Bayesian inference is a method of statistical inference in which Bayes' theorem is used to update the probability for a hypothesis as more evidence or information becomes available.
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H stands for any hypothesis whose probability may be affected by data (called evidence below). Often there are competing hypotheses, and the task is to determine which is the most probable.
Pr(H), the prior probability, is the estimate of the probability of the hypothesis H.
I agree. I think it is most convincing to present multiple angles. But it isn't (just) the sample space that is the issue. It is the probability space which includes: — JeffJo
If the initial set up calls for randomly assigning values for the two envelopes in the finite range ((1,2),(2,4),(4,8)) for instance, then, in that case, assuming the player knows this to be the initial set up (and hence uses it as his prior) then the posterior probability conditionally on observing any value of M that isn't either 1 or 8 (that is, conditionally on values 2 or 4 being observed) p will indeed be 1/2. — Pierre-Normand
You are misinterpreting what I said, what your link says and your source is Wikipedia. — Jeremiah
The point is that, in the correct version for your calculation E=($V/2)*P1 + ($2V)*P2, the probability P1 is not the probability of picking the larger value. It is the probability of picking the larger value, given that the larger value is $10. In my example, that 90%. In the OP, you do not have information that will allow you to say what it is. — JeffJo
But averaged over all possible values of V, there will be no expected gain. — JeffJo
I think the confusion comes when you switch from
E(other) = (larger)P(picked smaller) + (smaller)P(picked larger)
where the probabilities of picking smaller and larger are equal, to
E(other | picked = a) = (2a)P(picked smaller | picked = a) + (a/2)P(picked larger | picked = a)
because it's tempting to think these conditional probabilities are equal, just like the unconditional probabilities above, but this we do not know. — Srap Tasmaner
If you do look, THERE IS PROBABLY AN EXPECTED GAIN OR LOSS, but you have no information that would let you calculate it. This is different from knowing it is 0. — JeffJo
So since you don't know which case you are in after seeing Y and they are not equal you can't really calculate the expected value. Now if you never opened A and never saw Y, that is a different story. — Jeremiah
Where I differ from that perspective is that I reject the notion that there is such a thing as a 'real' probability (aka 'true', 'raw', 'correct', 'absolute' or 'observer independent' probability).No, it gives you a strategy that works on your assumed prior, not necessarily on reality. — JeffJo
Isn't this the same as: — Michael
I did read the thread. You did not read my replies. Like this one, where I said "you have no information that would let you calculate [the expected gain or loss]" and you replied with "you can't really calculate the expected value" as if I hadn't just said the same thing.If you do look, THERE IS PROBABLY AN EXPECTED GAIN OR LOSS, but you have no information that would let you calculate it. This is different from knowing it is 0. — JeffJo
So since you don't know which case you are in after seeing Y and they are not equal you can't really calculate the expected value. Now if you never opened A and never saw Y, that is a different story. — Jeremiah
You did not read the thread. — Jeremiah
I differ from that perspective is that I reject the notion that there is such a thing as a 'real' probability (aka 'true', 'raw', 'correct', 'absolute' or 'observer independent' probability). — andrewk
You did not read my replies. Like this one, where I said "you have no information that would let you calculate [the expected gain or loss]" and you replied with "you can't really calculate the expected value" as if I hadn't just said the same thing. — JeffJo
And if I see £10 then I stand to gain £10 and I stand to lose £5. — Michael
You should recognize that I have fixed the amount you will gain or lose by switching; you cannot know whether you chose the larger or the smaller, so you cannot know whether you will gain or lose that fixed amount by switching, so there is no reason either to switch or to stick. — Srap Tasmaner
If there's £10 in my envelope and I know that the other envelope contains either £5 or £20 because I know that one envelope contains twice as much as the other then I have a reason to switch; I want an extra £10 and am willing to risk losing £5 to get it. — Michael
You may entertain yourself by switching and call that a reason, but there is no expected gain from switching. — Srap Tasmaner
There is no basis whatsoever for either of these assumptions. — Srap Tasmaner
In broad terms I do not disagree with that characterisation. But there is often more than one way to represent uncertainty, and these lead to different probability spaces. I have referred previously to the observation that in finance many different, mutually incompatible probability spaces can be used to assign a value to a portfolio of derivatives. To try to mount an argument that a particular probability space is the sole correct probability space for analysing a problem, one would have to make a bunch of assumptions to start and, as we see from the length of this interesting thread, those assumptions are rarely uncontroversial.Probability is, essentially, a measure of our uncertainty about a result. — JeffJo
I am not an advocate for that expectation formula, so I don't see why you'd think I am avoiding those objections to it.And what you seem to be avoiding with that attitude, is that the expectation formula (v/2)/2 + (2v)/2 is already assuming: — JeffJo
What I don't understand is what your argument is against the alternative analysis. Which of these do you not accept?
1. The envelopes are valued at X and 2X, for some unknown X.
2. You have a 1/2 chance of picking the 2X envelope.
3. If you trade the 2X envelope, you lose X.
4. You have a 1/2 chance of picking the X envelope.
5. If you trade the X envelope, you gain X. — Srap Tasmaner
I accept all of them. I reject the implicit conclusion that the gain and loss are symmetric. If my envelope contains £10 then 3. and 5. are:
3. If I trade the 2X = 10 envelope, I lose X = 5.
5. If I trade the X = 10 envelope, I gain X = 10. — Michael
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