|
|
Scott H schrieb:
> According to Laplace's treatment of the sunrise problem, if the sun rises
> 2,000,000 times, its probability of rising tomorrow is exactly
>
> 2,000,001/2,000,002.
>
Informally, I'd argue like this:
If we assume that there is a total number n of sunrises and at some
(i.e. a random) moment we read out the "sunrise counter" and see the
number k, what can we infer?
Of course n>=k, that's easy.
But also the probability that we are very late in the sequence is low.
I.e. our random moment will see a counter >(1-eps)*n only with
probability ~eps.
BTW, it will also see a counter <eps*n only with probability ~eps.
Without other prior knowledge (!), we might conclude e.g. that the
event 0.005*n < k < 0.995*n happens with a probability of 0.99 (or
99%).
Therefore an observed k=2,000,000 allows us to say that with 99% that
2,010,050<n<400,000,000.
You have to elaborate the details a bit more precise from this to
obtain the exact fraction (k+1)/(k+2).
With the same right, a k=40 year old person might infer that if he will
die at age n then
0.1*n < k happens with a probability 90%, i.e. he is quite (but not
very) sure that he won't live longer than another 400 years.
We see from this that there are a few problems with this type of
reasoning:
Firstly, a 10% probability of surviving more than 400 years is
ridiculous (and don't say that there might be surprising medical
progress made in the next ten years or so). Our a priori knowledge
about typical life spans of contemporary humans rules this out, and the
formula is only valid in complete absence of such a priori knowledge.
One would have to play around with Bayes a bit, maybe.
Secondly, one has to be careful about the total timespan (the n).
A 40 year old man may think about this probabilistic argument, but the
moment when he does so is not a random moment in his life. For example,
he would not have been able to do the calculations in his first few
years.
Likewise, mankind and probability theorists didn't come to existence
with the first sunrise.
> Help me understand this a little. If a man's child is locked behind a door,
> and the man bangs twice, is there only a 3/4 chance that he will bang it a
> third time?
If we are only given the numbers and absolutely no a priori knowledge
at all, yes, the best guess for the probability is (k+1)/(k+2) -- no
matter what the event is.
If the man has not banged at all, there is a 1/2 chance that he will
bang -- either he does or he doesn't.
Also, there's a 1/2 chance that a purple meteor will hit the door and
destroy it -- either it happens or it doesn't.
Wait a minute, there's also a 1/2 chance for a green meteor to hit the
door and destroy it.
And since at most one meteor can destroy the door, the events are
exclusive. Thus it is almost sure that a (green or purple) meteor will
destroy the door.
Looks like one has to be careful ;)
Best guesses need not always be good guesses.
Formally, if X is a random variable denoting the total number of
sunrises and we have some a priori probabilities P0(X=n) for all
natural numbers n, and we make an observation ("k sunrises up to now"),
we can adjust and obtain a posteriori probabilities P1(X=n) according
to Bayes rules:
P(X>=k | X=n)=0 if n<k and P(X>=k | X=n)=(n+1-k)/n if n>=k.
Then P1(X=n) = 0 for n<k and P1(X=n) = c * (n+1-k)/n * P0(X=n) where
the constant c is adjusted so that the probabilities sum up to 1.
All we have to do is find a good a priori distribution, and that is
problematic if infinitely many values may be taken because for a
natural choice one would require P(X=n) not to depend on n.
OTOH, if one somehow had an a priori distribution (e.g. geometric or
whatsoever) it should at least be true that P(X=n) is approximately
equal to P(X=m) if n and m are approximately equal.
With this we have
P1(X=k) = c* 1/k *P0(X=k)
P1(X=k+1) = c * 2/(k+1) * P0(X=k+1) ~ 2 P1(X=k)
P1(X=k+2) = c * 3/(k+2) * P0(X=k+2) ~ 3 P1(X=k)
This would merely imply that P1(X=k) is approximately less than 1/6.
You can sharpen this inequality by going a few steps further, but that
depends on how long you would trust P(X=k+d) ~ P(X=k).
I don't know if there is a proper a priori distribution such that we
obtain
P1(X=k) = 1/(k+2)
exactly.
You might wish to work out the details and see what happens with the
following a priori distribution:
Assume P0(X=n)=1/N for n=0, ..., N-1 and P(X=n)=0 otherwise, where N is
much larger than k.
|
|