Cont: Proof of Immortality VIII

[jond]

- That is our observation -- but, we don't know that it has to be. That's what we're trying to find out, and the likelihood of the current existence of my self is 10^-100 if it depends upon my current, particular, body -- but much greater if it doesn't.

We can change every aspect of your self by altering your brain, physically or chemically. You cannot pretend that your self does not depend on your body in your current existence.

 

[MRC_Hans]

- That is our observation -- but, we don't know that it has to be. That's what we're trying to find out, and the likelihood of the current existence of my self is 10^-100 if it depends upon my current, particular, body -- but much greater if it doesn't.

Oh btw, did you explain how you got the 10^-100?

No, that is not what we are trying to do. YOU are trying to disprove the materialistic model. This is silly because disproving something is neigh impossible, but it was your own choice.

However there is indeed little else you can do as long as your competing hypothesis is pure assertion. What I don't understand is why you don't assert something more sure-fire, like "We are 100% certainly immortal." Now that would save you the trouble of inventing ridiculous small figures for H.

Hans

 

[Jabba]

- No. I shouldn't have agreed that it involves my body.

Sure you do. That is the evidence you are presenting. E is you, Jabba, telling us about your sense of self. You know, your sense of self which includes your memories, and your sensory inputs, and your "think therefore I am". Your body is part of that...

js,
- E is my experience of a self. E is not my body. We all agree that in order for a self to exist on the physical plane a body has to be involved. While H requires my current physical body, ~H does not.

Now, it could well be (however unlikely) that your E has as its cause some sort of soul that exists independent of your body, but you are providing E for consideration, not your soul...

- E is my experience of a self -- the self could be immortal, and therefor a soul. That's what we're trying to figure out...

- When the self exists on the "physical plane," it naturally involves A body -- but again, if reincarnation is true, my self doesn't require my current body...

...
Perhaps, but your evidence does.

- Just to make sure, to what specific evidence are you referring?

 

[Belz...]

- E is my experience of a self. E is not my body. We all agree that in order for a self to exist on the physical plane a body has to be involved. While H requires my current physical body, ~H does not.

No, that is false. You know that you need the body in order for the self to be able to make the observation. You said so yourself. E requires a body for both hypotheses.

 

[Jabba]

jt,
- I don't perceive the implication that you do. To me P(E|H) + P(E|~H) = 1...

WTF are you talking about? P(E|H) + P(E|~H) don't have to add to 1. They don't even add to 1 in your own so-called model...

- They do add to 1: i.e., 0 + 1.
- And as complimentary likelihoods, shouldn't they add to 1?

 

[Belz...]

- They do add to 1: i.e., 0 + 1.

Where did 10^-100 and 0.0062 go?

Stop changing the numbers with the wind. We know you're making them up as you go. You've already been explained that your numbers are nonsense.

And as complimentary likelihoods, shouldn't they add to 1?

"I don't read what people write, but could you write it again?"

 

[Jabba]

WTF are you talking about? P(E|H) + P(E|~H) don't have to add to 1. They don't even add to 1 in your own so-called model.

Your belief that P(E) = 1 contradicts your model, because P(E) = 1 implies that P(E|H) = P(E|~H) = 1. This has been demonstrated algebraically to you in multiple recent posts. Whether you accept it or not is irrelevant, except for what it reveals about your ability to do elementary algebra.
Again, you need to find numbers for P(E|H) and P(E|~H) that satisfy the following equation; otherwise, you have contradicted your own model.

1 = P(E|H)(.99) + P(E|~H)(.01) .

We look forward to seeing the two numbers that you pick.

jt,
- Please show me where you think that this has been demonstrated algebraically -- I don't think that it has.

 

[Belz...]

jt,
- Please show me where you think that this has been demonstrated algebraically -- I don't think that it has.

Stop asking people to do the same thing over and over. You've demonstrated the ability to go back through the thread when it suits your purposes, so YOU do it.

 

[Dave Rogers]

- They do add to 1: i.e., 0 + 1.

Only because you've made up numbers that add to 1. And, as we've all told you ad nauseam, your value of 0 for P(E|H) doesn't follow from your reasoning.

- And as complimentary likelihoods, shouldn't they add to 1?

No. This highlights your poor grasp of conditional probabilities.
Suppose I buy a cup of tea in a cafe on my way to work 99 days out of 100 if it's raining, and 83 days out of 100 when it isn't raining. We can call the probability of buying a cup of tea T, and of it raining R. We can therefore write that:
P(T|R)=0.99
P(T|~R)=0.83
We can therefore add them together (although there's no significance to the sum):
P(T|R)+P(T|~R)=1.82

Because, of course, there's no requirement that they sum to 1.

You really need to take an elementary statistics course; your understanding of conditional probabilities appears to be virtually non-existent.

Dave

 

[Belz...]

Only because you've made up numbers that add to 1. And, as we've all told you ad nauseam, your value of 0 for P(E|H) doesn't follow from your reasonong.



No. This highlights your poor grasp of conditional probabilities.
Suppose I buy a cup of tea in a cafe on my way to work 99 days out of 100 if it's raining, and 83 days out of 100 when it isn't raining. We can call the probability of buying a cup of tea T, and of it raining R. We can therefore write that:
P(T|R)=0.99
P(T|~R)=0.83
We can therefore add them together (although there's no significance to the sum):
P(T|R)+P(T|~R)=1.82

Because, of course, there's no requirement that they sum to 1.

You really need to take an elementary statistics course; your understanding of conditional probabilities appears to be virtually non-existent.

Dave

Jabba: "Did you really mean reasonong? I don't understand that word!"

 

[RoboTimbo]

- They do add to 1: i.e., 0 + 1.
- And as complimentary likelihoods, shouldn't they add to 1?

And since you've shown that H is 1, that leaves your ~H to be 0?

 

[Dave Rogers]

jt,
- Please show me where you think that this has been demonstrated algebraically -- I don't think that it has.

Jabba,

Since your only demonstrated ability in this thread has been to make up numbers that support your argument, how about making up a set that support your argument that 1 = P(E|H)(.99) + P(E|~H)(.01) can be true for some other values than P(E|H)=P(E|~H)=1? Let me give you a template for your answer:

P(E|H) = [insert first made-up number here]
P(E|~H) = [insert second made-up number here]

Then we can all check the calculation.

Dave

 

[Dave Rogers]

jt,
- Please show me where you think that this has been demonstrated algebraically -- I don't think that it has.

Actually, the proof's quite simple.

(1) P(H) + P(~H)=1, by definition.
(2) P(E|H)P(H) + P(E|~H)P(~H) = P(E), from Bayes' Theorem.
(3) P(E) = 1, as stipulated by Jabba.

Suppose P(E|H) < 1. Then,

P(E|H)P(H) < P(H) (multiplying both sides by P(H))

P(E|H)P(H) + P(~H) < P(H) + P(~H) (adding P(~H) to both sides)

P(E|H)P(H) + P(~H) < 1 (substituting from (1))

P(E|H)P(H) + P(~H) < P(E) (substituting from (3))

P(E|H)P(H) + P(~H) < P(E|H)(P(H) + P(E|~H)P(~H) (substituting from (1))

P(~H) < P(E|~H)P(~H) (subtracting P(E|H)P(H) from both sides)

1 < P(E|~H) (dividing both sides by P(~H))

But this implies a probability greater than 1, which is impossible. Therefore P(E|H) cannot be less than one.

As it is a probability, 0 <+ P(E|H) <= 1.

Therefore, P(E|H) = 1.

Now, we know from (1) and (2) that P(E|H)P(H) + P(E|~H)P(~H) = 1.

P(H) + P(E|~H)P(~H) = 1 (substituting for P(E|H))

P(H) + P(E|~H)P(~H) = P(H) + P(~H) (substituting from (1))

P(E|~H)P(~H) = P(~H) (subtracting P(H) from both sides)

P(E|~H) = 1 (dividing both sides by p(~H))

Therefore, P(E|H) = P(E|~H) = 1

QED.

Dave

 

[Jabba]

Oh btw, did you explain how you got the 10^-100?...

- Numerous times.

 

[Dave Rogers]

- Numerous times.

Please link to an explanation that reduces to something other than "I made it up."

Dave

 

[jsfisher]

- They do add to 1: i.e., 0 + 1.
- And as complimentary likelihoods, shouldn't they add to 1?

Wow. That is beyond wrong.

What is the likelihood/probability water is wet? P(W) = 1, right?

What is the likelihood/probability water is wet in Schenectady? P(W|S) = 1, right?

What is the likelihood/probability water is wet in any place other than Schenectady? P(W|~S) = 1, right?

 

[JayUtah]

E is my experience of a self.

It is your current experience of a self. It requires a body because you observe yourself right now as having a body. E is not what you imagine you would observe differently in a different time or place. You don't get to declare a priori what parts of the observation are significant and what aren't.

 

[JayUtah]

- Numerous times.

Then, given your demonstrated ability to scour the thread for posts from which to compose your anthologies, it should be easy for you to link to one of those numerous times. I've examined the thread and I don't find a single post that explains the calculation. What I find are numerous times when you were asked to tell how you calculated it, and all those posts went unanswered.

 

[RoboTimbo]

- Numerous times.

Jabba, I think you agree with me that this is a lie.

 

[Belz...]

- Numerous times.

Please show me where you think that this has been explained -- I don't think that it has.