The probability of a probability?

[alan_longor]

TEXT :
- We have an event called X with two possible outcomes that are outcome A and outcome B .
the probability for the outcome to be A is noted as p , where p is a real number in the range [0.0 ; 1.0].
we also note p' a real number representing the probability that p is in range [0.9 ; 1.0] .
- additional information "#" : the event X happened a billion times among which the outcome is always A.

QUESTIONS:

1)- without noting the additional information "#" , can a value be assigned to p' ?
(the student has to indicate the value and prove the existence of the value , if a value cannot be
assigned the student has to prove the non-existence of the value)

2)-
a)- if we note the additional information "#", does that change the previous value in anyway ?
b)- can we assign a new value to p' ?

 

[mfb]

What did you figure out so far?
Can you think of example systems that show this type of probability?
Have you heard of Bayesian statistics?

 

[alan_longor]

What did you figure out so far?
Can you think of example systems that show this type of probability?
Have you heard of Bayesian statistics?

thank you for the response sir .
well the answer to question one is 1/10 it's not that hard .
i am though unable to solve question 2 in anyway and i am unable to find a proper system to shape this problem . i know the bayes theorem we had that in probability , the probability of something given an other contributing factor , but i am unable to get that to serve here .

thank you .

 

[Stephen Tashi]

TEXT :
- We have an event called X with two possible outcomes that are outcome A and outcome B .
the probability for the outcome to be A is noted as p , where p is a real number in the range [0.0 ; 1.0].
we also note p' a real number representing the probability that p is in range [0.9 ; 1.0] .

This is either a very sophisticated problem involving concepts or a mis-statement of a problem involving Bayes' theorem that intends for the student to do some computations. As stated, the problem doesn't give any distribution for p. It only states the range of possible values of p. If the textbook author wants to pose a computation problem then he needs to give a distribution for p over its range - for example, he could say that p has a uniform distribution on [0,1].

- additional information "#" : the event X happened a billion times among which the outcome is always A.

1)- without noting the additional information "#" , can a value be assigned to p' ?
(the student has to indicate the value and prove the existence of the value , if a value cannot be
assigned the student has to prove the non-existence of the value)

That is a very sophisticated question - even if the distribution of p is given !

Given the distribution of p, one could compute the expected value of p', but can we call this expected value the unique value of p' ?

The sophisticated nature of the question may be unintentional. If this is a problem in an elementary textbook then I think the author would have said that p has a uniform distribution on [0,1] and want the student to compute the expected value of p' and declare it to be "the" value of p'.

To discuss the question in a sophisticated manner would be interesting because it involves the fundamentals of probability theory - sample spaces and sample spaces underlying conditional probabilities etc.

 

[alan_longor]

This is either a very sophisticated problem involving concepts or a mis-statement of a problem involving Bayes' theorem that intends for the student to do some computations. As stated, the problem doesn't give any distribution for p. It only states the range of possible values of p. If the textbook author wants to pose a computation problem then he needs to give a distribution for p over its range - for example, he could say that p has a uniform distribution on [0,1].That is a very sophisticated question - even if the distribution of p is given !

Given the distribution of p, one could compute the expected value of p', but can we call this expected value the unique value of p' ?

The sophisticated nature of the question may be unintentional. If this is a problem in an elementary textbook then I think the author would have said that p has a uniform distribution on [0,1] and want the student to compute the expected value of p' and declare it to be "the" value of p'.

To discuss the question in a sophisticated manner would be interesting because it involves the fundamentals of probability theory - sample spaces and sample spaces underlying conditional probabilities etc.

thank you sir ,
the problem is pointed at college students so i think the solution can be relatively complicated .
i wrote all of the information of the problem and there is no information about the distribution .
the most unusual thing though is that the book contains solutions for the problems , but that one in specific has no solution within the book .

 

[Stephen Tashi]

i wrote all of the information of the problem and there is no information about the distribution .

well the answer to question one is 1/10 it's not that hard .

Is 1/10 the answer given by the textbook ?

If so then I think the author intends the wording of the problem to imply that p is uniformly distributed on [0,1].

This thread is in the pre-calculus section, but is the problem from a college textbook that assumes the student knows how to do integration?

 

[alan_longor]

Is 1/10 the answer given by the textbook ?

If so then I think the author intends the wording of the problem to imply that p is uniformly distributed on [0,1].

This thread is in the pre-calculus section, but is the problem from a college textbook that assumes the student knows how to do integration?

the book is written for college students , and i was quite unable to understand most of what it contains , but this problem in specific seemed interesting to me , but there was no solution for it in the book . that's why i posted it here .
the '1/10' value is mine , and i think it's probably false .

 

[Delta2]

Usually the probability of an outcome is determined in a ... deterministic way, so we have that the (Probability of (probability of outcome A happening=p ))= 1 for some p in (0,1) and (Probability of (probability of outcome A happening=x ))= 0 for every other x in (0,1) except the p.

Anyway if we assume that p is not determined in a deterministic way and thus it has a distribution , and it is a uniform distribution of p, then for 1) it will be p'=1/10.

for 2) if we take into account that info, and from the definition of "a posteriori probability" we will have that the a posteriori probability of outcome A is very close to 1. Thus the new value for p' will also be close to 1.

 

[Ray Vickson]

thank you sir ,
the problem is pointed at college students so i think the solution can be relatively complicated .
i wrote all of the information of the problem and there is no information about the distribution .
the most unusual thing though is that the book contains solutions for the problems , but that one in specific has no solution within the book .

Using the Bayesian approach in this problem with a so-called uniform prior for $p$ gives an initial value of $p'= 0.1$, as you suggest. However, the posterior value of $p'$, given the experimental outcomes you describe, is quit a bit different: I get
$$ \text{posterior value of }\, p' = 1-0.2474955439 \times 10^{-45757490} $$
This is pretty close to 1: it is 0.999... with about 45,000,000 '9's after the decimal point.

You need calculus to get this, and a having access to results from a more advanced probability course would be helpful as well.

 

[Stephen Tashi]

the book is written for college students

Is it a mathematics textbook about probability ? - or a collection of varied mathematical problems? - or a collection of problems from all sorts of disciplines ?

I agree that the problem is interesting. I think elementary textbooks deliberately avoid posing problems that explicitly involve "the probability of a probability".

the '1/10' value is mine , and i think it's probably false .

If we assume p is uniformly distributed on [0,1] then I agree with your answer.

As to question 2), are you interested in a qualitative answer or in detailed computations?

 

[mfb]

The answer to problem 2 is easy: it is the same for all reasonable priors. The answer to problem 1 is more interesting.

If we assume p is uniformly distributed on [0,1] then I agree with your answer.

There is no reason to assume this, however.

Take a coin, the two sides are the outcomes A and B. What is your prior for its distribution of p?

 

[haruspex]

Is 1/10 the answer given by the textbook ?

Sounds to me as though @alan_longor thinks that is obviously the answer, but the textbook does not provide one.
If so, I suggest that no prior should be assumed and that the point of the first part is to prove that no value can be assigned.

 

[alan_longor]

my problem is the fact that this book seems to contain some absurd things that i have never heard of , here is an other problem i came across that is supposed to follow that one . this one got me confused even more .TEXT :

we have an event X that has two possible outcomes A and B . for the purpose of this problem we will set a definition , a "special" event means to us now an event that we can set a probability for it's outcome , which means that if X is a special event , there exists a real number p where p would be in range [0.0 ; 1.0] and where p represents the probability for the outcome of X to be A , if the event is not special we cannot assign a fixed real number as a probability for it's outcome .

> the following is only meant for demonstration purposes :
> let the event X happen N times , where n is the number of times that A is the outcome .
> if X is special then if N -> oo then n/N -> p , where p is a real fixed number in range [0.0 ; 1.0]
> if X is not special then if N -> oo then n/N is indefinite but always remains in range [0.0 ; 1.0]


now let p' be a real number in range [0.0 ; 1.0] that describes the probability that event X is special , which is the probability that p exists .

additional information # : event X happened a billion times among which the outcome is always A .

QUESTIONS :

1)- without noting the additional information "#" , can a value be assigned to p' ?
(the student has to indicate the value and prove the existence of the value , if a value cannot be
assigned the student has to prove the non-existence of the value)

2)-
a)- if we note the additional information "#", does that change the previous value in anyway (if it exists) ?
b)- can we assign a new value to p' (if it exists) ?

3)- if we note the additional information "#" , let p" be a real number in range [0,0 ; 1.0] where p" would represents the probability that the outcome of the next attempt is B ( attempt number 10^(9) + 1 ) (if p" exists)
a)- can p" exist for any value of p' (if p' exists) ?
b)- can we assign a value to p" ?

 

[haruspex]

my problem is the fact that this book seems to contain some absurd things that i have never heard of , here is an other problem i came across that is supposed to follow that one . this one got me confused even more .TEXT :

we have an event X that has two possible outcomes A and B . for the purpose of this problem we will set a definition , a "special" event means to us now an event that we can set a probability for it's outcome , which means that if X is a special event , there exists a real number p where p would be in range [0.0 ; 1.0] and where p represents the probability for the outcome of X to be A , if the event is not special we cannot assign a fixed real number as a probability for it's outcome .

> the following is only meant for demonstration purposes :
> let the event X happen N times , where n is the number of times that A is the outcome .
> if X is special then if N -> oo then n/N -> p , where p is a real fixed number in range [0.0 ; 1.0]
> if X is not special then if N -> oo then n/N is indefinite but always remains in range [0.0 ; 1.0]


now let p' be a real number in range [0.0 ; 1.0] that describes the probability that event X is special , which is the probability that p exists .

additional information # : event X happened a billion times among which the outcome is always A .

QUESTIONS :

1)- without noting the additional information "#" , can a value be assigned to p' ?
(the student has to indicate the value and prove the existence of the value , if a value cannot be
assigned the student has to prove the non-existence of the value)

2)-
a)- if we note the additional information "#", does that change the previous value in anyway (if it exists) ?
b)- can we assign a new value to p' (if it exists) ?

3)- if we note the additional information "#" , let p" be a real number in range [0,0 ; 1.0] where p" would represents the probability that the outcome of the next attempt is B ( attempt number 10^(9) + 1 ) (if p" exists)
a)- can p" exist for any value of p' (if p' exists) ?
b)- can we assign a value to p" ?

It's tough to get your head around, but it does not strike me as absurd. Not sure that I can explain it any more clearly, but if you care to indicate where you get lost and why then I'll try.

 

[Stephen Tashi]

TEXT :

we have an event X that has two possible outcomes A and B . for the purpose of this problem we will set a definition , a "special" event means to us now an event that we can set a probability for it's outcome , which means that if X is a special event , there exists a real number p where p would be in range [0.0 ; 1.0] and where p represents the probability for the outcome of X to be A , if the event is not special we cannot assign a fixed real number as a probability for it's outcome .

This doesn't sound like material from routine probability text! What is the title of this book?

The only interpretation of that text than I can make in terms of probability theory is that a "special" event is an event with a constant probability that can be subjected to an independent set of trials. An non-special event could be interpreted as an "event" in the sense of common speech - for example, "The traffic light at the intersection near my house was red when I got to it". We can conduct a set of trials of a non-special event, but the result doesn't have the same probability of occurrence on each of the trials.

> the following is only meant for demonstration purposes :
> let the event X happen N times , where n is the number of times that A is the outcome .
> if X is special then if N -> oo then n/N -> p , where p is a real fixed number in range [0.0 ; 1.0]
> if X is not special then if N -> oo then n/N is indefinite but always remains in range [0.0 ; 1.0]

Those statements illustrate the difference between "probability" and "observed frequency".

now let p' be a real number in range [0.0 ; 1.0] that describes the probability that event X is special , which is the probability that p exists .

Another way to put the same question is "Given an event X, let E be the event that X is a special event. Is E a special event or a non-special event?"

 

[haruspex]

An non-special event could be interpreted as an "event" in the sense of common speech - for example, "The traffic light at the intersection near my house was red when I got to it". We can conduct a set of trials of a non-special event, but the result doesn't have the same probability of occurrence on each of the trials

I don't think that would be sufficiently badly behaved. Over arbitrarily long trials, the frequency would still converge. The events are, at base, controlled by atomic events with fixed probabilities.
We can take the definition of nonspecial as being that the probability does not converge. Imagine e.g. a demon who is tracking the outcomes and adjusting the probability. If the frequency up to some point is f, the demon could then set the probability to a value f' that differs from f by 0.5. As the trials proceed, the frequency (over the entire sequence) approaches f', whereupon the demon adjusts it again.

 

[Stephen Tashi]

I don't think that would be sufficiently badly behaved. Over arbitrarily long trials, the frequency would still converge. The events are, at base, controlled by atomic events with fixed probabilities.

I'll disagree. We don't know that an arbitrarily chosen event is controlled by atomic events with fixed probabilities.

Defining a "repeatable event" (or repeatable experiment) is essentially a contradiction in terms because if we exactly repeated something, it would not be a "different" event or experiment. So when we define a "repeatable" event we specify a set of different things that will be regarded as "the same event". This necessary ambiguity in the definition of an repeatable event is sufficient to produce a situation where the probability of an event on a given trial is not constant. The probability that a result occurs on a given trial is distinct from the probability that the result occurs on a randomly selected trial.

We can take the definition of nonspecial as being that the probability does not converge. Imagine e.g. a demon who is tracking the outcomes and adjusting the probability. If the frequency up to some point is f, the demon could then set the probability to a value f' that differs from f by 0.5. As the trials proceed, the frequency (over the entire sequence) approaches f', whereupon the demon adjusts it again.

That's a good idea, but it only pushes the requirement of being non-special onto the demon. Can the demon be implemented by a deterministic algorithm?

If there was a demon, "Even Steven", that implemented the commonly held fallacy that a strings of heads in tosses of a coin should made a tail more likely then Even Steven could produce a frequency of heads near 0.5, but the event "head" wouldn't have the same probability on each trial. If we pick one of the tosses of Even Steven at random then the probability that the result is heads is about 0.5.

(I can see why the book didn't try to answer to this probem!)

 

[haruspex]

Can the demon be implemented by a deterministic algorithm?

Yes, I specified it, pretty much. We can make it simpler. The demon fixes the outcomes as 01100001111111100000000000000... Clearly the "average over the first n" gives a sequence that does not converge.

Even Steven could produce a frequency of heads near 0.5, but the event "head" wouldn't have the same probability on each trial.

This is a somewhat different issue, that of probability given what? The initial, description of special event doesn't specify whether the history of preceding outcomes is known. If it is unknown then Even Steven does achieve an evens chance on any given trial. But that description is not really a usable definition. The convergence definition provided later says that Even Steven is 'special'. In fact he would be extra special, converging more rapidly than independent coin tosses.

 

[Stephen Tashi]

Yes, I specified it, pretty much. We can make it simpler. The demon fixes the outcomes as 01100001111111100000000000000... Clearly the "average over the first n" gives a sequence that does not converge.

I see what you mean.

The convergence definition provided later says that Even Steven is 'special'. In fact he would be extra special, converging more rapidly than independent coin tosses.

I think it's a convergence theorem instead of a definition. It's also a garbled theorem because it omits reference to probability. It guarantees that n/N "approaches" p as N approaches infinity. It doesn't define what "approaches" means in that context. It isn't a statement of "The Law Of Large Numbers".

 

[haruspex]

I think it's a convergence theorem instead of a definition. It's also a garbled theorem because it omits reference to probability. It guarantees that n/N "approaches" p as N approaches infinity. It doesn't define what "approaches" means in that context. It isn't a statement of "The Law Of Large Numbers".

Yes, the notion of convergence in the context of random variables needs to be defined first.

 

[alan_longor]

i took the problems from a series called "Serie de problemes (Ecole normale de lyon)" it was printed from an original in a local university .
i sent the problems to a friend of a friend and he sent me back a paper in french , he was unable to solve the second problem but actually solved the first one , i hope i can translate this well :

We have the event X and we know nothing about it , we also know nothing about the uniformity or continuity of p . this system is relatively simple since it only has a couple of outcomes . it is safe to consider p a number between [0.0;1.0] if we don't the extra information . it can take any real value between 0 and 1 therefore if we are asked what's the probability that it takes a value between 0.9 and 1 we can just say 1/10 . i know this might seem absurd and blunt since we know nothing about the distribution of p , though that's the same reason that allows us to assign the value 0.1 in the first place since like i said , p to us is just a number for now since we know nothing about the probability relied to it , i will return to this point later .

now we are presented with the additional information , X happened a billion times and the outcome is always A . now this is why the notion of p' was presented in the first place , since if we don't know the probability of an event but we know the result of some of it's happenings we can in some way predict the upcoming results so the notion of p' is relatively logical .

now the question is how can we know the probability that p is in range [0.0 ; 0.9] , well to simplify this l'ets go to an other simpler case , what if the question was : we have an event X with outcomes A and B . and p is the probability of A , now p can either be 0.5 or 0.7 and nothing else . and let p' be the probability that
p is equal to 0.5 . and then X happens once and it's outcome is A . if p would be equal to 0.5 the probability for the previous event should be 0.5 and it's the same case for 0.7 . the probability for p to be equal to 0.7 is therefore higher than the probability for p to be equal to 0.5 since the event happened .
let p" be the probability that p=0.7 . we can then say p' + p" = 1 (two possibilities) . or due to proportionality we have : p' = 0.5/0.7p" and p" = 0.7/0.5 p'
then we have p' = 0.5 / (0.5 + 0.7) . now what if X happened a lot more than once , well l'ets say X happened N times ,and A was the outcome for n times .
we previous should shange to p' = ((0.5)ⁿ(1-0.5)^N-n)/((0.7)ⁿ(1-0.7)^N-n) * p" and we then have
p' = ((0.5)ⁿ(1-0.5)^N-n)/(((0.7)ⁿ(1-0.7)^N-n) + ((0.5)ⁿ(1-0.5)^N-n))
well now what if the event has 4 probable probabilities instead of two ? which means now that the probability p is either 0.1 0.2 0.3 or 0.4 for example .
then if we logically follow the previous we should have : ((( notation : P(p=X) is the probability that p is equal to X )))
so if X happens only once and the outcome is A we have :
we should then have P(p=0.1) = (0.1/0.2)P(p=0.2) = (0.1/0.3)P(p=0.3) = (0.1/0.4)P(p=0.4) and it so on for the other values
with the following P(p=0.1) + P(p=0.2) + P(p=0.3) + P(p=0.4) = 1.0 since p can only take 4 possible values , so we can deduce that .
so if p' would be P(p=0.1) we would have through a simple deduction that : p' = 0.1 / (0.1 + 0.2 + 0.3 + 0.4) and the same follows the previous if X happens N times among which the outcome is always A we have:p' = (0.1)^N / ( (0.1)^N + (0.2)^N + (0.3)^N + (0.4)^N ) . we should also note that P(p≥0.2) = P(p=0.2) + P(p=0.3) + P(p=0.4) , this one is quite simple to understand since p can only take four values for now .

now if we have that clear already . we can comeback to our question ... here though p can take any value between 0.0 and 1.0 and so there are an infinite
number of values p can take , so if X only happens once with A as outcome we have : P(p=0.1) = 0.1 / (0 + .... + 1) with the dots representing every single real number between 0 and 1 the thing is there is an infinite number of those , to make matters even worse , X has to happen N times and A is the outcome only for n times . to solve this we need to use an integral . first we declare ƒ(x)_N,n = (x)ⁿ(1-x)^N-n
so if X happens according to N and n we have P(p=0.1) = ƒ(0.1) / (ƒ(0) + ... + f(1)) .
for the problem we need p' , we have p' = P(0.9≤p≤1) which is also an infinite sum
p' = P(p=0.9) + ... + P(p=1.0) and since the denominator is the same (luckily)we have p' = (ƒ(0.9) + ... + ƒ(1)) / (ƒ(0) + ... + ƒ(1))
which is an infinite sum over an other . and which simply translates to the following,

p' = ∫_0.9¹ ƒ(x)_N,n dx / ∫₀¹ ƒ(x)_N,n dx
in our case N = n = 10⁹ so p' can simply be reduced to
p' = [x¹⁰⁹⁺¹]_0.9^1.0
so this concludes to p' = 1 - (0.9)¹⁰⁹⁺¹
now for me to return to that previous point , i should point out that if N = 0 which means X never happened it's safe to assume that p' = 1/10 . though if N>0 p=1/10 might no longer be valid

 

[Ray Vickson]

i took the problems from a series called "Serie de problemes (Ecole normale de lyon)" it was printed from an original in a local university .
i sent the problems to a friend of a friend and he sent me back a paper in french , he was unable to solve the second problem but actually solved the first one , i hope i can translate this well :

We have the event X and we know nothing about it , we also know nothing about the uniformity or continuity of p . this system is relatively simple since it only has a couple of outcomes . it is safe to consider p a number between [0.0;1.0] if we don't the extra information . it can take any real value between 0 and 1 therefore if we are asked what's the probability that it takes a value between 0.9 and 1 we can just say 1/10 . i know this might seem absurd and blunt since we know nothing about the distribution of p , though that's the same reason that allows us to assign the value 0.1 in the first place since like i said , p to us is just a number for now since we know nothing about the probability relied to it , i will return to this point later .

Saying that $p'=0.1$ is the initial estimate of $p'$ is not absurd, but it does assume that the prior distribution of $p$ is uniform (pr [pssibly has some other distribution that integrates to 0.1 from $p =0.9$ to $p= 1$. If you do assume a Bayesian approach with a uniform prior, then the posterior distribution of $p$, given $N$ observations of X in $N$ trials is given by Bayes agaiin:
$$f(p|N) = \frac{P(N|p) f_0(p)}{P_0(N)}, $$
where $f_0(p)$ is the prior density of $p$, $P(N|p)$ is the probability of the event $N$, given a value for $p$, and
$$ P_0(N) = \int_0^1 f_0(p) P(N | p) \, dp$$
is the "prior" estimate of the probability of event $N$. For a uniform prior $f_0(p) = 1$ we have
$$ P_0(N) = \int_0^1 p^N \, dp = \frac{1}{N+1}$$.
Thus, the posterior density of $p$ is
$$f(p|N) = (N+1) p^N, \; 0 \leq p \leq 1$$.
The posterior probability that $p \geq a$ is
$$ P(p \geq a|N) = \int_a^1 (N+1) p^N \, dp = 1 - a^{N+1}$$
For $a = 9/10$ and $N = 10^9$ we get the result I wrote in post #9. Of course, in this universe we might as well take $p' = 1$, since it is essentially ridiculous to give results with 45 million decimal-place precision when we are really dealing with an approximate real-world model at best.

 

[haruspex]

it can take any real value between 0 and 1 therefore if we are asked what's the probability that it takes a value between 0.9 and 1 we can just say 1/10

The argument here seems to be that in our general experience all probabilities, from 0 to 1, occur equally often. That is clearly not true. In particular, many processes are discrete and have rational probabilities; many processes are deterministic, i.e. have probabilities 0 or 1.
So I find it completely unconvincing and maintain my view that the questioner is expecting a proof that no probability can be assigned.

For the second question, as mfb posted (#11), this is not hard. Just consider the hypothesis p<0.9 and what that gives for the probability of the observation. Then apply Bayes' theorem.

 

[haruspex]

If you do assume a Bayesian approach with a uniform prior,

Big If. I see no justification for it when the nature of the event is completely unknown.

 

[alan_longor]

The argument here seems to be that in our general experience all probabilities, from 0 to 1, occur equally often. That is clearly not true. In particular, many processes are discrete and have rational probabilities; many processes are deterministic, i.e. have probabilities 0 or 1.
So I find it completely unconvincing and maintain my view that the questioner is expecting a proof that no probability can be assigned.

For the second question, as mfb posted (#11), this is not hard. Just consider the hypothesis p<0.9 and what that gives for the probability of the observation. Then apply Bayes' theorem.

interesting ... what about the second problem ? can it be solved ?
though i might agree with you that 1) of problem 1 needs a deeper proof , because the book says that these problems are designated for research purposes so the resolution might be complicated

 

[Ray Vickson]

interesting ... what about the second problem ? can it be solved ?
though i might agree with you that 1) of problem 1 needs a deeper proof , because the book says that these problems are designated for research purposes so the resolution might be complicated

If the problem is an "abstract" one, essentially without context and just given for practice, then, of course, there is no real argument to be made either way: a uniform prior is just as good (or just as bad) as any other. In fact, we are here in the area known roughly as "subjective probability", and it is perfectly OK and common for Mr. Jones to use one prior and Mr. Smith to use another.

In practice, if we are looking at an actual application area (whether physical, economic or social) it may well be that uniform priors are not appropriate: it may be that anybody with expertise in the subject will use a non-uniform prior. Still, even in that case, the priors of two different experts, Miller and Sandhu, may be different. There again, there is in principle no way to say "this is the correct prior and the other is wrong". It remains "personal". It may be informed by knowing about lots of past data about similar situations, but absent an example like yours (where a billion successes occur a billion trials) there will always be some uncertainty. That is just the nature of probability and statistics in the real world.

 

[haruspex]

what about the second problem ? can it be solved ?

I answered that:

For the second question, as mfb posted (#11), this is not hard. Just consider the hypothesis p<0.9 and what that gives for the probability of the observation. Then apply Bayes' theorem.

Edit: just realized you said second problem, not the second part if the first problem.
(This is one reason we encourage a new thread for a new problem.)

I'll study it later.

 

[alan_longor]

If the problem is an "abstract" one, essentially without context and just given for practice, then, of course, there is no real argument to be made either way: a uniform prior is just as good (or just as bad) as any other. In fact, we are here in the area known roughly as "subjective probability", and it is perfectly OK and common for Mr. Jones to use one prior and Mr. Smith to use another.

In practice, if we are looking at an actual application area (whether physical, economic or social) it may well be that uniform priors are not appropriate: it may be that anybody with expertise in the subject will use a non-uniform prior. Still, even in that case, the priors of two different experts, Miller and Sandhu, may be different. There again, there is in principle no way to say "this is the correct prior and the other is wrong". It remains "personal". It may be informed by knowing about lots of past data about similar situations, but absent an example like yours (where a billion successes occur a billion trials) there will always be some uncertainty. That is just the nature of probability and statistics in the real world.

ok , l'ets say we see a bright light in the sky that comes up exactly at 3pm every day for a billion days , we know nothing about the light nor where it comes from . so if i ask you on day number one billion what's the probability that the light will come up tomorrow to ... is that a meaningful question ?

 

[haruspex]

ok , l'ets say we see a bright light in the sky that comes up exactly at 3pm every day for a billion days , we know nothing about the light nor where it comes from . so if i ask you on day number one billion what's the probability that the light will come up tomorrow to ... is that a meaningful question ?

Sure, but you do not need a uniform prior for that. This what mfb meant when he mentioned 'any reasonable' prior distribution. You only need a prior which is not massively biased against the light coming up tomorrow.

 

[alan_longor]

Sure, but you do not need a uniform prior for that. This what mfb meant when he mentioned 'any reasonable' prior distribution. You only need a prior which is not massively biased against the light coming up tomorrow.

thank you , so what would the probability be ?

 

[haruspex]

thank you , so what would the probability be ?

Strictly speaking, it will depend on the prior. But suppose your prior is 1000:1 against. What posterior likelihood do you get?

 

[alan_longor]

Strictly speaking, it will depend on the prior. But suppose your prior is 1000:1 against. What posterior likelihood do you get?

well one of the conditions was that we know nothing about the light , the only thing we know is that it came up for one billion nights . now when i asked weather or not we can set a probability for it coming up the next day , that means weather or not the information we have is enough to set a probability . if something else has to be known then we cannot set it . in either cases logically the probability must be a number very close to 1 .

 

[haruspex]

well one of the conditions was that we know nothing about the light , the only thing we know is that it came up for one billion nights . now when i asked weather or not we can set a probability for it coming up the next day , that means weather or not the information we have is enough to set a probability . if something else has to be known then we cannot set it . in either cases logically the probability must be a number very close to 1 .

I assume you have been introduced to Bayes' theorem. This tells you that you cannot estimate the probability of anything from observations unless you start with a prior estimate.

 

[alan_longor]

I assume you have been introduced to Bayes' theorem. This tells you that you cannot estimate the probability of anything from observations unless you start with a prior estimate.

ok , then in this case we cannot set a probability for the appearance of the light on the next night .

 

[Stephen Tashi]

so if i ask you on day number one billion what's the probability that the light will come up tomorrow to ... is that a meaningful question ?

It may be a meaningful question, but it isn't a mathematical question unless you state a specific probability model.

Even if you state a probability model, it isn't a solvable mathematical question unless there is sufficient given information.

People who wish to demonstrate that mathematics can be applied in a way that agrees with "common sense" can reformulate the question in various ways by creating a probability model for it that gives sufficient information to create a solvable mathematical problem.