Timeline for Google question: In a country in which people only want boys
Current License: CC BY-SA 3.0
25 events
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| May 30, 2013 at 18:11 | history | edited | Steven Landsburg | CC BY-SA 3.0 |
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| May 17, 2013 at 13:05 | comment | added | Todd Trimble | Although Matthew has bowed out, I just want to say what a pleasure it is to read his comments generally, not only for their erudition but for their wonderful civility. Too bad not all comments under this question in particular are so civil... | |
| Jan 6, 2011 at 7:56 | comment | added | Emerton | Dear Douglas, As I wrote above, I've bowed out. Regards, Matthew | |
| Jan 6, 2011 at 5:52 | comment | added | Douglas Zare | Another way to look at it is as follows: Suppose each year, $1 billion will be split equally among the members of the population. If you are a 2-year-old boy, you expect to get a slightly greater share of this than a 2-year-old girl does, since you expect to split it with fewer others since the population includes no younger siblings. That your younger siblings affect the population size, and boys have no younger siblings, means that boys get weighted more than girls. | |
| Jan 6, 2011 at 5:44 | comment | added | Douglas Zare | @Emerton: It would be mathematically shocking if the expected proportion were less than 1/2 for the easily computable cases of k=1, k=2, k=10, but exactly 1/2 for country-sized numbers of families. And this doesn't happen, there is a bias for any k. If your intuition tells you that for large enough k, the expected value is not just close to 1/2 but equal to 1/2, then your intuition is wrong. Let me ask something which may help. Suppose there are unusually many girls born in one year. Is the population unusually large, or small, the next year? Ans: Large. That means girls get weighted less. | |
| Jan 6, 2011 at 1:56 | comment | added | Emerton | I don't think I want to try to model an actual country with this stopping policy, since I don't think I could write down anything like a realistic model for any kind of country-sized population, whatever the policy on reproduction is! So I might bow out for now. Thanks very much for taking the time to explain your point of view. Best wishes, Matt | |
| Jan 6, 2011 at 0:39 | comment | added | Emerton | Dear Steven, Thanks for your reply. I think after reading your example and your last comment, I understand your point, and presumably you understand what I am arguing. I agree that my model is idealized (e.g. the ratio will presumably never be exactly 1:1, just because even in a large sample population, the actual value of a random variable won't precisely equal the expected value, just be very close to it), but I think it gives a good way to think about the question (and, although I can't be sure of course, I think it explains the sense in which T. was not "assuming any specfic model"). | |
| Jan 5, 2011 at 19:02 | comment | added | Steven Landsburg | Emerton: I'm happy to keep discussing, but let me know if you think there's a better forum than these comments. Here is what I see as the main point: If I call my example a "country", then it's a country in which E(G/G+B) is not 1/2. You say this doesn't count as a country. I think (and I hope this doesn't too argumentative) that at this point it's incumbent on you to give an explicit example of a "country" in which the answer is 1/2 --- specifying things like number of families, mortality rates, fertility of subsequent generations, etc. I suspect you won't find this easy. | |
| Jan 5, 2011 at 18:47 | comment | added | Emerton | Dear Steven, I looked at your example, and of course I don't disagree with the mathematics; I guess my disagreement is one of interpretation (which might be what you mean by model). The thing is, I don't think you can call it a country. E.g. in scenario 1, when the outcome is B/B, the country dies off, so in the long run the population is 0 and there are neither boys nor girls. I am imagining a large country in which their are many births happening over a long period of time (hence my analogy to the pile of marbles and my claims about what happens "in the long run"). | |
| Jan 5, 2011 at 18:24 | comment | added | Emerton | (Also, I should add that I'm happy to keep discussing this, since I would like to clarify my misunderstandings, but if you would prefer not to, please just let me know.) | |
| Jan 5, 2011 at 18:23 | comment | added | Emerton | Dear Steve, I agree that in a single family the situation is different, but in a country, the dynamics of individual families might be very complicated. This is why I am think just in terms of the overall population. If we forget about children for a moment, and just think about marbles in a pile, am I right in arguing that if I add marbles to the pile with an even chance of red or black, that in the long run the proportion of red to black will be 1:1? Note: I am not asking about expected values, but about the actual proportions in the pile. | |
| Jan 5, 2011 at 18:18 | comment | added | Steven Landsburg | Emerton: PS--- the example I've posted here might help: landsburg.org/alt.txt | |
| Jan 5, 2011 at 18:14 | comment | added | Steven Landsburg | e) Therefore to conclude that E(G/G+B) = 1/2, one must make at least one additional assumption that is not stated in the problem. | |
| Jan 5, 2011 at 18:13 | comment | added | Steven Landsburg | Emerton: Your argument proves that E(G)=E(B). It does not address the question of E(G/G+B). It will help a lot to clarify things, I think, if you can tell me which of the following statements is the first one you disagree with: a) For a single family, E(G/G+B) = 1-log(2). b) Therefore for a country with just one family, E(G/G+B)= 1-log(2). c) Therefore, there exists a model in which, for the country, E(G/G+B) != 1/2. d) Therefore it is not true that in every model, for the country, E(G/G+B) = 1/2. (CONTINUED...) | |
| Jan 5, 2011 at 16:16 | comment | added | Emerton | "Isn't the same as" was supposed to read "Isn't it the same as" (and sorry for the slightly ungrammatical sentences that follow, as well as the other sundry typos and misspellings). | |
| Jan 5, 2011 at 16:13 | comment | added | Emerton | ... marbles don't die, so they are not being removed from the pile, whereas in a population, people are being removed. Is this what you are worried about? | |
| Jan 5, 2011 at 16:12 | comment | added | Emerton | Dear Steven, I don't see how the population is not 1/2 boy and 1/2 girl, but maybe I'm being dense. I imagine a population somewhere. Now children are being born. 1/2 the time the child is a boy, the other 1/2 its a girl. These children keep appearing over time. How does the population not (in the long run) stablize to be 1/2 boy and 1/2 girl? Isn't the same as imagining a big pile of marbles, some red, some black. Whatever is there to begin with, I start adding marbles, red half the time, black half the time. In the long run, the pile will be half red and half black. Of course, ... | |
| Jan 5, 2011 at 14:37 | comment | added | Steven Landsburg | Emerton: I am having trouble following your argument. 1) Certainly E(G) = E(B), whether the population is large or small. 2) Certainly it does not follow from this that "1/2 the population will be boy and half girl". If this means G/G+B = 1/2, well, G/G+B could in fact be just about anything at all. If it means E(G/G+B) = 1/2, that's a better conjecture but it's still false. 3) It's easy to write down a simple model where k is independent of the stopping rule and still E(G/G+B) != 1/2. In fact, it's quite difficult to construct a model in which E(G/G+B) is 1/2. | |
| Jan 5, 2011 at 14:22 | comment | added | Emerton | ... but this is only apparent, because $k$ is not a quantity which is independent of the stopping rule, but is in fact being influenced by the stopping rule (it is related to the growth of the population, which as Vipal note's will potentially be affected by the stopping rule). Does this make any sense? | |
| Jan 5, 2011 at 14:20 | comment | added | Emerton | Dear Steven, I think that T. is saying something similar to the second para. of Vipal Naik's answer, namely, that in a large polulation, a random birth (happening somewhere in the population) will be a girl 1/2 the time, and a boy the other 1/2 of the time. So 1/2 the population will be boy, and half girl. As Vipal then goes on to discuss, the growth of the population, for example, may be affected by stopping strategies of the kind discussed here. T.'s point then is that, if you look at how things depend on the number of families, i.e. on $k$, you can find an apparent asymmetry, ... | |
| Jan 4, 2011 at 14:58 | comment | added | Steven Landsburg | T.: If you're not assuming any specific model, I don't see how you can be getting a specific answer. I think it would help me understand your point if you could give an explicit example (specifying number of families, mortality rates, fertility of the chidren, etc) in which the answer is 1/2. | |
| Jan 4, 2011 at 7:20 | comment | added | T.. | I'm not assuming any specific model, but pointing out that differences between your answer and 1/2 arise from artificial (i.e., gender-asymmetric) conditioning of the problem. Assuming $k$ families as in Zare's model or your present suggestion, is equivalent to assuming "at most $k$ boys in population", or exactly $k$ boys if it is also assumed the families complete their reproduction. No such asymmetric conditioning was part of the Google problem. Your calculations show that a symmetrical distribution can be approximated by asymmetric ones, not that the Google problem is asymmetric. | |
| Jan 4, 2011 at 6:20 | comment | added | Steven Landsburg | T: I can't tell what assumptions you're making. Surely if we assume a fixed finite number of families, all of whose children are infertile, the calculation above is correct (or if it isn't, I hope you can point to the exact place where it goes wrong). If you are making some alternative assumptions, it would be good for you to state them. | |
| Jan 4, 2011 at 6:05 | comment | added | T.. | Steven, that is incorrect. The issue is whether the proportion of boys, denoted $f(B,G)$ above, is convex as a function of two variables so that $E[f(B,G)] > f(E[B],E[G]) = 1/2$. It isn't convex, as simple calculations demonstrate. (See the comments on convexity and Jensen's inequality under Douglas Zare's posting). It is convex if you condition on B (i.e., restrict f to lines B=constant), and concave if you condition on G. Such conditioning is foreign to the Google problem and imposed artificially in Doug's model. | |
| Dec 22, 2010 at 21:47 | history | answered | Steven Landsburg | CC BY-SA 2.5 |