Absence of evidence IS evidence of absence (Goji, get the fuck in here)

 
More Than Mortal
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This is the way the world ends. Not with a bang but a whimper.
EVERYBODY APART FROM GOJIRA (AND MAYBE TURKEY) SHOULD PROBABLY JUST READ THE RECAP AT THE BOTTOM.

Definitions:

1) X is any arbitrary event. (I was born in Germany.)

2) ¬X denotes "not X". (I was not born in Germany.)

3) E represents the existence of positive evidence that indicate X is correct. (My birth certificate is from Germany.)

4) ¬E denotes "not E", or the total absence of positive evidence.

5) P(X) denotes the probability of X.

6) P(X/E) denotes the conditional probability of X given E; this is the joint probability of X and E divided by the probability of E.

P(X/E) =  P(X^E) / P(E)

Assumption (1):
- If an event like X were to really have happened, then it very likely left some evidence of itself. In other words, the probability of E, given X, is greater than the probability of NOT E, given X.

P(E/X) > P(¬E/X)

1 - P(¬E/X) > P(¬E/X)

P(¬E/X) < 1/2

This isn't an unfair assumption, as most things of significance leave some sort of evidence.

Now we invoke Bayes' Theorem: P(¬E/X) = P(X/¬E)P(¬E) / P(X).

->  1/2 > P(X/¬E)P(¬E) / P(X)

P(X/¬E) < (1/2) P(X)/P(¬E)

Thinking about P(X) in the equation immediately prior, is X a likely or unlikely event?

Assumption 2:
- The event X is extraordinary. P(X) << 1. The probability of event X is very small.

Let X be an intersection of two statistically independent events, A and B. X = A^B. (I was born in Germany, and I love shitting on my neighbour's lawn.) However, the joint probability of A and B is always equal to or less than the probability of A or B. P(AB) = P(A)P(B). The more events which define X, the lower the probability.

Returning to the emboldened equation, consider the term P(¬E). What can we say about the likelihood of evidence for X?

Assumption 3:
- We have searched for evidence of X, but failed to find any. P(¬E) [approx.]= 1. The probability of no actual evidence for X is very high, and the more we search for E but fail, the closer this value approaches 1.

Bringing forward the emboldened equation again:

P(X/¬E) < (1/2) P(X)/P(¬E)

The more specific and extraordinary X, the closer P(X) is to 0. And the more we search for evidence, but fail, the closer ¬E is to 1. The ratio here [P(X)/P(¬E)] then, must be very small. As long as this ratio is less than one, the entire right-hand side of the inequality is less than one-half.

P(X/¬E) < 1/2. This inequality must hold. This implies: P(¬X/¬E) > 1/2.

We finally arrive at:

P(¬X/¬E) > P(X/¬E). In other words, given an absence of any evidence for X, the more likely event is that X did not, in fact, occur.

This is a demonstration of the epistemic principle known as the inference to the best explanation. Many things cannot be known with absolute certainty, but we can show which explanations are most preferable.

Recap:
1) An event like X should leave evidence.
2) All things being equal, X is unlikely.
3) We have searched for evidence of X, but failed.

From these premises, it mathematically follows that ¬X is a more likely event than X.

Let X be the following claim:
Quote
The Virgin Mary, upon being impregnated by Yahweh, gave birth to a half-blooded demigod named Jesus of Nazareth. During his life, Jesus performed many miracles that included healing the sick, raising the dead and turning water into wine. Jesus also took the aggregate sins of humanity upon himself, and gave his own life for us. Upon his execution by Roman authorities, Jesus rose from the dead and ascended to Heaven. All of these events were compiled into the record, with inerrancy, known as the New Testament.

And anybody who fails to believe this will spend an eternity in endless suffering.

If any of this is to be the case, we should find evidence beyond mere say-so.

P(E/X) > P(¬E/X).

X is a huge intersection of independent events, all competing with the various denominations and interpretations of Christian doctrine which is, in turn, competing with those belonging to other religions. Significant positive evidence for Christianity is yet to have been found, also.

P(X) [appox.]= 0.

P(¬E) [approx.]= 1.

P(¬X/¬E) [approx.]= 1. Quad erat demonstrandum.
Last Edit: February 04, 2015, 05:59:32 PM by Meta Cognition


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We knew the world would not be the same.
A few people laughed, a few people cried, most people were silent.
I remembered the line from the Hindu scripture, the Bhagavad-Gita.
Vishnu is trying to persuade the Prince that he should do his duty
and to impress him takes on his multi-armed form and says,
"Now, I am become Death, the destroyer of worlds.."
I suppose we all thought that one way or another.
Damn, I gotta grab my old notes on logic equations. I'm rusty.


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I need a tl:dr for that tl:dr...


 
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We knew the world would not be the same.
A few people laughed, a few people cried, most people were silent.
I remembered the line from the Hindu scripture, the Bhagavad-Gita.
Vishnu is trying to persuade the Prince that he should do his duty
and to impress him takes on his multi-armed form and says,
"Now, I am become Death, the destroyer of worlds.."
I suppose we all thought that one way or another.
I need a tl:dr for that tl:dr...
Here's your tl;dr:

Spoiler
gODS NOT REAL


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Quote
AND MAYBE TURKEY

Bruh, my major was basically probability on crack, I understood you just fine. What you did is a well established Bayesian principle already, and I agree with it. Absence of proof isn't proof of absence, though.



g💚jira | Heroic Unstoppable!
 
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A) This is Turkey's domain man, he's on another plane of existence when it comes to math. He's right on: while this is a well known principle [yes yes it's 1 AM gimme a break ;_;] I get the feeling it's not what you're looking for [proof != evidence].

B) But now, you're on MAT proof turf. Direct proof by algebra [exceedingly messy]:

Let X be a claim dependent on evidence Y

Want to show P(X|Y) > P(X) => P(X'|Y') > P(X').

{ Bayes' Thm [I have to give up my beloved square brackets for this ;_;]}

P(X|Y) = P(Y|X)[P(X)/P(Y)]

{ P(X|Y) > P(X), obvious as claim/evidence supersedes claim, bring in B Thm and substitute for P(X|Y)}

P(Y|X)[P(X)]/P(Y) > P(X)

{ Bring P(Y) over }

P(Y|X)[P(X)] > P(X)P(Y)

{ We know P(Y) = 1-P(Y') and P(Y|X) = 1-P(Y'|X) by definition, so }

[1-P(Y'|X)][P(X)] > [P(X)][1-P(Y')]

{ Expand to cancel P(X), then multiply by negative 1 [flips the inequality] }

[P(X)][P(Y')] > [P(X)][P(Y'/X)]

{ By Bayes' Thm, after bringing the denominator over, [P(X)][P(Y'|X)] = [P(X|Y')][P(Y')] }

[P(X)][P(Y')] > [P(X|Y')][P(Y')]

{ P(X) = 1-P(X') and P(X|Y') = 1-P(X'|Y') by definition }

[P(Y')][1-P(X')] > [P(Y')][1-P(X'|Y')]

{ Again, multiply it out, cancel P(Y') and negate }

[P(Y')][P(X'|Y')] > [P(X')][P(Y')]

{ Cancel P(Y') }

P(X'|Y') > P(X')



Q to the E to the D

Spoiler
Formatting this thing was the actual challenge holy hell
Last Edit: February 05, 2015, 12:07:11 AM by gogojirajira


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Goji, you da real MVP for that correct notation.


 
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This is the way the world ends. Not with a bang but a whimper.
Absence of proof isn't proof of absence, though.
Evidence, mang, evidence. Let's not throw the word proof around >.>


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i wish i were good enough at math to comprehend and appreciate this

so ill just clap my hands like the idiot i am


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Urgh we did a lot of this last year in statistics as part of a module in A level maths.


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Absence of proof isn't proof of absence, though.
Evidence, mang, evidence. Let's not throw the word proof around >.>

It's important to clarify, though. Evidence isn't proof.