Comments on: Absence of Evidence is Evidence of Absence

By: Mike

Mike — Thu, 05 Jan 2012 19:44:36 +0000

Yeah, I think we’re on the same page too. I do think they’re independent. It’s just the fact about the null hypothesis that necessitates the < result.

By: Mike

Mike — Thu, 05 Jan 2012 19:42:39 +0000

Only in absent evidence cases. That’s just because the null hypothesis must predict that there will be no positive evidence. The no ghosts hypothesis predicts that every investigated claim will turn up empty, right?

By: Hendy

Hendy — Thu, 05 Jan 2012 19:40:38 +0000

Sorry, posted my last bit before your analogy. I think we’re on the same case. I think I’m thinking of cases where the evidence is a bit more binary — either it implies the hypothesis or not — vs. your red/blue bucket example. That makes sense, and continuing to draw red balls would certainly mean that:

– Pr(many red balls drawn | red/blue mixed bucket) << Pr(many red balls drawn | ~red/blue bucket (red bucket))

Thanks for the example; that helped.

By: Hendy

Hendy — Thu, 05 Jan 2012 19:35:18 +0000

Gotcha. I get that it needs to be lower than 1 (purely statistical), but does that imply that:
— Pr(E│H) < Pr(E│¬H)?

Or you just mean that in cases of absence of evidence, it turns out that way? I don't think there's any a priori connection between Pr(E│H) and Pr(E│¬H); they're independent.

By: Mike

Mike — Thu, 05 Jan 2012 19:29:12 +0000

Let me elaborate a bit more. Say you have two buckets. One of them is filled entirely with red balls and the other is filled with a mixture of red and blue balls. Now, someone says they are going to blind fold you and place a random bucket in front of you. Your prior probability of the R/B bucket is 0.5 and the same for the R bucket. You pull out a ball and someone tells you it’s red (let’s assume they aren’t lying). If it’s red, then the likelihood for the R bucket is 100%, but the likelihood for the other bucket will match whatever the mixture is, maybe 50% if it’s evenly divided between red and blue.

That’s a pretty intuitive example and what I want to say is that absence of evidence is just like drawing that red ball. It is completely expected under ~H and only partially expected under H. That’s what I’m trying to say with that sentence. If the evidence is irrelevant, then I see no reason to have any difference in likelihood. However, if the evidence is relevant (and it is not found when expected), then Pr(H) will necessarily be lower than 1.

By: Mike

Mike — Thu, 05 Jan 2012 18:52:16 +0000

Ah, yes, it was a typo. Thanks. I’ll fix that later.

Re: the sentence, I may have worded it confusingly. I’m still talking about absence of evidence. So, if evidence is expected and not found, then the likelihood will necessarily be lower than 1.

By: Hendy

Hendy — Thu, 05 Jan 2012 17:18:55 +0000

You show:
— 0.5 * Pr(E│H) / [0.5 * Pr(E│H)]* [0.5 * Pr(E│¬H)]

Did you mean?

— 0.5 * Pr(E│H) / [0.5 * Pr(E│H)] + [0.5 * Pr(E│¬H)]

(Note the “+” instead of “*” for the denominator).

If I use the latter formulation, I get the same answer as you for your calculations. If I use the formula how it’s shown, I do not. For example (last equation in the post):

— 0.5 * 0.5 / [ (0.5 * 0.5) * (0.5 * 1) ] = 0.125 (0.5^2/0.5^3)
— 0.5 * 0.5 / [ (0.5 * 0.5) + (0.5 * 1) ] = 0.33 (0.5^2/0.75)

Lastly, could you clarify this statement:
,—–
| So, in most cases where the evidence is relevant to the hypothesis,
| Pr(E│H) will be lower than Pr(E│¬H)
`—–

I would have expected it to be the other way around. In other words, the probability of some evidence E, given our hypothesis H (which is supposed to be explanatory) is higher than the same evidence E if our hypothesis is false.