3 ms·

Your use of "of course" seems to imply that there's some statistical reason that the probability of the next bus being inbound vs outbound wouldn't be equal. Is

Your use of "of course" seems to imply that there's some statistical reason that the probability of the next bus being inbound vs outbound wouldn't be equal. Is there? If so, it seems like it must be a different reason than the one in the article. What am I missing...?

theoh 8y ago
It is impossible to catch more than one inbound bus on any given occasion, whereas any number of outbound buses might pass. BTW, on a slightly unrelated point, if there's no timetable, but the interval between buses is maintained reliably, the expected waiting time is uniformly distributed over that interval. If you have to get a second bus, you need to convolve two of those two uniform distributions to find out the distribution of overall journey times. This is a trapezoidal distribution, which is just about analytically manageable. But a journey with two transfers (3 buses in total) results in a likely overall time distributed according to a uniform distribution convolved with a trapezoidal distribution, which is a very weird non-smooth shape. You can see why people choose to model distributions with Gaussians, which are well-behaved (convolve two Gaussians, get another Gaussian). The Gaussian just lends itself ideally to recursive applications, hence recursive filtering (e.g. Kalman filters).
- LolWolf 8y ago
  Also, gaussians are great approximations for large n, too, since the convolution of any distribution with itself n times (for n "large enough") is close to gaussian (by the CLT. More generally, there are very nice error estimates for many distributions). I suspect this analysis can be carried out and yield quite good results in the gaussian case (a careful analysis might even yield error bounds on the result).
  - theoh 8y ago
    Yes. If you spend your whole life on one long multi-transfer bus journey, you'll end up with a gaussian. It's a bit less clear that gaussians should be used when e.g. fitting a coordinate to an astronomical feature, which might not actually be symmetrical. The other useful property that the gaussian has is its separability, in the 2D case. That is unique to the gaussian and counts for a lot.
    - LolWolf 8y ago
      Eh, I don’t think that many are required. Convergence to a Gaussian is pretty fast (you should check out page 299 of [0]), at four or five a Gaussian is already a quite good approximations. ——— [0] https://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter7.pdf https://www.dartmouth.edu/~chance/teaching_aids/books_articl...
dmurray 8y ago
It's not the next bus, it's the number of buses going the other way, which will on average be greater than 0.5 by the same reasoning given in the article. It's also possible for the result to be biased because of scheduling. If inbound buses pass every 10 minutes at 16.00, 16.10, 16.20,... and outbound buses at 16.01, 16.11, 16.21, ... you'll usually see an inbound bus first. Though I expect this was not the case here.
herodotus 8y ago
Because we might score -1, -2 or worse if 2 or 3 busses went in the other direction before ours came, but if ours came first, we score 1. We get on the bus and thus don’t know if another one or more arrives first on our side.
- repsilat 8y ago
  This reminds me of a mathematical paradox that makes me doubt your conclusion: "In this country, every couple wants to have one daughter. They keep having children until they have a daughter, and then they stop. What gender balance should we expect?" Couples can have any number of sons, and every couple has exactly one daughter. Still, the accepted mathematical solution is an equal gender ratio for the couples' children.
  - DalekBaldwin 8y ago
    The expected number of sons is 1, and the expected number of daughters is 1 (by the framing of the problem, in every possible scenario, there is exactly one daughter), but the expected value of the ratio is not 1:1. E[X]/E[Y] = E[X/Y] is not a valid identity. http://www.thebigquestions.com/2010/12/22/a-big-answer-2/ http://www.thebigquestions.com/2010/12/22/a-big-answer-2/
    - Pristina 8y ago
      I read that and it seems wrong. The question asked "what fraction of the pop is female" but his argument is that 3 families of 4 girls and 1 family of 12 boys make the fraction of girls in the average family 75% (the average of 100% x3 and 0% x1) which is non-sensical to me. >E[X]/E[Y] = E[X/Y] is not a valid identity. is completely irrelevant here because it is being used to point out that a non-answer is wrong.
      - DalekBaldwin 8y ago
        He starts with a simpler problem to elucidate the principle that is key to the original problem. Do continue reading.

theoh 8y ago

It is impossible to catch more than one inbound bus on any given occasion, whereas any number of outbound buses might pass. BTW, on a slightly unrelated point, if there's no timetable, but the interval between buses is maintained reliably, the expected waiting time is uniformly distributed over that interval. If you have to get a second bus, you need to convolve two of those two uniform distributions to find out the distribution of overall journey times. This is a trapezoidal distribution, which is just about analytically manageable. But a journey with two transfers (3 buses in total) results in a likely overall time distributed according to a uniform distribution convolved with a trapezoidal distribution, which is a very weird non-smooth shape. You can see why people choose to model distributions with Gaussians, which are well-behaved (convolve two Gaussians, get another Gaussian). The Gaussian just lends itself ideally to recursive applications, hence recursive filtering (e.g. Kalman filters).

LolWolf 8y ago
Also, gaussians are great approximations for large n, too, since the convolution of any distribution with itself n times (for n "large enough") is close to gaussian (by the CLT. More generally, there are very nice error estimates for many distributions). I suspect this analysis can be carried out and yield quite good results in the gaussian case (a careful analysis might even yield error bounds on the result).
- theoh 8y ago
  Yes. If you spend your whole life on one long multi-transfer bus journey, you'll end up with a gaussian. It's a bit less clear that gaussians should be used when e.g. fitting a coordinate to an astronomical feature, which might not actually be symmetrical. The other useful property that the gaussian has is its separability, in the 2D case. That is unique to the gaussian and counts for a lot.
  - LolWolf 8y ago
    Eh, I don’t think that many are required. Convergence to a Gaussian is pretty fast (you should check out page 299 of [0]), at four or five a Gaussian is already a quite good approximations. ——— [0] https://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter7.pdf https://www.dartmouth.edu/~chance/teaching_aids/books_articl...

dmurray 8y ago

It's not the next bus, it's the number of buses going the other way, which will on average be greater than 0.5 by the same reasoning given in the article. It's also possible for the result to be biased because of scheduling. If inbound buses pass every 10 minutes at 16.00, 16.10, 16.20,... and outbound buses at 16.01, 16.11, 16.21, ... you'll usually see an inbound bus first. Though I expect this was not the case here.

herodotus 8y ago

Because we might score -1, -2 or worse if 2 or 3 busses went in the other direction before ours came, but if ours came first, we score 1. We get on the bus and thus don’t know if another one or more arrives first on our side.

repsilat 8y ago
This reminds me of a mathematical paradox that makes me doubt your conclusion: "In this country, every couple wants to have one daughter. They keep having children until they have a daughter, and then they stop. What gender balance should we expect?" Couples can have any number of sons, and every couple has exactly one daughter. Still, the accepted mathematical solution is an equal gender ratio for the couples' children.
- DalekBaldwin 8y ago
  The expected number of sons is 1, and the expected number of daughters is 1 (by the framing of the problem, in every possible scenario, there is exactly one daughter), but the expected value of the ratio is not 1:1. E[X]/E[Y] = E[X/Y] is not a valid identity. http://www.thebigquestions.com/2010/12/22/a-big-answer-2/ http://www.thebigquestions.com/2010/12/22/a-big-answer-2/
  - Pristina 8y ago
    I read that and it seems wrong. The question asked "what fraction of the pop is female" but his argument is that 3 families of 4 girls and 1 family of 12 boys make the fraction of girls in the average family 75% (the average of 100% x3 and 0% x1) which is non-sensical to me. >E[X]/E[Y] = E[X/Y] is not a valid identity. is completely irrelevant here because it is being used to point out that a non-answer is wrong.
    - DalekBaldwin 8y ago
      He starts with a simpler problem to elucidate the principle that is key to the original problem. Do continue reading.