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From: Huang on 14 Jan 2010 23:24 On Jan 14, 7:22 pm, "Ostap S. B. M. Bender Jr." <ostap_bender_1...(a)hotmail.com> wrote: > On Jan 14, 6:19 am, Huang <huangxienc...(a)yahoo.com> wrote: > > > > > > > On Jan 14, 7:46 am, Huang <huangxienc...(a)yahoo.com> wrote: > > > > On Jan 14, 12:13 am, "preedmont" <nos...(a)spamless.com> wrote: > > > > > "Huang" <huangxienc...(a)yahoo.com> wrote in message > > > > >news:001d62db-a3e2-47da-b185-4b8d041b1e8d(a)p8g2000yqb.googlegroups.com... > > > > > > Consider 3 trials of a random variable { X | H, T } . > > > > > > The result is no different than a single trial of the random variable > > > > > { X | HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } . > > > > > the first has 3 events, the second has just one event. > > > > the first is has 2 states, the second has 8. > > > > > The results can be the same, does not imply the process is the same.. > > > > > The first can be a coin, the second is an 8 sided thingie. > > > > > > How do we get back and forth from one to the other ? > > > > > by moving eyes around from one to the other. > > > > > > How to we make > > > > > the conversion ? > > > > > trivial, you did it > > > > > >How to perform such a transform ? Obviously these > > > > > things are equivalent, but how to do it without the hand wave ? > > > > > they are not the same. > > > > True. These things are not the same. They differ in process, but the > > > result is only trivially different. In each case you will get a > > > squence of 3 outcomes, and both processes produce an identical outcome > > > space. They are just worded differently. > > > > It is true that the processes differ. But the difference is trivial. > > > It is very much like saying that > > > > 5 = 5 + 0 = 5 + 0 + 0 = 5 + 0 + 0 + 0 + 0 etc > > > > These things are different, but the difference is trivial. > > > > There should be a way to do this algebraically. Shoul be a way to show > > > symbolically that : > > > > 3 trials of a random variable > > > { X | H, T } > > > is no different than a single trial of the random variable > > > { X | HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } . > > > > I think that the way we can show this is by treating conservation as a > > > trivial operator. This operator would work equally well with discrete > > > and continuous random variables, because it is trivial it does not > > > care if it is multiplying or adding.- Hide quoted text - > > > > - Show quoted text - > > > Lets make a tentative symbol to represent the trivial operator, > > something like this : > > > [@] > > > Then, for discrete random variables, from the example above, we might > > write something like this > > > 3 [@] { X | H, T } = { X | HHH, HHT, HTT, THH, THT, TTT, HTH, > > TTH } > > > or possibly > > > { X | H, T } [@] { X | H, T } [@] { X | H, T } = { X | HHH, > > HHT, HTT, THH, THT, TTT, HTH, TTH } > > > and of course the operator [@] should behave the same whether we are > > considering continuous or discrete random variables, I dont have a > > continuous example at hand but it should work. > > > (It would'nt fit in the margin) > > > One very important thing to note, if we write > > 3 [@] { X | H, T } , > > then we are combining numbers with things which are not numbers. This > > is not neccesarily a criminal act at this point but definately an > > important consideration. > > > It does make sense for many reasons that conservation would be > > regarded as a genuine operator. Trivial, but genuine. > > Since this child play seems to never end, let me tell you that the > concept that you are looking for is called 'Cartesian Product'. It is > usually covered in the first month of any introductory course in > probability theory. You can read about it for example here: > > http://media.wiley.com/product_data/excerpt/56/04712597/0471259756.pdf > > 1 Experiments and Probability > > 1.2 COMBINED EXPERIMENTS > > Combined experiments play an important role in probability theory > applications. There are many ways we can combine experiments, > including cartesian products in which independent trials of the same > or different experiments can be described. > > 1.2.1 Cartesian Product of Two Experiments > > Consider the case of having two separate experiments specified by the > following: i{ : ( 5 '1,^1,^1(.)) and S'2 : (S2, ^ 2 , ^2(.)). The > sample spaces Sx and S2 are usually different sets, for example, > results of a coin toss and results of a die roll, but they could be > the same sets representing separate trials of the same experiment as > in repeated coin-tossing experiments. We can define a new combined > experiment by using the cartesian product concept as S $\®S'2, where > the new sample space S = S{ ® S2 is the cartesian product of the two > sample spaces expressible by the ordered pair of elements where the > first element is from S{ and the second is from S2. > > 1.2.2 Cartesian Product of n Experiments > > Consider the case of having n separate experiments specified by Sk : > (S/c, 3Fk, &k(.)) for k = 1, 2 , . . . , n. Define a new combined > experiment S : (S, ^ , ^(.)) as a cartesian product: <f = #'t ® ^ 2® * > * *® $n where the new sample space S = Sx ® S2® -® Sn is the > cartesian product of the n spaces and expressible by the ordered ^- > tuples of elements whose first element is from Sx and the second from > S2 . . . , the nth from Sn. The $k are, in general, different, but in > many cases the experiment could be formed from independent trials of > the same experiment.- Hide quoted text - > > - Show quoted text - Thank you for the excellent post and link. My only comment at the moment is that the cartesian product is a nontrivial operator. It is somewhat different than what we are talking about. Definately nontrivial. Consider 3 trials of a random variable { X | H, T } and a single trial of the random variable { X | HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } . The difference between the two things above is strictly trivial. And so too - here's another example of something trivial 5 = 5 + 0 = 5 = 0 + 0 = 5 + 0 + 0 + 0 + 0 etc The only difference in any of the above terms is trivial. While there is a trivial element in set theory (the empty set), the cartesian product is a nontrivial operation. My claim is that conservation can be regarded as an operator, a trivial operator. It is a process of composing things where the result is "no change", unlike addition or multiplication (disregarding the behaviour of identity elements).
From: Ostap S. B. M. Bender Jr. on 15 Jan 2010 01:07 On Jan 14, 8:24 pm, Huang <huangxienc...(a)yahoo.com> wrote: > On Jan 14, 7:22 pm, "Ostap S. B. M. Bender Jr." > > > > > > <ostap_bender_1...(a)hotmail.com> wrote: > > On Jan 14, 6:19 am, Huang <huangxienc...(a)yahoo.com> wrote: > > > > On Jan 14, 7:46 am, Huang <huangxienc...(a)yahoo.com> wrote: > > > > > On Jan 14, 12:13 am, "preedmont" <nos...(a)spamless.com> wrote: > > > > > > "Huang" <huangxienc...(a)yahoo.com> wrote in message > > > > > >news:001d62db-a3e2-47da-b185-4b8d041b1e8d(a)p8g2000yqb.googlegroups.com... > > > > > > > Consider 3 trials of a random variable { X | H, T } . > > > > > > > The result is no different than a single trial of the random variable > > > > > > { X | HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } . > > > > > > the first has 3 events, the second has just one event. > > > > > the first is has 2 states, the second has 8. > > > > > > The results can be the same, does not imply the process is the same. > > > > > > The first can be a coin, the second is an 8 sided thingie. > > > > > > > How do we get back and forth from one to the other ? > > > > > > by moving eyes around from one to the other. > > > > > > > How to we make > > > > > > the conversion ? > > > > > > trivial, you did it > > > > > > >How to perform such a transform ? Obviously these > > > > > > things are equivalent, but how to do it without the hand wave ? > > > > > > they are not the same. > > > > > True. These things are not the same. They differ in process, but the > > > > result is only trivially different. In each case you will get a > > > > squence of 3 outcomes, and both processes produce an identical outcome > > > > space. They are just worded differently. > > > > > It is true that the processes differ. But the difference is trivial.. > > > > It is very much like saying that > > > > > 5 = 5 + 0 = 5 + 0 + 0 = 5 + 0 + 0 + 0 + 0 etc > > > > > These things are different, but the difference is trivial. > > > > > There should be a way to do this algebraically. Shoul be a way to show > > > > symbolically that : > > > > > 3 trials of a random variable > > > > { X | H, T } > > > > is no different than a single trial of the random variable > > > > { X | HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } . > > > > > I think that the way we can show this is by treating conservation as a > > > > trivial operator. This operator would work equally well with discrete > > > > and continuous random variables, because it is trivial it does not > > > > care if it is multiplying or adding.- Hide quoted text - > > > > > - Show quoted text - > > > > Lets make a tentative symbol to represent the trivial operator, > > > something like this : > > > > [@] > > > > Then, for discrete random variables, from the example above, we might > > > write something like this > > > > 3 [@] { X | H, T } = { X | HHH, HHT, HTT, THH, THT, TTT, HTH, > > > TTH } > > > > or possibly > > > > { X | H, T } [@] { X | H, T } [@] { X | H, T } = { X | HHH, > > > HHT, HTT, THH, THT, TTT, HTH, TTH } > > > > and of course the operator [@] should behave the same whether we are > > > considering continuous or discrete random variables, I dont have a > > > continuous example at hand but it should work. > > > > (It would'nt fit in the margin) > > > > One very important thing to note, if we write > > > 3 [@] { X | H, T } , > > > then we are combining numbers with things which are not numbers. This > > > is not neccesarily a criminal act at this point but definately an > > > important consideration. > > > > It does make sense for many reasons that conservation would be > > > regarded as a genuine operator. Trivial, but genuine. > > > Since this child play seems to never end, let me tell you that the > > concept that you are looking for is called 'Cartesian Product'. It is > > usually covered in the first month of any introductory course in > > probability theory. You can read about it for example here: > > >http://media.wiley.com/product_data/excerpt/56/04712597/0471259756.pdf > > > 1 Experiments and Probability > > > 1.2 COMBINED EXPERIMENTS > > > Combined experiments play an important role in probability theory > > applications. There are many ways we can combine experiments, > > including cartesian products in which independent trials of the same > > or different experiments can be described. > > > 1.2.1 Cartesian Product of Two Experiments > > > Consider the case of having two separate experiments specified by the > > following: i{ : ( 5 '1,^1,^1(.)) and S'2 : (S2, ^ 2 , ^2(.)). The > > sample spaces Sx and S2 are usually different sets, for example, > > results of a coin toss and results of a die roll, but they could be > > the same sets representing separate trials of the same experiment as > > in repeated coin-tossing experiments. We can define a new combined > > experiment by using the cartesian product concept as S $\®S'2, where > > the new sample space S = S{ ® S2 is the cartesian product of the two > > sample spaces expressible by the ordered pair of elements where the > > first element is from S{ and the second is from S2. > > > 1.2.2 Cartesian Product of n Experiments > > > Consider the case of having n separate experiments specified by Sk : > > (S/c, 3Fk, &k(.)) for k = 1, 2 , . . . , n. Define a new combined > > experiment S : (S, ^ , ^(.)) as a cartesian product: <f = #'t ® ^ 2® * > > * *® $n where the new sample space S = Sx ® S2® -® Sn is the > > cartesian product of the n spaces and expressible by the ordered ^- > > tuples of elements whose first element is from Sx and the second from > > S2 . . . , the nth from Sn. The $k are, in general, different, but in > > many cases the experiment could be formed from independent trials of > > the same experiment.- Hide quoted text - > > > - Show quoted text - > > Thank you for the excellent post and link. My only comment at the > moment is that the cartesian product is a nontrivial operator. It is > somewhat different than what we are talking about. Definately > nontrivial. > The concept of "triviality" depends on the context. If you are not familiar with the concept of n-tuples and Cartesian coordinates - then I suppose the Cartesian Product is "nontrivial". However, the cure for this is not to re-invent the wheel but instead to study the basic foundations of mathematics. I mean, Descartes is hardly a newcomer. http://en.wikipedia.org/wiki/Descartes René Descartes ((31 March 1596 11 February 1650), also known as Renatus Cartesius (Latinized form),[2] was a French philosopher, mathematician, physicist, and writer. The invention of Cartesian coordinates in the 17th century by René Descartes revolutionized mathematics by providing the first systematic link between Euclidean geometry and algebra. I saw you use the term "random variable" correctly. Thus, you do have familiarity with the terminology of probability theory. So, look up the way in probability theory, the Cartesian Product captures independent trials. It's really not rocket science. > > Consider 3 trials of a random variable > { X | H, T } > and a single trial of the random variable > { X | HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } . > > The difference between the two things above is strictly trivial. And > so too - here's another example of something trivial > > 5 = 5 + 0 = 5 = 0 + 0 = 5 + 0 + 0 + 0 + 0 etc > > The only difference in any of the above terms is trivial. > > While there is a trivial element in set theory (the empty set), the > cartesian product is a nontrivial operation. > Well, if you restrict yourself only to trivial constructs and empty sets, your results will remain trivial and empty. Garbage in - garbage out. > > My claim is that conservation can be regarded as an operator, a > trivial operator. It is a process of composing things where the result > is "no change", unlike addition or multiplication (disregarding the > behaviour of identity elements).- Hide quoted text - > > - Show quoted text -
From: Huang on 15 Jan 2010 07:19 On Jan 15, 12:07 am, "Ostap S. B. M. Bender Jr." <ostap_bender_1...(a)hotmail.com> wrote: > On Jan 14, 8:24 pm, Huang <huangxienc...(a)yahoo.com> wrote: > > > > > > > On Jan 14, 7:22 pm, "Ostap S. B. M. Bender Jr." > > > <ostap_bender_1...(a)hotmail.com> wrote: > > > On Jan 14, 6:19 am, Huang <huangxienc...(a)yahoo.com> wrote: > > > > > On Jan 14, 7:46 am, Huang <huangxienc...(a)yahoo.com> wrote: > > > > > > On Jan 14, 12:13 am, "preedmont" <nos...(a)spamless.com> wrote: > > > > > > > "Huang" <huangxienc...(a)yahoo.com> wrote in message > > > > > > >news:001d62db-a3e2-47da-b185-4b8d041b1e8d(a)p8g2000yqb.googlegroups.com... > > > > > > > > Consider 3 trials of a random variable { X | H, T } . > > > > > > > > The result is no different than a single trial of the random variable > > > > > > > { X | HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } . > > > > > > > the first has 3 events, the second has just one event. > > > > > > the first is has 2 states, the second has 8. > > > > > > > The results can be the same, does not imply the process is the same. > > > > > > > The first can be a coin, the second is an 8 sided thingie. > > > > > > > > How do we get back and forth from one to the other ? > > > > > > > by moving eyes around from one to the other. > > > > > > > > How to we make > > > > > > > the conversion ? > > > > > > > trivial, you did it > > > > > > > >How to perform such a transform ? Obviously these > > > > > > > things are equivalent, but how to do it without the hand wave ? > > > > > > > they are not the same. > > > > > > True. These things are not the same. They differ in process, but the > > > > > result is only trivially different. In each case you will get a > > > > > squence of 3 outcomes, and both processes produce an identical outcome > > > > > space. They are just worded differently. > > > > > > It is true that the processes differ. But the difference is trivial. > > > > > It is very much like saying that > > > > > > 5 = 5 + 0 = 5 + 0 + 0 = 5 + 0 + 0 + 0 + 0 etc > > > > > > These things are different, but the difference is trivial. > > > > > > There should be a way to do this algebraically. Shoul be a way to show > > > > > symbolically that : > > > > > > 3 trials of a random variable > > > > > { X | H, T } > > > > > is no different than a single trial of the random variable > > > > > { X | HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } . > > > > > > I think that the way we can show this is by treating conservation as a > > > > > trivial operator. This operator would work equally well with discrete > > > > > and continuous random variables, because it is trivial it does not > > > > > care if it is multiplying or adding.- Hide quoted text - > > > > > > - Show quoted text - > > > > > Lets make a tentative symbol to represent the trivial operator, > > > > something like this : > > > > > [@] > > > > > Then, for discrete random variables, from the example above, we might > > > > write something like this > > > > > 3 [@] { X | H, T } = { X | HHH, HHT, HTT, THH, THT, TTT, HTH, > > > > TTH } > > > > > or possibly > > > > > { X | H, T } [@] { X | H, T } [@] { X | H, T } = { X | HHH, > > > > HHT, HTT, THH, THT, TTT, HTH, TTH } > > > > > and of course the operator [@] should behave the same whether we are > > > > considering continuous or discrete random variables, I dont have a > > > > continuous example at hand but it should work. > > > > > (It would'nt fit in the margin) > > > > > One very important thing to note, if we write > > > > 3 [@] { X | H, T } , > > > > then we are combining numbers with things which are not numbers. This > > > > is not neccesarily a criminal act at this point but definately an > > > > important consideration. > > > > > It does make sense for many reasons that conservation would be > > > > regarded as a genuine operator. Trivial, but genuine. > > > > Since this child play seems to never end, let me tell you that the > > > concept that you are looking for is called 'Cartesian Product'. It is > > > usually covered in the first month of any introductory course in > > > probability theory. You can read about it for example here: > > > >http://media.wiley.com/product_data/excerpt/56/04712597/0471259756.pdf > > > > 1 Experiments and Probability > > > > 1.2 COMBINED EXPERIMENTS > > > > Combined experiments play an important role in probability theory > > > applications. There are many ways we can combine experiments, > > > including cartesian products in which independent trials of the same > > > or different experiments can be described. > > > > 1.2.1 Cartesian Product of Two Experiments > > > > Consider the case of having two separate experiments specified by the > > > following: i{ : ( 5 '1,^1,^1(.)) and S'2 : (S2, ^ 2 , ^2(.)). The > > > sample spaces Sx and S2 are usually different sets, for example, > > > results of a coin toss and results of a die roll, but they could be > > > the same sets representing separate trials of the same experiment as > > > in repeated coin-tossing experiments. We can define a new combined > > > experiment by using the cartesian product concept as S $\®S'2, where > > > the new sample space S = S{ ® S2 is the cartesian product of the two > > > sample spaces expressible by the ordered pair of elements where the > > > first element is from S{ and the second is from S2. > > > > 1.2.2 Cartesian Product of n Experiments > > > > Consider the case of having n separate experiments specified by Sk : > > > (S/c, 3Fk, &k(.)) for k = 1, 2 , . . . , n. Define a new combined > > > experiment S : (S, ^ , ^(.)) as a cartesian product: <f = #'t ® ^ 2® * > > > * *® $n where the new sample space S = Sx ® S2® -® Sn is the > > > cartesian product of the n spaces and expressible by the ordered ^- > > > tuples of elements whose first element is from Sx and the second from > > > S2 . . . , the nth from Sn. The $k are, in general, different, but in > > > many cases the experiment could be formed from independent trials of > > > the same experiment.- Hide quoted text - > > > > - Show quoted text - > > > Thank you for the excellent post and link. My only comment at the > > moment is that the cartesian product is a nontrivial operator. It is > > somewhat different than what we are talking about. Definately > > nontrivial. > > The concept of "triviality" depends on the context. If you are not > familiar with the concept of n-tuples and Cartesian coordinates - then > I suppose the Cartesian Product is "nontrivial". > > However, the cure for this is not to re-invent the wheel but instead > to study the basic foundations of mathematics. I mean, Descartes is > hardly a newcomer. > > http://en.wikipedia.org/wiki/Descartes > > René Descartes ((31 March 1596 11 February 1650), also known as > Renatus Cartesius (Latinized form),[2] was a French philosopher, > mathematician, physicist, and writer. The invention of Cartesian > coordinates in the 17th century by René Descartes revolutionized > mathematics by providing the first systematic link between Euclidean > geometry and algebra. > > I saw you use the term "random variable" correctly. Thus, you do have > familiarity with the terminology of probability theory. So, look up > the way in probability theory, the Cartesian Product captures > independent trials. It's really not rocket science. > > > > > > > > > Consider 3 trials of a random variable > > { X | H, T } > > and a single trial of the random variable > > { X | HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } . > > > The difference between the two things above is strictly trivial. And > > so too - here's another example of something trivial > > > 5 = 5 + 0 = 5 = 0 + 0 = 5 + 0 + 0 + 0 + 0 etc > > > The only difference in any of the above terms is trivial. > > > While there is a trivial element in set theory (the empty set), the > > cartesian product is a nontrivial operation. > > Well, if you restrict yourself only to trivial constructs and empty > sets, your results will remain trivial and empty. Garbage in - garbage > out. > > > > > > > My claim is that conservation can be regarded as an operator, a > > trivial operator. It is a process of composing things where the result > > is "no change", unlike addition or multiplication (disregarding the > > behaviour of identity elements).- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text - I'll take another look at dependence & independence and tell you what I think. But my comment that cartesian product is nontrivial, basically what that means is that the only way it can act trivially is if it operates over trivial elements. Just like addition or multiplication or anything else in math, the operators that we use are always nontrivial. Currently I do not think that mathematics recognizes a trivial operator. I think it's a shame, because it makes sense for many reasons, including some subtle philosophical considerations regarding conservation.
From: Ostap S. B. M. Bender Jr. on 15 Jan 2010 07:38
On Jan 15, 4:19 am, Huang <huangxienc...(a)yahoo.com> wrote: > On Jan 15, 12:07 am, "Ostap S. B. M. Bender Jr." > > > > <ostap_bender_1...(a)hotmail.com> wrote: > > On Jan 14, 8:24 pm, Huang <huangxienc...(a)yahoo.com> wrote: > > > > On Jan 14, 7:22 pm, "Ostap S. B. M. Bender Jr." > > > > <ostap_bender_1...(a)hotmail.com> wrote: > > > > On Jan 14, 6:19 am, Huang <huangxienc...(a)yahoo.com> wrote: > > > > > > On Jan 14, 7:46 am, Huang <huangxienc...(a)yahoo.com> wrote: > > > > > > > On Jan 14, 12:13 am, "preedmont" <nos...(a)spamless.com> wrote: > > > > > > > > "Huang" <huangxienc...(a)yahoo.com> wrote in message > > > > > > > >news:001d62db-a3e2-47da-b185-4b8d041b1e8d(a)p8g2000yqb.googlegroups.com... > > > > > > > > > Consider 3 trials of a random variable { X | H, T } . > > > > > > > > > The result is no different than a single trial of the random variable > > > > > > > > { X | HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } . > > > > > > > > the first has 3 events, the second has just one event. > > > > > > > the first is has 2 states, the second has 8. > > > > > > > > The results can be the same, does not imply the process is the same. > > > > > > > > The first can be a coin, the second is an 8 sided thingie. > > > > > > > > > How do we get back and forth from one to the other ? > > > > > > > > by moving eyes around from one to the other. > > > > > > > > > How to we make > > > > > > > > the conversion ? > > > > > > > > trivial, you did it > > > > > > > > >How to perform such a transform ? Obviously these > > > > > > > > things are equivalent, but how to do it without the hand wave ? > > > > > > > > they are not the same. > > > > > > > True. These things are not the same. They differ in process, but the > > > > > > result is only trivially different. In each case you will get a > > > > > > squence of 3 outcomes, and both processes produce an identical outcome > > > > > > space. They are just worded differently. > > > > > > > It is true that the processes differ. But the difference is trivial. > > > > > > It is very much like saying that > > > > > > > 5 = 5 + 0 = 5 + 0 + 0 = 5 + 0 + 0 + 0 + 0 etc > > > > > > > These things are different, but the difference is trivial. > > > > > > > There should be a way to do this algebraically. Shoul be a way to show > > > > > > symbolically that : > > > > > > > 3 trials of a random variable > > > > > > { X | H, T } > > > > > > is no different than a single trial of the random variable > > > > > > { X | HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } . > > > > > > > I think that the way we can show this is by treating conservation as a > > > > > > trivial operator. This operator would work equally well with discrete > > > > > > and continuous random variables, because it is trivial it does not > > > > > > care if it is multiplying or adding.- Hide quoted text - > > > > > > > - Show quoted text - > > > > > > Lets make a tentative symbol to represent the trivial operator, > > > > > something like this : > > > > > > [@] > > > > > > Then, for discrete random variables, from the example above, we might > > > > > write something like this > > > > > > 3 [@] { X | H, T } = { X | HHH, HHT, HTT, THH, THT, TTT, HTH, > > > > > TTH } > > > > > > or possibly > > > > > > { X | H, T } [@] { X | H, T } [@] { X | H, T } = { X | HHH, > > > > > HHT, HTT, THH, THT, TTT, HTH, TTH } > > > > > > and of course the operator [@] should behave the same whether we are > > > > > considering continuous or discrete random variables, I dont have a > > > > > continuous example at hand but it should work. > > > > > > (It would'nt fit in the margin) > > > > > > One very important thing to note, if we write > > > > > 3 [@] { X | H, T } , > > > > > then we are combining numbers with things which are not numbers. This > > > > > is not neccesarily a criminal act at this point but definately an > > > > > important consideration. > > > > > > It does make sense for many reasons that conservation would be > > > > > regarded as a genuine operator. Trivial, but genuine. > > > > > Since this child play seems to never end, let me tell you that the > > > > concept that you are looking for is called 'Cartesian Product'. It is > > > > usually covered in the first month of any introductory course in > > > > probability theory. You can read about it for example here: > > > > >http://media.wiley.com/product_data/excerpt/56/04712597/0471259756.pdf > > > > > 1 Experiments and Probability > > > > > 1.2 COMBINED EXPERIMENTS > > > > > Combined experiments play an important role in probability theory > > > > applications. There are many ways we can combine experiments, > > > > including cartesian products in which independent trials of the same > > > > or different experiments can be described. > > > > > 1.2.1 Cartesian Product of Two Experiments > > > > > Consider the case of having two separate experiments specified by the > > > > following: i{ : ( 5 '1,^1,^1(.)) and S'2 : (S2, ^ 2 , ^2(.)). The > > > > sample spaces Sx and S2 are usually different sets, for example, > > > > results of a coin toss and results of a die roll, but they could be > > > > the same sets representing separate trials of the same experiment as > > > > in repeated coin-tossing experiments. We can define a new combined > > > > experiment by using the cartesian product concept as S $\®S'2, where > > > > the new sample space S = S{ ® S2 is the cartesian product of the two > > > > sample spaces expressible by the ordered pair of elements where the > > > > first element is from S{ and the second is from S2. > > > > > 1.2.2 Cartesian Product of n Experiments > > > > > Consider the case of having n separate experiments specified by Sk : > > > > (S/c, 3Fk, &k(.)) for k = 1, 2 , . . . , n. Define a new combined > > > > experiment S : (S, ^ , ^(.)) as a cartesian product: <f = #'t ® ^ 2® * > > > > * *® $n where the new sample space S = Sx ® S2® -® Sn is the > > > > cartesian product of the n spaces and expressible by the ordered ^- > > > > tuples of elements whose first element is from Sx and the second from > > > > S2 . . . , the nth from Sn. The $k are, in general, different, but in > > > > many cases the experiment could be formed from independent trials of > > > > the same experiment.- Hide quoted text - > > > > > - Show quoted text - > > > > Thank you for the excellent post and link. My only comment at the > > > moment is that the cartesian product is a nontrivial operator. It is > > > somewhat different than what we are talking about. Definately > > > nontrivial. > > > The concept of "triviality" depends on the context. If you are not > > familiar with the concept of n-tuples and Cartesian coordinates - then > > I suppose the Cartesian Product is "nontrivial". > > > However, the cure for this is not to re-invent the wheel but instead > > to study the basic foundations of mathematics. I mean, Descartes is > > hardly a newcomer. > > >http://en.wikipedia.org/wiki/Descartes > > > René Descartes ((31 March 1596 11 February 1650), also known as > > Renatus Cartesius (Latinized form),[2] was a French philosopher, > > mathematician, physicist, and writer. The invention of Cartesian > > coordinates in the 17th century by René Descartes revolutionized > > mathematics by providing the first systematic link between Euclidean > > geometry and algebra. > > > I saw you use the term "random variable" correctly. Thus, you do have > > familiarity with the terminology of probability theory. So, look up > > the way in probability theory, the Cartesian Product captures > > independent trials. It's really not rocket science. > > > > Consider 3 trials of a random variable > > > { X | H, T } > > > and a single trial of the random variable > > > { X | HHH, HHT, HTT, THH, THT, TTT, HTH, TTH } . > > > > The difference between the two things above is strictly trivial. And > > > so too - here's another example of something trivial > > > > 5 = 5 + 0 = 5 = 0 + 0 = 5 + 0 + 0 + 0 + 0 etc > > > > The only difference in any of the above terms is trivial. > > > > While there is a trivial element in set theory (the empty set), the > > > cartesian product is a nontrivial operation. > > > Well, if you restrict yourself only to trivial constructs and empty > > sets, your results will remain trivial and empty. Garbage in - garbage > > out. > > > > My claim is that conservation can be regarded as an operator, a > > > trivial operator. It is a process of composing things where the result > > > is "no change", unlike addition or multiplication (disregarding the > > > behaviour of identity elements).- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text - > > I'll take another look at dependence & independence and tell you what > I think. > > But my comment that cartesian product is nontrivial, > Look, if you are unfamiliar with the concept of 'independence' then clearly you have never studied probability at all, because this is by far the most important concept in probability, especially introductory. Probability is one of the oldest branches of mathematics, going back to people like Pascal. Why do you want to talk about probability without reading at least an introduction to it - is beyond me. Why do you literally want to re- invent the wheel? > > basically what > that means is that the only way it can act trivially is if it operates > over trivial elements. > Yes, everything about triviality is trivial. But why do you want to devote your life to triviality? > > Just like addition or multiplication or > anything else in math, the operators that we use are always > nontrivial. Currently I do not think that mathematics recognizes a > trivial operator. > We do. But we find it too trivial. Few people are interested in trivialities, except maybe modern philosophers and politicians. > > I think it's a shame, because it makes sense for many reasons, > including some subtle philosophical considerations regarding > conservation. > In any case, probability theory is non-trivial, and if you study it, you will exclude it from the list of trivial subjects. |