3 ms·

> How do you select f and g? That's the difficulty. You see no problems because you're happy playing with your Borel measurable thingies. Borel measurablility

by graycat 12y ago

> How do you select f and g? That's the difficulty. You see no problems because you're happy playing with your Borel measurable thingies. Borel measurablility is important work; it sounds like you have contempt for it. There is no need or justification for contempt. We should mention Borel measurability, as I did, if we are carefully considering the Kolmogorov foundations, but in practice Borel measurability means essentially nothing since cooking up a function, such as f or g in what I wrote, that is not Borel measurable is so tricky that essentially any function anyone would select for f or g will be Borel measurable. The usual example of a function not Borel or Lebesgue measurable uses the axiom of choice -- we're talking tricky stuff won't see in SPSS, SAS, R, Mathematica, Matlab, big data, machine learning, etc. For picking f or g, let's see, from our definition of p above, p = P( X_i = 0 ) = P( heads ) and 1 - p = P( X_i = 1 ) So, for Bob's case, we can just set Z = g(X_1, X_2, X_3) = 1 - (1/3) (X_1 + X_2 + X_3) which is what anyone would guess anyway. Or, for more, for i = 1, 2, 3, E[X_i] = 0 P(X_i = 0) + 1 P(X_i = 1) = 1 - p so that E[Z] = E[ 1 - (1/3) (X_1 + X_2 + X_3) ] = 1 - (1/3) E[ X_1 + X_2 + X_3 ] = 1 - (1/3) ( 3(1 - p) ) = p so that our estimator Z = g(X_1, X_2, X_3) = 1 - (1/3) (X_1 + X_2 + X_3) is an unbiased estimator of p. We expected something else? This was difficult? Here we specified a function g and showed that it gives an unbiased estimator of p and did this without mentioning the trial little omega, the sample space big Omega, or the sigma algebra of events script F. So, what we did is just an elementary part of standard junior level introductory mathematical statistics, and so would be responses to your other questions. And, again, we never mentioned frequentism or Bayesianism. Yet again, when considering the mathematical foundations of probability, I see no need to consider either frequentism or Bayesianism. Again, the Kolmogorov foundations work just fine for long standard probability, applied probability, mathematical statistics, and applied statistics. No worries. For Alice, a different derivation is required. My point is that from Kolmogorov we have some rock solid foundations for probability and, then, do not have to consider either frequentism or Bayesianism. So, students struggling over frequentism or Bayesianism can relax and just f'get about these two.

kgwgk 12y ago
> So, for Bob's case, (....) so that our estimator Z = g(X_1, X_2, X_3) = 1 - (1/3) (X_1 + X_2 + X_3) is an unbiased estimator of p. We expected something else? This was difficult? (....) For Alice, a different derivation is required. A different derivation that you tried (I saw your comment appear briefly). You proposed Z=1/N, which is (as before) the maximum-likelihood estimator. And the likelihood function is the same for Alice and for Bob, so it is not surprising that we get the same estimate p=1/3 in both cases. But after "proving" that "again our estimator Z is unbiased" I guess you noticed that this was not in fact correct. Did you expect something else? What was the difficulty? > My point is that from Kolmogorov we have some rock solid foundations for probability and, then, do not have to consider either frequentism or Bayesianism. Both are based on probability, how is probability going to replace them? > So, students struggling over frequentism or Bayesianism can relax and just f'get about these two. Sure, they can avoid thinking about the different approaches to inference... and just do it in the frequentist way. Assuming that the unknown parameter is fixed, that the estimator is a random variable, that the criteria to select an estimator are the unbiasedness, consistency or asymptotic distribution... all of these are frequentist considerations (even if you say that you "do not have to consider either frequentism or Bayesianism"). You never answered my questions about Alice and Bob getting different interval estimates (even though the outcome of their experiments is identical and their model for the loaded coin as well). We've seen that they agree on their (MLE) point estimate, but their confidence intervals will be different. I imagine you accept the frequentist idea of confidence intervals, and agree that they will be different because the distribution of potential experiment outcomes is different. Do you think that Alice and Bob can get different conclusions from the same model and the same data? Carol performs another experiment. She starts by rolling a die to see if she'll do it like Alice (even) or like Bob (odd). So with 50% probability she will throw the loaded coin 3 times and with 50% probability she will do it until she gets a tail. She gets 5 on the die, she throws the loaded coin three times and gets 'HTH'. How will you select your estimator for p? Do you use the results you obtained for Bob? Do you repeat you analysis considering the mixture of both experiments?

by graycat 12y ago

kgwgk 12y ago

> So, for Bob's case, (....) so that our estimator Z = g(X_1, X_2, X_3) = 1 - (1/3) (X_1 + X_2 + X_3) is an unbiased estimator of p. We expected something else? This was difficult? (....) For Alice, a different derivation is required. A different derivation that you tried (I saw your comment appear briefly). You proposed Z=1/N, which is (as before) the maximum-likelihood estimator. And the likelihood function is the same for Alice and for Bob, so it is not surprising that we get the same estimate p=1/3 in both cases. But after "proving" that "again our estimator Z is unbiased" I guess you noticed that this was not in fact correct. Did you expect something else? What was the difficulty? > My point is that from Kolmogorov we have some rock solid foundations for probability and, then, do not have to consider either frequentism or Bayesianism. Both are based on probability, how is probability going to replace them? > So, students struggling over frequentism or Bayesianism can relax and just f'get about these two. Sure, they can avoid thinking about the different approaches to inference... and just do it in the frequentist way. Assuming that the unknown parameter is fixed, that the estimator is a random variable, that the criteria to select an estimator are the unbiasedness, consistency or asymptotic distribution... all of these are frequentist considerations (even if you say that you "do not have to consider either frequentism or Bayesianism"). You never answered my questions about Alice and Bob getting different interval estimates (even though the outcome of their experiments is identical and their model for the loaded coin as well). We've seen that they agree on their (MLE) point estimate, but their confidence intervals will be different. I imagine you accept the frequentist idea of confidence intervals, and agree that they will be different because the distribution of potential experiment outcomes is different. Do you think that Alice and Bob can get different conclusions from the same model and the same data? Carol performs another experiment. She starts by rolling a die to see if she'll do it like Alice (even) or like Bob (odd). So with 50% probability she will throw the loaded coin 3 times and with 50% probability she will do it until she gets a tail. She gets 5 on the die, she throws the loaded coin three times and gets 'HTH'. How will you select your estimator for p? Do you use the results you obtained for Bob? Do you repeat you analysis considering the mixture of both experiments?