -bayes.pdf



This is an assumption, and I know that Carrier can be criticized for it (and I have done so), since it is left implicit.

The general form of the assumption of conditional independence is:

P(X,Y|Z) = P(X|Z)P(Y|Z)

The assumptions here are:

P( e₁, e₂, e₃, e₄ | h ) = P ( e₁ | h ) * P( e₂ | h ) * P( e₃ | h ) * P( e₄ | h )
P( e₁, e₂, e₃, e₄ | ~h ) = P ( e₁ | ~h ) * P( e₂ | ~h ) * P( e₃ | ~h ) * P( e₄ | ~h )

Bayesian Calculation



And its corollary:



P( h | e ) =
P( h | e₁, e₂, e₃, e₄ ) =
P( e₁, e₂, e₃, e₄ | h ) * P ( h ) / ( P( e₁, e₂, e₃, e₄ | h ) * P ( h ) + P( e₁, e₂, e₃, e₄ | ~h ) * P ( ~h ) )

Now:

P( e₁, e₂, e₃, e₄ | h ) = P ( e₁ | h ) * P( e₂ | h ) * P( e₃ | h ) * P( e₄ | h ) = 0.4608 * 0.72 * 1 * 1 = 0.331776
P( e₁, e₂, e₃, e₄ | ~h ) = P ( e₁ | ~h ) * P( e₂ | ~h ) * P( e₃ | ~h ) * P( e₄ | ~h ) = 1 * 1 * 1 * .34722 = 0.34722

So:

P( h | e ) = 0.331776 * 0.33333 / ( 0.331776 * 0.33333 + 0.34722 * 0.66667 ) = 0.110592 / ( 0.110592 + 0.231814 ) = 0.32298

(With a slight difference from rounding error.) Carrier's own approximation to the third decimal place reaches the same result of 32.3%.

[Bernard Muller]

to Peter,

Bernard wrote:

But even if we have 4 cases in favor of h, each one rated at only 30 %, the total will come to a probability of 76 %.

Yikes... This is just absurd. We've gone round in a circle, and your understanding is none improved for the trip.

Play with dices. You'll get to the same result. OK, dices have 6 sides. let's say that 1 and 6 are TRUE. That's 33.33333% to be TRUE.
Roll four dices together many times: 80% of the time, one of the four dices will come with a 1 or a 6.
That's according to the addition law of probability, or the sum rule.

OK, let's say you trust someone at 50% to tell the truth on a particular happening.
(let say he/she saw the suspect attacking the victim but it was from a long distance away, so the 50%)
You trust another person at 50% to tell the truth on the same matter.
(ditto)
These two persons do not know each other.
What are the chance of knowing the truth from these two persons. I say 0.50 + (0.50 x 0.50) = 0.75 => probability = 75%
Carrier would say: the odds for each is 1/1. (1/1 x 1/1) = 1/1 => probability = 50%

If we add a third person with 20% of saying the truth on the same thing (same circumstance but has his vision seriously deficient, so the 20%).
For me, that adds 5% (0.25 x 0.20) to the 75% for probability of 80%.
For Carrier 1/1 x 1/4 = 1/4 => probability of 20%

Cordially, Bernard

[Peter Kirby]

to Peter,

Bernard wrote:

But even if we have 4 cases in favor of h, each one rated at only 30 %, the total will come to a probability of 76 %.

Yikes... This is just absurd. We've gone round in a circle, and your understanding is none improved for the trip.

Play with dices. You'll get to the same result. OK, dices have 6 sides. let's say that 1 and 6 are TRUE. That's 33.33333% to be TRUE.
Roll four dices together many times: 80% of the time, one of the four dices will come with a 1 or a 6.
That's according to the addition law of probability, or the sum rule.

OK, let's say you trust someone at 50% to tell the truth on a particular happening.
(let say he/she saw the suspect attacking the victim but it was from a long distance away, so the 50%)
You trust another person at 50% to tell the truth on the same matter.
(ditto)
These two persons do not know each other.
What are the chance of knowing the truth from these two persons. I say 0.50 + (0.50 x 0.50) = 0.75 => probability = 75%
Carrier would say: the odds for each is 1/1. (1/1 x 1/1) = 1/1 => probability = 50%

If we add a third person with 20% of saying the truth on the same thing (same circumstance but has his vision seriously deficient, so the 20%).
For me, that adds 5% (0.25 x 0.20) to the 75% for probability of 80%.
For Carrier 1/1 x 1/4 = 1/4 => probability of 20%

Cordially, Bernard

h = testimony is accurate, e₁ = first witness, e₂ = second witness, e₃ = third witness

P( e₁ | h ) = 1
P( e₁ | ~h ) = 0.5

P( e2 | h ) = 1
P( e₂ | ~h ) = 0.5

P( e₃ | h ) = 1
P( e₃ | ~h ) = 0.8

P( h ) = 0.5
P( ~h ) = 0.5

An assumption that testimonies e1, e₂, and e₃ are conditionally independent of each other given h and given ~h:

P( e₁, e₂, e₃ | h ) = P( e₁ | h ) * P( e₂ | h ) * P( e₃ | h )
P( e₁, e₂, e₃ | ~h ) = P( e₁ | ~h ) * P( e₂ | ~h ) * P( e₃ | ~h )

Bayes' Theorem:



P( h | e₁, e₂, e₃ ) =
P( e₁, e₂, e₃ | h ) * P( h ) / ( P( e₁, e₂, e₃ | h ) * P( h ) + P( e₁, e₂, e₃ | ~h ) * P( ~h ) )

P( e₁, e₂, e₃ | h ) = P( e₁ | h ) * P( e₂ | h ) * P( e₃ | h ) = 1 * 1 * 1 = 1
P( e₁, e₂, e₃ | ~h ) = P( e₁ | ~h ) * P( e₂ | ~h ) * P( e₃ | ~h ) = 0.5 * 0.5 * 0.8 = 0.2

P( h | e₁, e₂, e₃ ) =
1 * 0.5 / ( 1 * 0.5 + 0.2 * 0.5 ) = 0.5 / 0.6 = 0.83333

The posterior probability of the hypothesis h, given the assumptions above, is thus 5 out of 6 or 83.333%.

Carrier would say: the odds for each is 1/1. (1/1 x 1/1) = 1/1 => probability = 50%

For Carrier 1/1 x 1/4 = 1/4 => probability of 20%

No, a Bayesian analysis would have arrived at 5 out of 6 or 83.333%.

[Bernard Muller]

to Peter,

P( h ) = 0.33333
P( ~h ) = 0.66667

P( e1 | h ) = 0.4608
P( e2 | h ) = 0.72
P( e3 | h ) = 1
P( e4 | h ) = 1

P( e1 | ~h ) = 1
P( e2 | ~h ) = 1
P( e3 | ~h ) = 1
P( e4 | ~h ) = .34722

The problem I see here is if P( h ) = 0.33333 then P( ~h ) = 0.66667
In my example about the two witnesses, if they are deemed 50% right, that does not mean they are deemed wrong at 50%. The other 50% is just inconclusiveness, incertitude, not that these two witnesses saw others than the suspect and the victim.

Cordially, Bernard

[Peter Kirby]

In my example about the two witnesses, if they are deemed 50% right, that does not mean they are deemed wrong at 50%. The other 50% is just inconclusiveness, incertitude, not that these two witnesses saw others than the suspect and the victim.

Regarding:

P( e₁ | h ) = 1
P( e₁ | ~h ) = 0.5

The first statement is saying that, of the first testimony, it is 100% likely to be affirming h on the hypothesis h.

The second statement is saying that, of the first testimony, it is 50% likely to be affirming h (saying it happened) on the hypothesis ~h (that it did not happen).

These are conditional probabilities, and actually nothing is said here regarding whether h happened or not happened. Nothing at all.

Rather, what we are talking about is how likely this testimony would be the way it is, on the respective hypothesis h and ~h.

More simply, P( e₁ | ~h ) is the likelihood that the first witness could be giving false testimony (somehow, anyhow).

We could of course use a different estimate, if we felt that a 50% likelihood of forming a false testimony is too high. For example, you could give it a 25% likelihood, with this particular understanding: 25% chance to be wrong by making an unreliable report that happens to be false (thus giving us the false testimony), 25% chance to be right by making an unreliable report that happens to be true (thus giving us a true testimony), and a 50% chance that the witness' testimony is a reliable report (which would give a true testimony) and thus could not have arisen on the hypothesis ~h. We might even get these numbers by reviewing past examples where we have full information and find that 1 out of 2 witnesses gave a strictly reliable report, 1 out of 4 witnesses were correct by some kind of guess, and 1 out of 4 witnesses were wrong. The 25% is the 1 out of 4 that turned out to be wrong. Whatever the estimate we use, we use that.

In my example about

This thead is about "Carrier's numbers and math in OHJ," as I recall. He wrote the book, so we should be trying to understand the Bayesian mathematical method, a valid and well-known technique that is employed in the book, and possibly criticizing its particular application here. There's a time and a place for everything. Carrier never claimed to do Mullerian mathematics. He claimed to do Bayesian mathematics.

More specifically, Carrier's consequent probabilities do not employ your categories of certitude VS noncertitude, conclusiveness VS nonconclusiveness. If you are going to deal in the categories of certitude VS noncertitude or conclusiveness VS nonconclusiveness, you are either going to misinterpret the original argument or to impose your own completely unrelated scheme. Either way it is invalid as a criticism.

Carrier is simply using the mathematical concept of a conditional probability. No more, no less. That is what Bayes' theorem involves.

Last but not least, I'm still not sure if you understand an expression such as P( e | h ). It has nothing to do with "how much we believe that e conclusively shows h." Not at all. Not at all. Not at all. Let me repeat myself:

http://www.earlywritings.com/forum/view ... 7&start=80

The probability P(e | h) is not stating the odds that a piece of evidence secures the conclusion of the hypothesis, which is how you are understanding it.

Such a concept is not used anywhere by Carrier. It is not the way Bayesian arguments work.

The probability P(e | h) is stating the conditional probability that "e" would occur given that a hypothesis "h" happened.

For example, let's say that the "e" is "the sidewalk is wet now" and that the "h" is "it rained considerably on the sidewalk five minutes ago."

P(e | h) = 0.999

If it rained considerably on the sidewalk five minutes ago, there's 999 out of 1000 odds that we would find the sidewalk wet now.

We would be completely retarded if we then use that to say that there's a 99.9% chance that it rained considerably on the sidewalk five minutes ago, even though that is exactly what you are doing when you misunderstand the meaning and structure of the mathematical argument.

Maybe now you see where you fit into that? Good grief.

[Peter Kirby]

P( h ) = 0.33333
P( ~h ) = 0.66667

The problem I see here is if P( h ) = 0.33333 then P( ~h ) = 0.66667

These are the prior probabilities assigned by Carrier. They must add up to 1. Mathematically must. No exceptions.