tag:blogger.com,1999:blog-6894866515532737257.post2480337987852261404..comments2024-04-22T21:33:32.590-07:00Comments on Probably Overthinking It: Bayes's Theorem is not optionalAllen Downeyhttp://www.blogger.com/profile/01633071333405221858noreply@blogger.comBlogger17125tag:blogger.com,1999:blog-6894866515532737257.post-22010083889966409672017-11-17T09:28:23.626-08:002017-11-17T09:28:23.626-08:00I wonder if you are still following and interested...I wonder if you are still following and interested in this thread. <br /><br />I was trying to map the example into other examples---eg you have a city of people and take a sample of 1, 2, and then 3 and test them for drugs . the drug test is only 2/3rds accurate ( if they have drugs in their system then 2/3rds of the time the test says Yes they do, and 1/3rd No they don't . Suimialrily if they dont have drugs in them, then 2/3rds of the time they come out 'negative', and 1/3rd positive. <br /><br />You can consider cases like 'maybe i dont know the size of the city--is it 1, 2, 3 or a million people'? To apply Bayes theorem you might need a 'prior' like what % of people in that city are on drugs. <br /><br />I'm really interested in a 'recursion' so you can go from 1 sample response (in your example, one friend), to 2, to 3, etc. <br /><br />I actually think of this in terms of the ising model and spin glasses of statistical physics. <br /><br />Also in term s of Venn diagrams--in you case the big circle is Seattle, and its cut in 2 by a line saying on one side of the circle (CITY) its raining and other side its not. Then you sample somone in the city but you dont know if they are on the rainy or sunny side, and also you only know they are telling the truth 2/3rds of the time when you ask them if its raining. (or you could ask them if they are on drugs, or did the crime.) <br /><br /> m impression is bayesianism reduced to frequentism if P(Rain) is unknown (just 50/50). <br /> also if you ask 3 people do you use (1/3rd +1/3rd +1/3rd)/3 or (1/3)^3 (times probabilities) . i think those are 2 different slightly different assumptions---one independent random variables (multiply) versus additive --which reduces to same thing sometimes. <br /> <br />i just view bayes theorem or rule as a version of the definition of conditional probability. <br /><br /> Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-76794178087946521662017-06-02T22:27:17.513-07:002017-06-02T22:27:17.513-07:00Hi Allen,
Thank you for posting this problem. It ...Hi Allen,<br /><br />Thank you for posting this problem. It seems very interesting. I have a quick question: why do we compute the probability of YYY? This information is already given in the problem ("All 3 friends tell you that "Yes" it is raining."), so P{YYY} should be 1?<br /><br />Thank you!Anonymoushttps://www.blogger.com/profile/10621469573786218341noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-36946818009232125082016-10-13T06:06:30.186-07:002016-10-13T06:06:30.186-07:00Hi. I just added a worksheet that shows how to so...Hi. I just added a worksheet that shows how to solve this problem using Bayes's Theorem.Allen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-56656754707304130282016-10-13T05:50:47.838-07:002016-10-13T05:50:47.838-07:00Hi Emile, I'm glad it is making sense.
But I...Hi Emile, I'm glad it is making sense.<br /><br />But I want to clarify the point of my article. I am not saying that there are two answers to this question, one Bayesian and one frequentist, and the Bayesian one is right.<br /><br />I am saying that there is only one answer to this question, and it is neither Bayesian nor frequentist. It is just a consequence of the laws of probability.<br /><br />The issue Dimiter raised is called the reference class problem: https://en.wikipedia.org/wiki/Reference_class_problem<br /><br />And while the reference class problem is relevant to the problem, it is a general problem for all of probability, and not specifically Bayesian or frequentist, either.Allen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-25120124062450485922016-10-12T18:28:14.111-07:002016-10-12T18:28:14.111-07:00I deleted my initial comment because I was probabl...I deleted my initial comment because I was probably overthinking it. I grappled with your explanation and the other posts, but now understand your statement that only a Bayesian approach can lead to the right answer. <br /><br />We want Prob(raining|YYY).<br />This is equal to Prob(raining and YYY)/Prob(YYY). We don't know Prob(raining and YYY).<br /><br />However Prob(raining and YYY) also equals Prob(raining)*Prob(YYY|raining)<br /><br />So we need Prob(raining)but don't know this. So as already mentioned by Dimiter, there are infinitely many answers in the absence of information. <br />However, we do know Prob(YYY|raining)=8/27<br /><br />Furthermore, <br />Prob(YYY) = Prob(YYY|raining)*Prob(raining) + Prob(YYY|not raining)* Prob (not raining)<br /><br />We also know Prob(YYY|not raining) = 1/27. <br /><br />The frequentist solution is really a relative odds ratio: Prob(YYY|raining)/{ Prob(YYY|raining) + Prob(YYY|not raining)}. It is not really a probability at all. <br /><br />Now when Prob(raining)=1/2 ( the best guess if you have no information) then the frequentist relative odds will equal the Bayesian probability. <br /><br />If instead of Seattle the trip was to Los Angeles, and the Prob(raining) =0.01, then having three friends saying "Yes it's raining" will result in Prob(raining|YYY) =~0.075.<br /><br />However if you asked 14 of your mostly truthful friends, and all 14 said,"yes dude, it's raining", then the conditional probability increase to ~0.993 (frequentist relative odds = ~0.99994). <br /><br />This makes sense.<br /> <br />Thanks for posting this.<br /><br />Anonymoushttps://www.blogger.com/profile/06682927722971051966noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-71304155594207886632016-10-11T23:24:50.902-07:002016-10-11T23:24:50.902-07:00Hello Allen, I'm having difficulty solving thi...Hello Allen, I'm having difficulty solving this problem using Bayes Theorem but have no idea where I'm going wrong. Could you please shed me a light?<br /><br />We have:<br /><br />P(rain|YYY) = P(YYY|rain)*P(rain) / P(YYY)<br /><br />P(YYY|rain) = 8/27 ?<br />P(rain) = 0.1<br />P(YYY) = 8/27 ?<br /><br />Are the previous values correct? If they are then P(rain|YYY) = 0.1*8/27 / 8/27 = 0.1<br /><br />This would imply that P(rain|YYY) is not dependent on P(YYY) at all. What am I doing wrong?Thicassahttps://www.blogger.com/profile/15299634454831159125noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-52070592745979605252016-10-10T08:18:47.642-07:002016-10-10T08:18:47.642-07:00This comment has been removed by the author.Anonymoushttps://www.blogger.com/profile/06682927722971051966noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-76766455963870996912016-10-04T16:03:50.474-07:002016-10-04T16:03:50.474-07:00Frankly, I have long thought of the bayesian appro...Frankly, I have long thought of the bayesian approach as one that is best digested as a of morning joe. Frankly, the frequentists are more like a protein shake with shopped up celery and spinach.<br /><br />However, non-parametric methods are the best. I love basic arithmetic. <br /><br />Sir Thomas was a priest after all..Thomas Baymanhttps://www.blogger.com/profile/08031681715056987599noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-60050865019228855542016-09-28T00:09:19.649-07:002016-09-28T00:09:19.649-07:00Allen: I think the probability question, as stated...Allen: I think the probability question, as stated, has not one and not two, but infinitely many answers. Why?<br />1) If we stick to the problem as stated, there is no information whatsoever that we can use to pick a prior probability of rain. So any prior must be as good as any other. By implication, that any posterior probability of rain can take any value between 0 and 1 (including these). So in the absence of any further information/ assumptions, the correct answers is probability of rain = [0;1]. We can say that the probability has significantly increased on hearing the signal from the friends, but we cannot see to how much, since we do not know the prior.<br />2) if we assume that if the probability was known with certainty in advance there would be no need for asking, we can exclude the boundaries of the interval, so p = (0, 1)<br />2b) if, in addition, we assume that probability is measured with finite precision (sey as whole percentages), the minimum prior probability becomes 0.01, so the interval for the posterior becomes p = (0.07, 1).<br />3) If we make the alternative assumption that the traveler would only ask for the solicited information if it would have a chance to change the decision (interpreted as moving the posterior above or below 0.5), the (posterior) probability becomes p = (0.5, 1). The friends send a signal that is as strong as it gets, so in the most adversarial prior it must just about move the probability above 0.5 so that in can satisfy the assumption of having a chance to influence the decision.<br />4) Now, if we want to make even further assumptions about what is known and what can be known in the framework of the problem, we can find reasons to pin the prior and, by implication the posterior, to any number within this range but a) this is not obviously warranted by the setup of the problem, and b) we enter the decision-theoretic problem mentioned in my previous post.<br />In sum, there might be one correct way to update probabilities based on prior information and new data. But in the absence of prior information and assumptions, any posterior probability must be correct.Anonymoushttps://www.blogger.com/profile/09098718685262606935noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-71351507574032455592016-09-27T17:32:47.682-07:002016-09-27T17:32:47.682-07:00And I didn't mean to sound snarky towards you....And I didn't mean to sound snarky towards you. Thanks for your comments!Allen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-19752113461864377832016-09-27T17:18:23.267-07:002016-09-27T17:18:23.267-07:00[Allen, I didn't mean for my comment to sound ...[Allen, I didn't mean for my comment to sound snarky towards you! My cheekiness was directed at the hypothetical interviewer asking this question :) ]<br /><br />Indeed, the question ends by asking for a probability. I totally agree with your solution to treating this as a toy probability puzzle. I agree there's no Bayes-vs-frequentist difference there. And I agree it's very important to distinguish Bayes' theorem from Bayesian inference.<br /><br />On the other hand, the question *starts* with "You want to know if you should bring an umbrella."<br />Let's take this seriously as an interview question, meant to help the interviewer decide which candidates would bring the most value to the company.<br /><br />What are they really getting at?<br />If they were asking:<br />"We want to know if our company should take action U. It only makes sense to invest in performing U if there's at least 50% chance that R is true. We can (just barely) afford to run 3 expensive tests. Each test independently has 2/3 chance of correctly identifying whether R is true. If a sensible prior on U is 10%, and all 3 tests come back True, what will be the probability that R is indeed true?"<br /><br />...then who would you rather hire?<br />Candidate A, who is content to stop after getting an answer of 47%?<br />Or Candidate B, who goes on to say:<br />"Look, even if all 3 tests agree in claiming that R is true, our estimated probability that R is true will *still* be under 50%. In all other cases, it'll be even lower. You're wasting money by running these 3 tests. If we can't afford more tests, let's just skip them and spend that money somewhere useful."<br /><br />The spirit of frequentism is to step back from the given data and understand the operating characteristics of your statistical procedures---not just the analyses but the data-collection (design) too. It's a useful habit of mind, whether your final inferences/analyses end up Frequentist or Bayesian. And yep, sometimes that means refusing to answer a particular question, when that question isn't what's actually needed.Jerzy Wieczorekhttps://www.blogger.com/profile/03611849167717252118noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-14353809901579712992016-09-27T12:57:40.183-07:002016-09-27T12:57:40.183-07:00Thanks, Dimiter. You make some great points and I...Thanks, Dimiter. You make some great points and I won't address them all, but I want to clarify one. The goal of my article is not to demonstrate the "added value of a Bayesian vs frequentist answer". The point I am trying to make is that the probability question, as stated, does not have two answers, one Bayesian and one frequentist answer. It has one answer that can be computed without any commitment to a Bayesian or frequentist interpretation of probability, and without any commitment to Bayesian or frequentist inference.<br /><br />It does, as you point out, require the choice of a reference class, but that is a general difficulty with many probability problems; it is not a special difficulty for Bayesianism or frequentism.Allen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-11569655580038207642016-09-27T12:48:10.263-07:002016-09-27T12:48:10.263-07:00two long related comments: 1) Why not use the prio...two long related comments: 1) Why not use the priors for the particular days you will be visiting (I guess the chance of rain varies significantly over the year)? But then why not use as a prior the conditional probability of rain in Seattle given the atmospheric conditions at the moment? Or even, why not commission specific research to inform your prior: why stop at consulting the Western Regional Climate Center or the weather forecast? You can say - not worth it for the problem at hand, but this requires a separate analysis of how much effort it is worth spending on establishing a good prior for this problem. In our case, the answer is probably 'close to zero', as the costs of taking an umbrella are negligible. But then for somebody with zero knowledge about the weather in Seattle a flat prior of rain/no rain, or equivalently a frequentest analysis, would seem as justified as any other. 2) Which brings me to the second point. The problem is introduced as a decision-theoretic one (bring an umbrella or not) but then it asks for a probability that, however defined and computed, is not sufficient to answer the decision-theoretic motivating problem. And it seems to me that the added value of a Bayesian vs. a frequentist answer to the probability question cannot be demonstrated outside of a decision-theoretic setup in which the costs of establishing a prior are compared to the benefits of increased precision of the answer. (And you cannot just say, oh but everybody knows the prior chance of rain in Seattle is 0.5 or 0.1 or whatever, as this info is not provided in the set-up of the problem). Anonymoushttps://www.blogger.com/profile/09098718685262606935noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-28998916468175189542016-09-27T11:48:02.294-07:002016-09-27T11:48:02.294-07:00Thanks, Jerzy. The question asks for a probabilit...Thanks, Jerzy. The question asks for a probability; I think your analysis answers a different question.<br /><br />But refusing to answer the question is certainly in the spirit of frequentism.Allen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-79765744841293372322016-09-27T11:19:18.817-07:002016-09-27T11:19:18.817-07:00Perhaps a more Frequentist-spirited answer would b...Perhaps a more Frequentist-spirited answer would be to discuss the study design:<br /><br />Your prior prob. of rain is under 50%, and in fact it's so low (at 10%) that *nothing* your 3 friends say could convince you the (posterior) prob. of rain is over 50%, even when they all agree it is raining.<br /><br />In other words, your study has no power to change your mind! (from the prior decision that it's probably not raining.)<br /><br />So why did you hassle your friends by asking them in the first place? Maybe that's why they lie to you 2/3 of the time :)Jerzy Wieczorekhttps://www.blogger.com/profile/03611849167717252118noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-36817919654265621822016-09-27T09:14:51.665-07:002016-09-27T09:14:51.665-07:00Hi Russ. A helpful reader submitted the following...Hi Russ. A helpful reader submitted the following explanation, which I accidentally rejected instead of publishing. So, with apologies to the helpful reader:<br /><br />my test blog has left a new comment on your post "Bayes's Theorem is not optional": <br /><br />If you look at the answer linked to in the post: https://www.glassdoor.com/Interview/You-re-about-to-get-on-a-plane-to-Seattle-You-want-to-know-if-you-should-bring-an-umbrella-You-call-3-random-friends-of-y-QTN_519262.htm<br /><br />and subsitute 10% chance of rain for 25%, you should get the answer listed here:<br />0.1*(8/27) / ( 0.1*8/27 + 0.9*1/27 )<br />8/270 / 8/270 + 9/270<br />8/17<br />47.06%<br /><br />In Downey's version, he uses both odds and probabilities, which makes the calculations in this case easier but, maybe, harder to follow. <br /><br />Here's the top answer from the linked post:<br /><br />Bayesian stats: you should estimate the prior probability that it's raining on any given day in Seattle. If you mention this or ask the interviewer will tell you to use 25%. Then it's straight-forward:<br /><br />P(raining | Yes,Yes,Yes) = Prior(raining) * P(Yes,Yes,Yes | raining) / P(Yes, Yes, Yes)<br /><br />P(Yes,Yes,Yes) = P(raining) * P(Yes,Yes,Yes | raining) + P(not-raining) * P(Yes,Yes,Yes | not-raining) = 0.25*(2/3)^3 + 0.75*(1/3)^3 = 0.25*(8/27) + 0.75*(1/27)<br /><br />P(raining | Yes,Yes,Yes) = 0.25*(8/27) / ( 0.25*8/27 + 0.75*1/27 )<br /><br />**Bonus points if you notice that you don't need a calculator since all the 27's cancel out and you can multiply top and bottom by 4.<br /><br />P(training | Yes,Yes,Yes) = 8 / ( 8 + 3 ) = 8/11<br /><br />But honestly, you're going to Seattle, so the answer should always be: "YES, I'm bringing an umbrella!"<br />(yeah yeah, unless your friends mess with you ALL the time ;)<br /><br />Interview Candidate on Sep 12, 2013 Allen Downeyhttps://www.blogger.com/profile/01633071333405221858noreply@blogger.comtag:blogger.com,1999:blog-6894866515532737257.post-32960971274446582392016-09-26T22:15:26.763-07:002016-09-26T22:15:26.763-07:00I was hoping to understand how to use Bayesian rea...I was hoping to understand how to use Bayesian reasoning, but I was completely lost by the Bayesian argument. Would you mind elaborating the reasoning behind the segment of the post that starts at "A base rate of 10 ... " and ends at "Probability(8 Odds(p))". (I don't even understand how to read that final bit of notation!) Thanks.Russ Abbotthttps://www.blogger.com/profile/15431389045571531450noreply@blogger.com