[Merged] Immortality & Bayesian Statistics

[Jabba]

- I received another email from Dr. Hoerl. He'll be traveling for a week, and doesn't expect to participate with us for a couple of weeks, but he does sound interested (to me).
- I'd post his email, but he hasn't said yet that I can quote him.

 

[Giordano]

JS,
- I think you're talking about what is to me the only weak link in my argument -- that I can use the likelihood of me, in particular, existing rather than the likelihood of someone/anyone existing.
- Do you accept my paraphrasing?

No! Because the correct math predicts the previous probability of anyone, not just you, existing. You are incorrectly calculating the previous probability of Jabba in particular existing, not just anyone existing. The probability of someone/anyone existing is nearly identical to the probability of a brain existing, since selves are emergent and are nearly inevitable results of forming a brain.

This is not a weak link in your argument- it is a fatal flaw. But I am thrilled that you finally are wiling to admit this concern!

Again- the previous probability of their being a Jabba in particular is low, but the previous probability of someone/anyone developing from a fertilized human zygote is quite high. You are here, so your current likelihood is 1, so in your case you became a Jabba. But under very slightly different circumstances, you would have become someone else.

We have presented lottery tickets, cards, and even VWs as examples to explain the flaws in your argument: you can re-read those. But also just think about your body, your particular face, and your particular fingerprints! The previous probability of you having just those physical traits is very small: in advance what was the probability that you in particular would have the unique fingerprints that you have? Tiny? No one else has your fingerprints (at least in theory). Are your fingerprints different in your proof than your self? Aren't your fingerprints reincarnated by the same proof that you use for your "self"? Aren't infinite VWs reincarnated by the same proof? If not, why not?

In SM, as you call it, fingerprints and "selves" are almost inevitable consequences of developing physically into a human being. No magic occurs, and none need to, in either case.

 

[jt512]

Tit for tat. Conditional probability does not work that way. The problem is A and Me are both independent. As such P(me,A) that is the probability of Me given that we are mortal and do not continue to exists after death (or whatever hypotheses A has Jabba) is the same as Probability Me. It does not prove anything on A.

Your choice of notation is horrible. P(X,Y) always means the joint probability of X and Y. The conditional probability of X given Y is always written P(X|Y). You should stick to conventional notation. Probability is confusing enough without everybody making up their own notation.

It would be the same as saying probability Me is the probability of having a black swan if you point at swan at random. Hypotheses A is swan are mortal and live only once. Non A is everything else including swan reincarnate , or swan mind live forever or whatever. Well yuppee do. it is quite obvious that the P(Me,A)=P(Me) and P(A,Me)=P(A) (ETA thanks Jfisher). A and Me are statistically independent.

I disagree that A and "me" are independent. Let "me" be the probability of seeing a particular black swan. It seems reasonable to say that the probability of seeing that swan is greater if it were, in some sense, pre-existing (which is, as far as I can tell, is Jabba's non-A hypothesis) than if it were the result of a random physical process (which, as far as I can tell, is Jabba's A hypothesis).

It would also be worth establishing how if P(me)=1 (Jabba's existence is a given, after all) then how P(me|x) can be other than 1 for all possible x (i.e. P(x) > 0).

No matter what P(me) is, by the law of total probability,

P(me) = P(me|A₁) + P(me|A₂) + . . . ,​

so it cannot be the case, that if P(me)=1, that P(me|x)=1 for all x (unless, of course, there is just one x).


However, I disagree that the P(me) should be considered 1. In the context of this thread, P(me), as the above equation indicates, is the sum of the conditional probabilities of "me" given each possible model. Those probabilities do not have to sum to 1, even if the event "me" has occurred.

Consider a simpler problem where there is only one model under consideration. Let A be the model that the results of one flip of a coin is determined by a random process where

P(H)=P(T)=1/2.​

You flip the coin and it lands heads. What is P(H)? Well, we just wrote it on the previous line. P(H)=1/2, not 1, because P(H) is the probability under the assumed model of H. Under the model, it doesn't matter whether the outcome of any particular experiment is H or T, because under the model the other event could just as easily have occurred. So it is perfectly reasonable—indeed correct—for Jabba to claim that P(me) < 1, even though he appears to actually exist, provided his existence was not guaranteed by the joint set of hypotheses under consideration.

 

[Sideroxylon]

- I received another email from Dr. Hoerl. He'll be traveling for a week, and doesn't expect to participate with us for a couple of weeks, but he does sound interested (to me).
- I'd post his email, but he hasn't said yet that I can quote him.

Send him a link to the thread.

 

[jsfisher]

No matter what P(me) is, by the law of total probability,

P(me) = P(me|A₁) + P(me|A₂) + . . . ,​

so it cannot be the case, that if P(me)=1, that P(me|x)=1 for all x (unless, of course, there is just one x).

You left out a really, really important bit:

P(me) = P(me|A₁)P(A₁) + P(me|A₂)P(A₂) + . . .​

 

[godless dave]

4. This is the inevitable red herring you were destined to attempt to slip in. The real H1 assumption is roughly analogous to the assumption that the sentience you are experiencing is the result of a gazillion-sided die being rolled, and a particular side having turned up. Not some arbitrary sentience. The particular sentience you are experiencing. Not some arbitrary side of the die. A particular side of the die.

My sentience is some arbitrary sentience. There is nothing special about it.

You are criticizing xtifr for not using the fallacy of special pleading.

 

[Garrette]

My sentience is some arbitrary sentience. There is nothing special about it.

You are criticizing xtifr for not using the fallacy of special pleading.

I think this is the key bit. No existing sentience was pre-selected or predicted; they are what happened. Perhaps they do not realize it, but both Jabba and Toontown are assuming specialness without justifying the assumption.

 

[jt512]

You left out a really, really important bit:

P(me) = P(me|A₁)P(A₁) + P(me|A₂)P(A₂) + . . .​

Oy. You're right. How stupid of me. And that correction shows that you were right: If P(me)=1 then P(me|A_i)=1 for all i such that P(A_i)>0. However, I still think that P(me) is not 1 just because Jabba exists. P(me) is still a probability under a model, or a set of models.

 

[jsfisher]

Oy. You're right. How stupid of me.

S'ok.


But now, I think you'll agree that if P(me) is 1, then P(me|x) for all possible x must be 1 as well.

There is an big "if", though, if P(me) is 1. You objected to it and countered with your coin flipping analogy. In that universe, a coin toss occurs with equal likelihood of a heads or a tails outcome. P(H) = P(T)= 0.5.

I toss a coin, and it lands up heads. Ok, so what is P(H)? It is still 0.5, of course, but P(H) was then and still is now the probability for a future event. The coin I tossed, the one that landed up heads and still sitting where it landed with heads up has a probability of it being heads of 1.

Jabba is like the already tossed coin. His existence is already determined.

 

[Loss Leader]

- I think you're talking about what is to me the only weak link in my argument

Hahahahahahahahahaha.

You're joking, right?

 

[Aepervius]

In layman's terms, is what you are saying that Jabba is proposing a Bayesian analysis but using a Frequentist interpretation? (In case it's not clear, this is a serious question , even if it is perhaps an uneducated one).

I don't think I'm the first to post this, but I recalled it and looked it up, and it helped crystallize it a bit for me. The question is did it crystallize correctly?

At this point i think jabba saw a nice formula, then decided to assign some value to each term to fits his wishes and belief, and he saw that if he put some "infinite" somewhere he could get a 1 to what he wished to have. That's about it. he has no priors except his wishes, he has no distribution on his probability, nothing approaching anything remotely sensible, and he continuously uses "infinity". I am not sure frequentist versus bayasian matters at this point. Neitehr recognize the usage of infinity.

 

[Jabba]

S'ok.


But now, I think you'll agree that if P(me) is 1, then P(me|x) for all possible x must be 1 as well.

There is an big "if", though, if P(me) is 1. You objected to it and countered with your coin flipping analogy. In that universe, a coin toss occurs with equal likelihood of a heads or a tails outcome. P(H) = P(T)= 0.5.

I toss a coin, and it lands up heads. Ok, so what is P(H)? It is still 0.5, of course, but P(H) was then and still is now the probability for a future event. The coin I tossed, the one that landed up heads and still sitting where it landed with heads up has a probability of it being heads of 1.

Jabba is like the already tossed coin. His existence is already determined.

- But, it can be useful to look back and see what the likelihood was of me existing given a particular hypothesis, or worldview. This has implications re that worldview. That's what Bayesian statistics is all about. The posterior probability of A|me is proportional to the likelihood of me|A.

 

[Aepervius]

Your choice of notation is horrible. P(X,Y) always means the joint probability of X and Y. The conditional probability of X given Y is always written P(X|Y). You should stick to conventional notation. Probability is confusing enough without everybody making up their own notation.

Yeah sorry, bad lazy habbits.

I disagree that A and "me" are independent. Let "me" be the probability of seeing a particular black swan. It seems reasonable to say that the probability of seeing that swan is greater if it were, in some sense, pre-existing (which is, as far as I can tell, is Jabba's non-A hypothesis) than if it were the result of a random physical process (which, as far as I can tell, is Jabba's A hypothesis).

P(A) says nothing about pre existence it says only something after it died. In fact the probability of a particular swan , is conditioned on pure physical process (DNA structure) and not whether a similar swan died previously or not. Just like is P(Me).

 

[jsfisher]

JS,
- I think you're talking about what is to me the only weak link in my argument -- that I can use the likelihood of me, in particular, existing rather than the likelihood of someone/anyone existing.
- Do you accept my paraphrasing?

Labeling it a weak link trivializes its importance, Jabba. Also, it isn't the likelihood of someone/anyone existing, it is the likelihood of a specific someone existing where the specifics are determined in advance.

If I may belabor the lotto analogy: You have approached this immortality proof thing as if you are trying to now win the NYS Lottery by knowing that last night's numbers were 2 -7 - 24 - 36 - 53 - 57.

 

[jsfisher]

- But, it can be useful to look back and see what the likelihood was of me existing given a particular hypothesis, or worldview. This has implications re that worldview. That's what Bayesian statistics is all about. The posterior probability of A|me is proportional to the likelihood of me|A.

Brother Bayes tells us that P(A|me)P(me) = P(me|A)P(A) (and as long as P(me)>0, we can express that as P(A|me) = P(me|A)P(A) / P(me)).

Since P(me) and P(me|A) are both 1, the above reduces to P(A|me) = P(A), reconfirming the independence of P(A) and P(me).

 

[jt512]

S'ok.


But now, I think you'll agree that if P(me) is 1, then P(me|x) for all possible x must be 1 as well.

Yes. I was just editing my last post to include that when you made your post.

There is an big "if", though, if P(me) is 1. You objected to it and countered with your coin flipping analogy. In that universe, a coin toss occurs with equal likelihood of a heads or a tails outcome. P(H) = P(T)= 0.5.

I toss a coin, and it lands up heads. Ok, so what is P(H)? It is still 0.5, of course, but P(H) was then and still is now the probability for a future event.

No. I disagree. P(H) is the probability of H under the assumed model, namely, that that the results of coin flips are random with a uniform distribution. After the coin is flipped and the results are known, P(H) is the probability under the model that that would have happened: .5. This is the very basis of statistical inference. We do an experiment, observe the outcome, calculate the probability of the observed outcome (or some function of it) under some assumed probability model, and then use that probability to judge the plausibility of the model.

The coin I tossed, the one that landed up heads and still sitting where it landed with heads up has a probability of it being heads of 1.

Not under the relevant model, which was that it was a random event. P(H!H) = 1, but that is never a probability we care about. If we want to perform a statistical test of whether a coin is fair, we flip the coin N times and calculate the probability of the actual result we get (or a function of it, like a p-value) under, say, the null hypothesis of fairness, not the probability of the outcome given that the outcome occurred. That would be useless.

Similarly, P(me|me), which is obviously 1, is uninteresting. What is potentially interesting is the probability that "me" would have been observed if "me" came about by different hypothesized processes. That is what Jabba is attempting to ascertain, and to use to judge the plausibilities of those hypotheses.

 

[PixyMisa]

Your problem is that you have no problem believing you've beaten giganogargantuan odds.

The probability that there will be a face showing when the die has been rolled is 1. Those are easy odds to beat.

I do not assume that I am a special snowflake. Nor do I assume that it is relevant that I am a special snowflake.

Your other problem is that it has not occurred to you to ask "Then why do I exist?", given that proposition A (which you firmly believe is unconditionally true) unambiguously requires an eternity of nonexistence

Nope. The Universe is finite.

Also not shown to be relevant.

Also answered adequately by a form of the anthropic principle: Only those who exist can ask that question.

 

[jt512]

Brother Bayes tells us that P(A|me)P(me) = P(me|A)P(A) (and as long as P(me)>0, we can express that as P(A|me) = P(me|A)P(A) / P(me)).

Since P(me) and P(me|A) are both 1, the above reduces to P(A|me) = P(A), reconfirming the independence of P(A) and P(me).

JS, in Bayesian inference for a hypothesis we combine data with the prior probability of the hypothesis to obtain an updated probability for the hypothesis, using Bayes theorm:

P(H|D) = P(D|H)P(H) / P(D) ,​

where H is the hypothesis and D is the data. The data, D, is always an event that has already occurred when we perform the analysis. According to you, that implies that P(D)=1, and therefore that P(D|H)=1. But if that were true, then P(H|D) would always equal P(H), which would imply that it is impossible to do Bayesian inference. No amount of data could ever affect our opinion about the plausibility of a hypothesis.

 

[PixyMisa]

- But, it can be useful to look back and see what the likelihood was of me existing given a particular hypothesis, or worldview. This has implications re that worldview. That's what Bayesian statistics is all about. The posterior probability of A|me is proportional to the likelihood of me|A.

Let's instead take Baby X, who if born will have thus-and-so genes and will one day ask that same question. We have now sidestepped the problem with P(me) that "me" already exists.

According to genetics, P(BabyX) is fairly low. But how are Baby X's genes relevant? How does it matter what shade of green Baby X's eyes are?

The probability of Somebaby X, where genes are ignored, and only the likelihood of some day questioning its own existence, is so close to 1 as to be indistinguishable.

P(SomebabyX) ≈ 1, or as I said earlier, P(someone) ≈ 1.

And all of these are the same given your proposition A, which as far as we can tell is entirely true. P(SomebabyX|A) ≈ 1.

Since your particular genes are irrelevant to the question, P(me|A) = P(someone|A), so P(me|A) in your formula ≈ 1. You are trying to peg it at ≈ 0. Obviously these approaches give very different outcomes.

This is why your argument produces an answer that contradicts all available evidence.

 

[PixyMisa]

JS, in Bayesian inference for a hypothesis we combine data with the prior probability of the hypothesis to obtain an updated probability for the hypothesis, using Bayes theorm:

P(H|D) = P(D|H)P(H) / P(D) ,​

where H is the hypothesis and D is the data. The data, D, is always an event that has already occurred when we perform the analysis. According to you, that implies that P(D)=1, and therefore that P(D|H)=1. But if that were true, then P(H|D) would always equal P(H), which would imply that it is impossible to do Bayesian inference. No amount of data could ever affect our opinion about the plausibility of a hypothesis.

Yes, you're right, and this is where I got sidetracked earlier. The problem is not that P(Jabba) = 1 because Jabba is known to exist; it's that P(Jabba) has not been shown to be relevant to the formula.