3 ms·

This is all mixed up. Alice has real random variables X_1 (borrowing TeX notation for a subscript), X_2, X_3. We assume that {X_i|i = 1, 2, 3} is independent.

by graycat 12y ago

This is all mixed up. Alice has real random variables X_1 (borrowing TeX notation for a subscript), X_2, X_3. We assume that {X_i|i = 1, 2, 3} is independent. We assume that for some number p in [0,1], P(X_i = 0) = p -- Alice uses 0 for H and 1 for T. Alice observes that X_1 = 0, X_2 = 0, and X_3 = 1. Now, for the set of real numbers, R, Alice has a Borel measurable function f: R^3 --> [0,1] and lets f(X_1, X_2, X_3) be her estimate of p. Okay, if that is what you meant. Bob does much the same with real random variables Y_i, i = 1, 2, 3 and function g: R^3 --> [0,1] and lets his estimate of p be g(Y_1, Y_2, Y_3). And Bob observes Y_1 = 0, Y_2 = 0, and Y_3 = 1. If functions f and g are the same, then, in this case, that is, with the data Alice and Bob observed, Alice and Bob get the same estimate for p. Else if f and g are different, then, even if the data they observe is the same, they might get different estimates. Even if f = g, since each of Alice and Bob is flipping the coin for themselves, they need not get the same estimate for p. Note: Since f is Borel measurable, we can set real random variable Z = f(X_1, X_2, X_3) and ask for E[Z], etc. Sometimes this step is useful. That is, our estimator of p is also a random variable. We might like to have E[Z] = p; in this case, Z is an 'unbiased' estimator of p. Also we can ask for the variance of Z, say, Var(Z), and maybe we want Var(Z) to be small. If we can show that Var(Z) is the smallest among all Borel measurable f, then the choice Alice made for f is a 'minimum variance' estimator. If Alice gets charged money for being wrong, then we can try to minimize the expected value of what Alice gets charged, and here we have a case of 'statistical decision theory'. Since order statistics are always sufficient, with the assumptions we have, Alice need only be told 2 zeros and 1 1 and can f'get about the rest. I see no surprises or difficulties here. But the sample space Omega is the same for both Alice and Bob. And, for both Alice and Bob, there is only one trial, that is, only one point little omega, in Omega involved. That is, in more detail, X_1 is a function, the function X_1: Omega --> R, and in our case for our trial little omega we have X_1(little omega) = 0. That is, usually in the notation we suppress little omega. Alice and Bob are both using the same trial little omega and the same sigma algebra script F on the same sample space Omega. We have no reason to believe that random variables X_1: Omega --> R and Y_1: Omega --> R are equal. Thus, given a Borel subset K of R, the events X_1^{-1}(K) in the sigma algebra script F on Omega and Y_1^{-1}(K) also in the sigma algebra script F on Omega need not be the same. That's a little of how applied probability based on 'modern probability' works. I see no problems and no need to consider frequentism or Bayesianism. Questions?

kgwgk 12y ago
"We might like to have E[Z]=p","maybe we want Var[Z] to be small"... 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. Try to go further in the 'statistical decision' field and see how longer can you avoid frequentist considerations. What do you think of the "likelihood principle"? Alice and Bob get the same data and the same likelihood function. Should they make the same inference? Do you think unbiasedness is an important property for an estimator? What makes an estimator admissible? Do you see a problem with an estimator that provides negative values for non-negative variables? By the way, I don't think you analysis is correct: Alice doesn't have random variables X_1, X_2, X_3. The possible events for her are T,HT,HHT,HHHT,HHHHT,... (she stops when she get tails, but not before). EDIT: In case it's not yet clear: saying that "our estimator of p is also a random variable" and looking at its sampling distribution is frequentism.
- deleted 12y ago
  [deleted]
- 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.

by graycat 12y ago

kgwgk 12y ago

"We might like to have E[Z]=p","maybe we want Var[Z] to be small"... 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. Try to go further in the 'statistical decision' field and see how longer can you avoid frequentist considerations. What do you think of the "likelihood principle"? Alice and Bob get the same data and the same likelihood function. Should they make the same inference? Do you think unbiasedness is an important property for an estimator? What makes an estimator admissible? Do you see a problem with an estimator that provides negative values for non-negative variables? By the way, I don't think you analysis is correct: Alice doesn't have random variables X_1, X_2, X_3. The possible events for her are T,HT,HHT,HHHT,HHHHT,... (she stops when she get tails, but not before). EDIT: In case it's not yet clear: saying that "our estimator of p is also a random variable" and looking at its sampling distribution is frequentism.

deleted 12y ago
[deleted]
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.