More on the stopped clock theorem
You look at your watch and see the time is 10am. You then see a stopped clock, which shows that the time is X ≠ 10am. Using Bayes' Rule, to what do you now revise your opinion of what time it is?
It's still 10am, isn't it?
Furthermore:
Lemma: All stopped clocks are independent of each other.
Therefore:
(1) Either: the conditional probability (T=X| [A stopped clock says that T=X]) = 0
(2) Or: there is some, possibly infinite, vector V of stopped clocks showing times [X1, X2 ...], such that if you saw V when your watch said 10am, then you would change your mind about the time.
The second can't be right, so the conditional probability that T=X given that a stopped clock says that T=X is zero; ie, "A stopped clock is never right" QED.
Whatcha think about that? Answer in comments.
Tangentially semi-related link
IF: You thought that I am probably still playing the same foolish game with the concept of a limit (or that I have still not bothered to look up Lebesque integrals)
ReplyDeleteYOU ARE A MATHEMATICIAN.
IF: You immediately started thinking about a possible world on which a sudden sonic blast or electromagnetic pulse had destroyed every timepiece on earth simultaneously, and my watch was showing 10:05 when all the other clocks in the world were showing 10:00
ReplyDeleteYOU ARE A PHILOSOPHER
IF: Your answer is the same as 2), but you gratuitously also assumed that the entire population of the earth had been killed in the sonic blast.
ReplyDeleteYOU ARE A MORAL PHILOSOPHER.
IF: You asked "what about when the clocks go forward"?
ReplyDeleteYOU ARE AN ECONOMIST.
If you read the above and think "g is nowhere near q on a keyboard" then you are a very bored and very pedantic mathematician who thinks that conditioning on events of null measure should be punishable by a slap with a wet haddock.
ReplyDeleteAssuming that the conditional probability (T=X| [A stopped clock says that T=X]) = 0, wouldn't it follow that there is at least one stopped clock about which, if your watch said 10 am, seeing it (the clock) would make you change your mind? Specifically, a stopped clock that says 10 am? Because then you would know your watch was wrong.
ReplyDeleteSince the time shown by a stopped clock is independent of the actual time, the conditional probability ([A stopped clock says that T=X] | T=X) must equal the probability [A stopped clock says that T=X]. Therefore, by Baye's Rule, if (T=X| [A stopped clock says that T=X]) = 0, then the probability that T=X = 0. So unless it is never any time at all, the vector V you describe must exist, its rightness notwithstanding.
ReplyDeleteI think, "n == 1; what's for lunch?"
ReplyDeleteIf you know the time is 10 am, and you look at a stopped clock that shows 10 am, P(T=X|Stopped clock shoes T=X) = 1.
ReplyDeleteThat stopped clock is in the vector V. [Square]
Shoes should be shows.
ReplyDeleteI just think that "11.59" would have been a much better Blondie track to accompany your measure-theoretic ruminations.
ReplyDeleteYou look at your watch and see the time is 10am
ReplyDeleteO RLY?
You look at your watch, and see that your watch says the time is 10am. What do the EPISTEMOLOGISTS have to say about that?
What about if you look at your watch and the clock and then consider you need at least one more source of data to verify which timepiece is stopped?
ReplyDeleteAn excellent theorem.
ReplyDeleteLet me point out an interesting corollary: it is never 10 AM.
Proof.
Get a clock, take out the battery, and set it 10 AM. By D2's theorem, this clock must be wrong. Therefore it is not 10AM. QED.
The set of all computable numbers (being the numbers which can be computed to arbitrary but finite position in arbitrary but finite time) is countable (proof by lexical induction over programs); countable sets are of measure zero on the real line (proof by covering with intervals of size given by powers of epsilon).
ReplyDeleteTherefore any time anyone makes any claim at all involving actual numbers I simply wag my tail and dismiss it as irrelevant cherry picking.
So unless it is never any time at all, the vector V you describe must exist, its rightness notwithstanding
ReplyDeleteI take the other fork; it is never any time (or at least, it is never any specific time like 10am). So while there might exist a clock which showed the correct time (a "working clock"), it would have to be such that you never looked at it, or it didn't have hands or something, because if it ever told you a specific time, then it would be wrong.
I therefore propose a reform in horology such that all watches and clocks have slightly forked hands, thus indicating that the correct time is between the two points of the fork. More accurate watches could have the tines of the fork closer together - the Swiss watchmakers could compete on this front.
ReplyDeleteThen the proverb would be "A stopped clock is correct for a certain interval every day equal to the span of time covered by the fork in its minute hand, although there are degenerate cases in which this 'rightness' is trivial"
I conclude that the only correct clock is one with no hands, citing Zizek (who I think was in turn citing Freud) who noted that an unaddressed letter always arrives at its destination.
Des: 11:59 would have been the wrong choice (or at least, the correc choice for a period of infinitesimal duration). "Call Me" contains the line "Anytime, any place, anywhere", and thus correctly describes the conditions under which a stopped clock is not right.
p(T=X | [A stopped clock says that T=X]) = p([A stopped clock says that T=X] | T=X)p(T=X) / p([A stopped clock says that T=X]) (by Bayes' Theorem)
ReplyDeleteBut p([A stopped clock says that T=X] | T=X) = p([A stopped clock says that T=X]) (By your Lemma)
So p(T=X | [A stopped clock says that T=X]) = p(T=X)
Which is more or less what you'd expect.
Note, however, that since we are assuming T is continuous, p is a "probability density" rather than "probability", and so p(T=X) > 0.
For clarity: do we see a stopped clock or do we see a stopped clock and see that it is stopped?
ReplyDeleteI think that your contrarianism is getting the better of you.
ReplyDeleteSo a stopped clock is never right because it is a clock, rather than because it is stopped.
ReplyDeleteI kind of like that, actually.
I therefore propose a reform in horology such that all watches and clocks have slightly forked hands, thus indicating that the correct time is between the two points of the fork.
ReplyDeleteI counter-propose a more modest reform, that all watches and clocks have hands with non-zero width at their outer terminating edges, thereby indicating that the correct time is along the width of the outside edge. I have already converted my wristwatch to such a state, and my feeling of added security in its reliability is indescribable. I am now an expert in this process, and offer to perform a similar conversion on any clock or watch for a modest fee.
I suppose we could save the expense (or the commission to Marc) by adopting a) digital watches and b) the convention suggested in the last stopped clocks thread that a digital clock showing 10:00 means "no earlier than 10:00 and no later than 10:00+x^0", where x is the least significant updating digit on the clock face. Obviously this does raise the question of what time it is at the bounds of the sets so described, but since this never happens (proof: see above), I doubt it's important.
ReplyDeleteI would then save my proposition by saying that when applied to a stopped clock, the least significant updating digit does not exist, so it is always wrong.
By the way, I realise that this is all the most tremendous bullshit, but I'd be interested in any criterion which would both explain why the discussion of whether a stopped clock is ever right is bullshit, but which would classify more than say 50% of a given modern epistemology journal as "not-bullshit". (Note: if your criterion is "contains loads of fallacious and/or unrigorous mathematics", I doubt I'll accept).
In the spirit of various such questions over the years, I will award X points for such a criterion and 128X points for the opposite proof - that a discussion (not necessarily this one) of the proposition "A stopped clock is right twice a day" could be other than valueless.
By the way, I realise that this is all the most tremendous bullshit, but I'd be interested in any criterion which would both explain why the discussion of whether a stopped clock is ever right is bullshit, but which would classify more than say 50% of a given modern epistemology journal as "not-bullshit"
ReplyDeleteAre you familiar with David Stove: http://web.maths.unsw.edu.au/~jim/wrongthoughts.html
Yes I am! Good on jokes, good on Hume and causation, really bad on Darwin IIRC.
ReplyDeleteIf you know that a clock is stopped, then, yes, it is never right and you should never use it to determine the time.
ReplyDeleteBut 'stopped' is an exact model for the clock's behavior for all time, and can never be proved from observation-- to use that model, you must have certain knowledge of how the clock works (or doesn't). Without that certain knowledge, you must use the stopped clock's readings. The error you get from using the stopped clock's readings is just a part of the price you pay for trying to tell time.
If you thought this was an attempt to shift the discussion from something DD doesn't understand (probability spaces) to something that he does, while maintaining a patina of intellectual curiosity and breadth
ReplyDeleteYOU ARE ACCURATE
Seriously, you like to talk about probability, so why not actually learn about it?
Goldberg’s Law: The man with one watch always knows the time. The man with two watches is never quite sure.
ReplyDeletehttp://www.dailyhowler.com/dh111009.shtml
FWIW, this is where I’d come from on this. Dan says:
ReplyDeletethe conditional probability (T=X | [A stopped clock says that T=X]) = 0
and attempts to base a conclusion on it. I think this is wrong as an argumentative strategy. Let’s say we believe that T=X with probability 0.9. Why wouldn’t we then say that we believe that (T=X given that a stopped clock says that T=X) with probability 0.9? After all, we might as well say that we believe that (T=X given that the Kraken are rising) with probability 0.9.
Alternatively, consider:
(1) T=X
(2) a stopped clock says that T=X
Which is all fine, except that the referent of ‘T=X’ in (1) isn’t the same sort of thing as the referent of ‘T=X’ in (2), this is because we just wouldn’t say of a stopped clock that it tells the time; not if we thought carefully about what we were saying, at any rate. Instead, a stopped clock just tells some number, or symbol, or whatever. After all, time is what working clocks agree on; when we say of a particular clock that it tells the time, this is to say that we believe it to be one of a group of clocks that tell the time. A clock isn’t counted as a member of that group if we believe that it’s stopped. Anyway, this is where I was going with the previous comment.
if your criterion is "contains loads of fallacious and/or unrigorous mathematics", I doubt I'll accept it
ReplyDeleteThis seems an entirely unreasonable position.
I have learned measure-theoretic probability, twice, (once from "An Introduction to Rigorous Probability", once from "Weighing the Odds". It was not fun either time. Barring a radical career change I am never going to have to learn it again, and barring a radical personality change I am unlikely to do so for fun. Instead, I will continue to post superficially plausible but actually ludicrous probabilistic arguments on my blog, because the pain caused to mathematicians on reading them gives me a slight quasi-sexual thrill.
ReplyDeleteOh all right, I've done it properly now (or at least I think I have). Pete, above, is correct, by the way.
ReplyDeleteI really admire this, I mean it really looks interesting! Very nice research. Thanks to the author.
ReplyDeleteI admire this article for the well-researched content and excellent wording. I got so involved in this material that I couldn’t stop reading. I am impressed with your work and skill. Thank you so much. skull wall clocks
ReplyDelete