Cont: Proof of Immortality VIII

[jond]

Monza,
- If reincarnation is true, my self awareness is brought back to life. It's like having amnesia, and not knowing who you are, but you "wake up" as a fetus(?). and no one expects you to know who you are.

How, then, would you differentiate between your self being reincarnated and a process generated by the brain? If the new person self idenifies as Sam Jones, how could you tell it’s really Jabba?

 

[Belz...]

Hilarious. In your model, your "self-awareness" isn't actually self-aware except when incarnated, and then only for the duration of the incarnation. In that respect it's functionally indistinguishable from materialism.

Quoted for truth.

 

[jt512]

Jabba has disproven his own argument, so I'm wondering why anyone is still arguing with him about anything else. It just gives him room to avoid addressing his own disproof. Here it is in more detail.

Jabba has stated that P(H) > 0, P(~H) > 0 and P(E) = 1. Now, the denominator in Bayes' Theorem is

P(E) = P(E|H)P(H) + P(E|~H)P(~H) .

Now, one can see by inspection that for P(E) to be 1 and P(H) and P(~H) to both be non-zero, P(E|H) and P(E|H) must both be 1 (which I've been saying all along).

Plugging these values into Bayes' Theorem,

P(H|E) = P(E|H)P(H) / P(E)
P(H|E) = (1)P(H) / 1
P(H|E) = P(H) .

Likewise, (P~H|E) = P(~H) .

That is, the posterior probabilities of H and ~H just equal their prior probabilities. Jabba's "evidence" E is not evidence for ~H over H, or H over ~H. It doesn't discriminate between H and ~H at all.

So, as long as he maintains that P(E) = 1 and that both H and ~H are non-zero, his Bayes argument is mathematically wrong. So, it seems to me, to be counterproductive to argue with him about any other aspect of his argument, unless and until he can fix this problem or finally admits that he's been wrong and gives up.

 

[Jabba]

Irrelevant. If the existence of your brain is, as you have stated, a given, then the likelihood of your existence under H is 1. The likelihood of your existence under ~H cannot be greater than this.

Mojo,
- The likelihood of my current existence under H is not 1 -- even though my brain and current existence are both givens...
- The likelihood of my current existence is not based upon actuality -- it's based upon the specifics of a particular hypothesis. In this case, it's based upon OOFLam. And, if OOFLam is true, my current existence is EXTREMELY unlikely. My current existence is much more likely if reincarnation is true.

 

[Mojo]

Mojo,
- The likelihood of my current existence under H is not 1 -- even though my brain and current existence are both givens...
- The likelihood of my current existence is not based upon actuality -- it's based upon the specifics of a particular hypothesis. In this case, it's based upon OOFLam. And, if OOFLam is true, my current existence is EXTREMELY unlikely. My current existence is much more likely if reincarnation is true.

Whatever the likelihood of your brain is, that is what the likelihood of your current existence is under H. The likelihood of your current existence under the hypothesis that you have an immortal soul is the likelihood of your brain multiplied by the likelihood of your soul occupying your brain.

Are you claiming that the likelihood of your soul occupying your body is greater than 1?

 

[jt512]

Whatever the likelihood of your brain is, that is what the likelihood of your current existence is under H. The likelihood of your current existence under the hypothesis that you have an immortal soul is the likelihood of your brain multiplied by the likelihood of your soul occupying your brain.

Are you claiming that the likelihood of your soul occupying your body is greater than 1?

At this point the skeptics are the ones keeping Jabba's argument alive. See my last post.

 

[Mojo]

I thought we were dancing on its grave.

 

[RoboTimbo]

One of these...

- The [U]likelihood[/U] of my current existence under H is [U]not[/U] 1

is not like the other.

-- even though my brain and current existence are both givens...

 

[jond]

Mojo,
- The likelihood of my current existence under H is not 1 -- even though my brain and current existence are both givens...
- The likelihood of my current existence is not based upon actuality -- it's based upon the specifics of a particular hypothesis. In this case, it's based upon OOFLam. And, if OOFLam is true, my current existence is EXTREMELY unlikely. My current existence is much more likely if reincarnation is true.

No. Given the materialist model, the likelihood of your current existence is the likelihood of your brain. In your ~H, the likelihood of your current existence is the likelihood of your brain X the likelihood of your soul existing X the likelihood of your soul connecting with your brain. This cannot be more likely than your the existence of your brain alone. You have lost, Jabba.

 

[jsfisher]

Mojo,
- The likelihood of my current existence under H is not 1 -- even though my brain and current existence are both givens...
- The likelihood of my current existence is not based upon actuality -- it's based upon the specifics of a particular hypothesis. In this case, it's based upon OOFLam. And, if OOFLam is true, my current existence is EXTREMELY unlikely. My current existence is much more likely if reincarnation is true.

Please stop misusing that term. P(E|H) is the probability of E occurring given H. It is a function of E.

It can also be viewed as a likelihood function of H for fixed E. In that case it becomes a function of H.

The probability of two heads in succession with a fair coin is P(HH|F) = 0.25. The likelihood of the coin being fair having observed two heads in succession is 0.25.

If E is "given", that is, P(E) = 1, then P(E|H) = 1. You don't get to throw out a smoke screen saying you are talking about likelihood and not probability. That is just nonsense. P(E|H) = 1.

 

[Belz...]

Jabba has disproven his own argument, so I'm wondering why anyone is still arguing with him about anything else. It just gives him room to avoid addressing his own disproof.

What else are we supposed to do at this point?

 

[Belz...]

Mojo,
- The likelihood of my current existence under H is not 1 -- even though my brain and current existence are both givens...

How do you multiply 1 and 1 to get 10^-100?

 

[jt512]

What else are we supposed to do at this point?

He has refuted his own argument. Just tell him that. It's over. Arguing with him about anything else is just letting him off the hook.

 

[Belz...]

He has refuted his own argument. Just tell him that. It's over.

Yeah, I know. I was there.

Arguing with him about anything else is just letting him off the hook.

I want him to admit that he pulled that number out of his ass. He's lost but I want him to realise that, too.

 

[jt512]

Yeah, I know. I was there.

I want him to admit that he pulled that number out of his ass. He's lost but I want him to realise that, too.

Well, like I said, at this point it is the skeptics who are perpetuating the argument.

 

[JoeMorgue]

I don't get why having a bunch of Woo Slingers on soap boxes we're all ignoring is all of a sudden so many people's goal.

We've ignored Jabba before. It doesn't stop him.

 

[Humots]

Jabba has disproven his own argument, so I'm wondering why anyone is still arguing with him about anything else. It just gives him room to avoid addressing his own disproof. Here it is in more detail.

Jabba has stated that P(H) > 0, P(~H) > 0 and P(E) = 1. Now, the denominator in Bayes' Theorem is

P(E) = P(E|H)P(H) + P(E|~H)P(~H) .

Now, one can see by inspection that for P(E) to be 1 and P(H) and P(~H) to both be non-zero, P(E|H) and P(E|H) must both be 1 (which I've been saying all along).

<snipped for brevity>

I don't follow. Would you please show this in more detail?

 

[Belz...]

Well, like I said, at this point it is the skeptics who are perpetuating the argument.

Well, if you think we're done, why are you still here?

 

[jsfisher]

I don't follow. Would you please show this in more detail?

The key is that for any two numbers greater than 0 but less than 1, their product is less than either of the two numbers. 1/2 x 3/4, for example, is 3/8, and 3/8 is less than either of 1/2 or 3/4.

Consider the first term, P(E|H) x P(H). For any value of P(E|H) < 1, the product will be less than P(H). The second term, P(E|~H) x P(~H), can be at most P(~H), so we'd have two products to add. One is strictly smaller than P(H) and the other can be at most P(~H).

We know that P(H) + P(~H) = 1, and so if the first term, P(E|H) x P(H), is smaller than P(H), and the second, P(E|~H) x P(~H), is at most P(~H), the sum must be less than 1.

Therefore, P(E|H) must be 1 for the sum to be 1.

The same argument applies to the second term, so P(E|~H) must also be 1.

 

[jt512]

I don't follow. Would you please show this in more detail?

P(E) = P(E|H)P(H) + P(E|~H)P(~H) = P(E|H)P(H) + P(E|~H)[1 – P(H)] ,

since P(H) and P(~H) are probabililites that sum to 1. Looking at the above equation, we can see that P(E) is the weighted average of P(E|H) and P(E|~H), with weights P(H) and P(~H), respectively (which Jabba says are both non-zero). But P(E|H) and P(E|~H), being probabilities, can be no greater than 1. The only way that a weighted average of two numbers can be 1, if neither number can be greater than 1, is if both of them are exactly 1. That is, for any 0 < P(H) < 1,

P(E|H)P(H) + P(E|~H)[1 – P(H)] = (1)P(H) + (1)[1 – P(H)] = P(H) + 1 – P(H) = 1 .

But unless P(E|H) = P(E~|H) = 1, then

P(E|H)P(H) + P(E|~H)[1 – P(H)] < 1 .

If you still don't see it, just try out some values of P(E|H) and P(E|~H) for yourself.