tag:blogger.com,1999:blog-6894866515532737257.comments2015-11-24T12:46:05.868-08:00Probably Overthinking ItAllen Downeyhttps://plus.google.com/111942648516576371054noreply@blogger.comBlogger559125tag:blogger.com,1999:blog-6894866515532737257.post-44571342286096523002015-11-24T12:46:05.868-08:002015-11-24T12:46:05.868-08:00Interesting idea: I've been thinking about it ...Interesting idea: I've been thinking about it all day. The reasons I haven't done it yet are:<br /><br />1) Partly for simplicity: I find it hard enough to explain single-level models.<br /><br />2) Partly because I don't really have a theory about what explanatory factors to include at the higher level.<br /><br />3) And partly because the subset models answer the question I am interested in (whether the association between Internet use and religiosity is consistent across countries).<br /><br />4) A multilevel model might answer different questions, like what national factors influence the size of the Internet effect. But with the data I have, I'm not confident I would find anything reliable.<br /><br />If you look at the IPython notebook for part four, I've got some scatterplots of (x) effect size of Internet use and (y) a few national factors. There are some weak correlations, but nothing that seems super meaningful.<br /><br />Thanks for the suggestion!Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-85478227113834832152015-11-23T22:29:52.482-08:002015-11-23T22:29:52.482-08:00Why not run a multilevel model using country inste...Why not run a multilevel model using country instead of running models on subsets of the data?<br />fdwwhttp://www.blogger.com/profile/04022435459606207630noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-28130909296032294822015-11-18T01:26:34.001-08:002015-11-18T01:26:34.001-08:00What is Intercept standing for? for neutral (rlgbl...What is Intercept standing for? for neutral (rlgblg, rlgdgr)?<br />and what is its coef.? Justinhttp://www.blogger.com/profile/12953665795281989168noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-90111378250879583962015-11-16T08:24:08.027-08:002015-11-16T08:24:08.027-08:00Sounds like a good thing to investigate. Let me k...Sounds like a good thing to investigate. Let me know if you find anything interesting.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-2155621420131653402015-11-16T08:23:12.504-08:002015-11-16T08:23:12.504-08:00Sam Mason sent the following comment, "I'...Sam Mason sent the following comment, "I'm wondering whether the logistic regression is imposing too much structure on the data. It almost looks (from the Independent estimates plot) as though there could be a step function between groups 1 and 2 with little further change. I was therefore wondering whether you'd attempted any non-parametric analysis?<br /><br />Gaussian processes are great for this sort of application as they put less constraints on the shape of the function they're inferring while still allowing you to share information between groups."Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-57528605579878732802015-11-14T04:03:17.922-08:002015-11-14T04:03:17.922-08:00Yes, I think that would be a reasonable option, an...Yes, I think that would be a reasonable option, and if the relationship were non-monotonic, it might be the best choice.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-45038859855282931342015-11-13T20:12:15.498-08:002015-11-13T20:12:15.498-08:00In terms of smoothing data by combining adjacent s...In terms of smoothing data by combining adjacent scores - couldn't you do a windowed approach to avoid the unfair lumping? Eg, take a group and its 2 neighbors? You could also do a non-uniform window and count the neighbors less than the group in question. Looks like you'd be able to trade independence from other groups for smoothing. This does have an issue for the end groups, but the smoothed curve might inform a reasonable extrapolation.Boris Dieseldorffhttp://www.blogger.com/profile/11818300387729526308noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-30909504693229373082015-11-12T08:26:15.076-08:002015-11-12T08:26:15.076-08:00There are three events of interest to a "thir...There are three events of interest to a "thirder" SB when she is awake: H1=Today is Monday and Heads flipped, T1=Today is Monday and Tails flipped, and T2=Today is Tuesday and Tails flipped.<br /><br />White called his event H, but he really was treating it as H1. His expression for P_(H|W) is more accurately called P_(H1|W) even though they represent the same thing. That's why the term (c/2) is used: the prior probability of being awake on Monday after Heads is c, and the prior probability of Heads is 1/2. He multiplied them to get his (c/2).<br /><br />Note that the denominator of his equation [P_(H) P_(W|H) + P_(~H) P_(W|~H)] has nothing to do with the fact that he is calculating P(H), it is merely an expression of the Law of Total Probability using the partition {H,~H}. So the same denominator can he used in a similar calculation for P_(T1|W) and P_(T2|W). Since the numerators for these calculations are P(H1)*P_(H1|W) = P_(T1)*P_(W|T1) = P_(T2)*P_(W|T2) = (c/2), you get the same result for each calculation. Like I said.<br /><br />I know you disagree with this approach. But it is what a thirder would use. What White accused a thirder of using is not.<br /><br />+++++<br />What you, and White, ignore is that H2=Today is Tuesday and Heads flipped *is* *also* *an* *event* *in* *the* *experiment*. SB can't observe it, but it is still an event. On Sunday night, it has the same status, and prior probability, as H1, T1, and T2.<br /><br />Her "new information" when she is awake is that H2 is not her current state.<br /><br />The partition of SB's posterior event space contains four independent events, not two. Since she samples twice, with amnesia in between, the sum pf the prior probabilities for these four events is 200%, not 100%. The laws of probability work just fine on these events.JeffJohttp://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-657207830364742422015-11-11T12:57:31.885-08:002015-11-11T12:57:31.885-08:00Thanks for your thoughts, Jeff.Thanks for your thoughts, Jeff.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-84587916463279975952015-11-11T11:04:59.703-08:002015-11-11T11:04:59.703-08:00No. His result is for P(H | SB didn't sleep th...No. His result is for P(H | SB didn't sleep through the experiment).<br /><br />Monday and Tuesday have nothing to do with it. SB doesn't know whether it is Monday or Tuesday, because it could be either day. All she knows is that she didn't sleep through the entire experiment.<br /><br />The credence for heads actually becomes P(H)=1/3 (the Thirder's answer), but only in the limit that it is extremely unlikely for her to have ever wake at all -- i.e., c->0.Brian Mayshttp://www.blogger.com/profile/13962229896535398120noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-48221413887639888972015-11-11T09:16:19.329-08:002015-11-11T09:16:19.329-08:00There are two "definitions of the problem.&qu...There are two "definitions of the problem." One is what the experiment director sees (and SB sees on Sunday), and a different one is what SB sees when she is awake. That one exactly matches the situation that H2 finds herself in, where the answer is 1/3. Which should she base her confidence on? I've shown you several ways why it must be 1/3, and all you do is repeat the other solution.<br /><br />So again, I am less interested in seeing your solution that gets 1/2, then knowing why you think she can treat Monday and Tuesday as the same day. Which is what White did.<br /><br />Digging out my notes from years ago (and I'm not re-deriving it), I see that White stated that P(H) = (c/2)/[(c/2+(1-(1-c)^2)/2] = 1/(3-c). But his event "H" really is "H&Mon." You can use his exact same logic to get P(T&Mon)=P(T&Tue)=1/(3-c). These should add up to 1, and they don't.JeffJohttp://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-80838081445465780952015-11-11T08:03:10.196-08:002015-11-11T08:03:10.196-08:00"The question Brian refuses to address is, wh..."The question Brian refuses to address is, which is more appropriate to the original question?"<br /><br />Nonsense! I have said over and over that the one that is more "appropriate" is the one that corresponds to the definition of the problem. X2 depends on the outcome of X1. That is explicitly stated without any ambiguities.<br /><br />By the way, you still haven't addressed White's "generalized Sleeping Beauty Problem." Google it if you want to find the original article published in <em>Analysis</em> in 2006. Or you can click on the link that Creosote provided above. It probably has a link that will take you there.Brian Mayshttp://www.blogger.com/profile/13962229896535398120noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-79190030919130049072015-11-11T07:34:38.971-08:002015-11-11T07:34:38.971-08:00Okey-dokey, I'm done. I look forward to other...Okey-dokey, I'm done. I look forward to other Thirders attempting the "beans experiment" in future, and will be interested to see what conclusion (if any) is eventually reached at www.sleepingbeautyproblem.org .Creosotehttp://www.blogger.com/profile/02571941331642177202noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-89515331165509920462015-11-11T07:14:43.702-08:002015-11-11T07:14:43.702-08:00@JeffJo @Creosote @BrianMays My general policy is ...@JeffJo @Creosote @BrianMays My general policy is that I'll publish a comment as long as it is on point and not abusive. But I encourage you to keep the tone civil. And if it looks like you can't reach consensus, you might have to agree to disagree.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-88338670347070456842015-11-11T06:40:33.401-08:002015-11-11T06:40:33.401-08:00In the original problem, say a six-sided die is al...In the original problem, say a six-sided die is also rolled on Sunday Night after SB is asleep. If it comes up even, an alarm is programed to go off at noon on Monday. If odd, it will go off at noon on Tuesday, but she may not be awake to hear it. SB is instructed to enter her confidence in Heads into a computer at precisely 12:30PM.<br /> <br />On Sunday Night, she recognizes that the coin and the die represent independent events, and the alarm is equally likely to go off under any of the four possible situations {H,Mon}, {T,Mon}, {H,Tue}, and {T,Tue}. Of course, she will not be awake to hear it under {H,Tue}. Since she is not assured of hearing it, if she does it constitutes “new information” and allows her to update her confidence in Heads to 1/3. But not hearing it is also “new information,” and also allows her to update her confidence in Heads to 1/3.<br /><br />What if SB is, unknown to the experiment director, unable to hear certain frequencies and so can’t hear the alarm? She can still update her confidence to 1/3 because the alarm was in only one state of going off, or not going off, and both allow the same update. Essentially, this is now a form of Bertrand’s Box Paradox. She can update her confidence to 1/3 under every possible state of the alarm, so her confidence must be 1/3 without considering the alarm. But please, please, please note that this answer only applies when the status of one day is all that is known.<br /><br />The point I didn’t want to get into when Brian mentioned independence, is that independence changes as a result of the amnesia drug. On Sunday Night, being awake under T1 and T2 represent the same future, so they are dependent events. Brian’s solution treats them as dependent and gets 1/2. As such, it is addressing the problem from Sunday’s (or Wednesday’s) viewpoint only; that is, a state where all possible awake days are considered. But when SB is awake, she has no information about the possible “other” day, T1 and T2 represent different presents, and they are independent contrary to how Brian treats them. The Thirder solution treats them as independent to get 1/3. It address the problem from the viewpoint of either Monday or Tuesday, but not both as Brian does.<br /><br />Both the Alarm-Clock version, and my four-volunteer version, are designed to isolate state where T1 and T2 are independent events. Yes, it is a different problem than the one Brian solves when he treats them as dependent. The question Brian refuses to address is, which is more appropriate to the original question?JeffJohttp://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-42550410954070269352015-11-11T05:06:57.462-08:002015-11-11T05:06:57.462-08:00"What part of 'that's the last time I..."What part of 'that's the last time I try to explain it to you' do you not understand?!!" The fact that your explanation does not address my issue. It dodges it. The issue isn't whether my solution is to a different problem than yours, we agree it is. The issues is which corresponds to the SB problem.<br /><br />H2 is identical to SB. H2's answer to the same question is 1/3. <br /><br />I'll avoid ad hominem attacks.JeffJohttp://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-59017046971523036452015-11-11T04:27:31.962-08:002015-11-11T04:27:31.962-08:00"Beauty must make her decision based on her c..."Beauty must make her decision based on her credence for the jars being filled the Heads way or the Tails way,."<br /><br />No, she really shouldn't. If the coin landed Tails, she will pick two beans, not one. So there is a difference between her confidence that *this* bean is safe, and that the set of beans she will consume on Wednesday will be safe.<br /><br />But Beauty is not asked for what her confidence was on Sunday, or will be on Wednesday (which is what you are solving for). She is asked for her confidence *NOW*. This is exactly the question my variation - the one you ignore because its answer is 1/3 - solves for.<br /><br />AGAIN: In my variation, H2 finds herself awake and in the exact situation SB is in. The other two awake volunteers are in symmetric situations. Exactly one of their labels matches the coin. The confidence any one of them should have, that it is herself, must be 1/3.JeffJohttp://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-73920826145928284022015-11-11T00:16:24.967-08:002015-11-11T00:16:24.967-08:00For what appears to be an excellent and far-reachi...For what appears to be an excellent and far-reaching survey of the available literature, readers might enjoy perusing http://www.sleepingbeautyproblem.org/tiki-index.php?page=Project%20Portal . Seems like it's a work in progress, so should get even better over time.Creosotehttp://www.blogger.com/profile/02571941331642177202noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-31991851326097473892015-11-10T18:01:41.526-08:002015-11-10T18:01:41.526-08:00"So I repeat: Brian, what do you think is dif..."So I repeat: Brian, what do you think is different about the volunteer labeled H2, and the original Sleeping Beauty?"<br /><br />What part of "that's the last time I try to explain it to you" do you not understand?!!<br /><br />I've already explained it to you, again and again. Since you cannot accept my explanation, you counter with:<br /><br />1) "'In your different problem, X1 and X2 are completely independent random quantities.' No, they are not."<br /><br />Oh really? No reason given.<br /><br />2) "When you consider the joint distribution for X1 and X2 with any specific value of X3, the distribution is either identical to the one you describe for SBP (for X3=H2), or symmetric with it."<br /><br />At this point, I seriously doubt you even know what a joint distribution is, but ... (sigh) ... giving you the benefit of the doubt, if the joint distributions were identical then we wouldn't be having this discussion -- would we? We would agree on the result.<br /><br />3) "I can tell you what is wrong with White's analysis as well, but I don't want to expand the issue further."<br /><br />In other words, White's analysis is way over your head.<br /><br />4) "Using 1/3 properly -- and I am too rushed at the moment to do it -- will only tell you the probability that *this* jelly bean will be poison ..."<br /><br />At this point, you have become a parody of yourself. Creosote is correct: you are by definition a Halfer.<br /><br />Please leave us alone now. You are the last person who should be demanding anything from anybody.Brian Mayshttp://www.blogger.com/profile/13962229896535398120noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-12498737163707548652015-11-10T14:56:25.012-08:002015-11-10T14:56:25.012-08:00Beauty must make her decision based on her credenc...Beauty must make her decision based on her credence for the jars being filled the Heads way or the Tails way, i.e. her credence for the coin flip being Heads or Tails. For Thirders, those credences are 1/3 and 2/3 respectively.<br /><br />If you insist that Beauty makes her choice based on credence 1/2 for the Heads way of filling the jars then you are by definition a Halfer.Creosotehttp://www.blogger.com/profile/02571941331642177202noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-30365807200042940412015-11-10T14:04:57.369-08:002015-11-10T14:04:57.369-08:00Fair enough. I think we might be enduring a bit of...Fair enough. I think we might be enduring a bit of a verbal dispute about what it takes to obtain a genuine single-case probability. This seems closely connected, though, to a possibly substantive issue about the strength of the commitment to the reference class decision. I'm not completely sold, but I think you've brought some valuable clarity to a really interesting issue.Jeff Kasserhttp://www.blogger.com/profile/16847118814783859674noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-43101093154080512482015-11-10T12:39:59.513-08:002015-11-10T12:39:59.513-08:00I understand why you are troubled, and I think you...I understand why you are troubled, and I think you are not alone. I tried to address this in #2 in my Q&A section. After you choose a reference class, it is tempting to also include specific information about the individual. But if you do that, you are effectively challenging the choice of the reference class, implying that it would be better to use a narrower class based on more attributes. And that might be true, but it goes to the issue of choosing the reference class, not applying the results to an individual. If we agree about the best choice for the reference class, I think part of that agreement is the commitment to apply the results to each individual in the class.Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-1306984501233236732015-11-10T12:30:40.312-08:002015-11-10T12:30:40.312-08:00That's helpful, though I'm not yet untroub...That's helpful, though I'm not yet untroubled. My being troubled is not necessarily your problem, of course. We can assign a probability to a trait in a reference class. If we know nothing else about a given individual besides the fact that he was chosen randomly from that reference class, we should assign to him to probability for the trait in the reference class. But when we designate an individual, we know more about him than that he was chosen from the reference class at random, and we don't know that this additional information is irrelevant, do we? That's why the "that move is part of the modeling decision" doesn't yet seal the deal for me. Does that make sense?Jeff Kasserhttp://www.blogger.com/profile/16847118814783859674noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-79160774915399533062015-11-10T12:15:52.135-08:002015-11-10T12:15:52.135-08:00Maybe this will help clarify what I'm trying t...Maybe this will help clarify what I'm trying to say:<br /><br />1) Brain Mays solves the SB problem with a solution with an answer of 1/2. Call this solution B, for Brian.<br />2) I presented a variation of the SB problem that I say, and I think Brian agreed with, has the answer 1/3. Call its solution J.<br />3) Brian insists that solutions B and J address different problems. I agree.<br />4) What the "four volunteers" example is supposed to show, is that the original Sleeping Beauty finds herself in exactly the same set of circumstances as one of the four volunteers, and symmetrically equivalent circumstances to the other three. Brian has not contested this, he has only said it needs a different solution than solution B.<br />5) Those circumstances are addressed by solution J.<br /><br />So I repeat: Brian, what do you think is different about the volunteer labeled H2, and the original Sleeping Beauty? Because she says the probability that the coin matches her label - i.e., landed Heads - is 1/3. A agree that this changes the problem from the one you solved - I'm saying that you solved the wrong problem.JeffJohttp://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-22770609900357081652015-11-10T11:18:17.950-08:002015-11-10T11:18:17.950-08:00Ah, I think I see where we are diverging. The pro...Ah, I think I see where we are diverging. The problem is that "Smith" is ambiguous: it might refer to an individual or it might refer to a random member of reference class C. I am arguing that it is only in the second sense that Smith has a probability of recidivism.<br /><br />If Smith (the individual) commits a crime, that gives me evidence about Smith (as a member of C). And that's the basis of my claim that it is meaningful to assign a probability to Smith (as a member of C), because future data will support or refute that claim (contrary to the charge that it is a meaningless claim because future data could neither support nor refute).<br /><br />The second part of the argument is that it is appropriate to apply a probability that pertains to Smith (as a member of C) to Smith (the individual), because that move is part of the modeling decision to treat/regard Smith as a member of a reference class.<br /><br />That last part is, I think, implied by frequentist notion(s) of probability. So that gets me to "single-case probabilities are not a special problem under frequentism".<br /><br />How's that?Allen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.com