The obvious way to pursue an analogy is that masses are probabilities, but that seems to not work—the distances have no meaning, and there’s no advantage over adding and subtracting probabilities of disjoint events.
So what if we make the distances probabilities. Now we can have P(A) be some distance AA’, and P(A|B) be AB… No, no good… or if we make the probability the position of a point on the middle, is there anything interesting we can do with the weights of P(A) and P(B) to make P(AB)… So if you fix one endpoint to have mass 1, and the position to be P(A), the value of the other point will be the odds ratio P(A)/P(not-A). But there’s no convenient way to multiply values of points...
Not sure there’s any way to use all the parts of this buffalo, sorry.
It’s just that… I mean, I know the next bit is the opposite of rigorous and all that, but since my job is basically to teach kids how to play with shiny toys, I should at least see if they are child-friendly, right?:) so I am asking you as someone who probably has played more.
Suppose we have several—say, 9 - ‘tests’ we’ve run daily for two months, covering the turn of spring into summer, and the outcomes of all of them should reasonably change as the seasons progress, but not quite as straightforwardly as the development of foliage, fruit, etc.* Like, test A is ‘how much acetone will evaporate from a 40 ml bottle in an open shady space between 12.00 and 12.50?’, test B - ‘how long will it take for a soaking wet handkerchief to dry to constant weight if hung at 12.15?’, …, test I - ‘what percentage of shepherd’s purse plants in this patch is blooming?’ All the tests are but weakly connected to each other, but it is possible, generally speaking, to assign some measure of ‘distance’ - for example, A and B are clearly closer to each other than each of them is to I (although it seems that 5 or 6 tests are more realistic to juggle). Then, we draw a nonagon (or 5-, or 6-...) Such that the distances between the vertices equal the ‘distances between tests’.
The masses of the vertices are the probabilities that the given outcomes support the hypothesis that the day is Day X. (The question is, of course, ‘what day is it?‘) We are told that X is either 9, or 24, or 35. Now we can find the respective centers of mass for each combination of probabilities for all three hypotheses. Then we obtain a ‘cloud’ of centers of mass, each of which is ‘more or less in favor’ of one hypothesis (except, perhaps, one where the probabilities are equal). If we divide the cloud into three layers (one for each H), and have the masses of the points be the p(H), we can find the center of masses for each day. When we plot the ‘final day points’, we can now see how close they are to which test. Which means...well, in our example, nothing...but if it were something grown-up, perhaps that there are different views on the question depending on what experiments a researcher trusts best, even if they all share the same data from the same set of experiments...
Anyway, sorry for so many words.
*I am going to have the kids track the vegetation development before the ‘more random’ stuff. Generally, the question will remain ‘what day is it?’, but the answer will be expressed as an interval, without probabilities assigned.
Huh, MPG seems like an interesting trick.
The obvious way to pursue an analogy is that masses are probabilities, but that seems to not work—the distances have no meaning, and there’s no advantage over adding and subtracting probabilities of disjoint events.
So what if we make the distances probabilities. Now we can have P(A) be some distance AA’, and P(A|B) be AB… No, no good… or if we make the probability the position of a point on the middle, is there anything interesting we can do with the weights of P(A) and P(B) to make P(AB)… So if you fix one endpoint to have mass 1, and the position to be P(A), the value of the other point will be the odds ratio P(A)/P(not-A). But there’s no convenient way to multiply values of points...
Not sure there’s any way to use all the parts of this buffalo, sorry.
Well that is rather what I think, too.
It’s just that… I mean, I know the next bit is the opposite of rigorous and all that, but since my job is basically to teach kids how to play with shiny toys, I should at least see if they are child-friendly, right?:) so I am asking you as someone who probably has played more.
Suppose we have several—say, 9 - ‘tests’ we’ve run daily for two months, covering the turn of spring into summer, and the outcomes of all of them should reasonably change as the seasons progress, but not quite as straightforwardly as the development of foliage, fruit, etc.* Like, test A is ‘how much acetone will evaporate from a 40 ml bottle in an open shady space between 12.00 and 12.50?’, test B - ‘how long will it take for a soaking wet handkerchief to dry to constant weight if hung at 12.15?’, …, test I - ‘what percentage of shepherd’s purse plants in this patch is blooming?’ All the tests are but weakly connected to each other, but it is possible, generally speaking, to assign some measure of ‘distance’ - for example, A and B are clearly closer to each other than each of them is to I (although it seems that 5 or 6 tests are more realistic to juggle). Then, we draw a nonagon (or 5-, or 6-...) Such that the distances between the vertices equal the ‘distances between tests’.
The masses of the vertices are the probabilities that the given outcomes support the hypothesis that the day is Day X. (The question is, of course, ‘what day is it?‘) We are told that X is either 9, or 24, or 35. Now we can find the respective centers of mass for each combination of probabilities for all three hypotheses. Then we obtain a ‘cloud’ of centers of mass, each of which is ‘more or less in favor’ of one hypothesis (except, perhaps, one where the probabilities are equal). If we divide the cloud into three layers (one for each H), and have the masses of the points be the p(H), we can find the center of masses for each day. When we plot the ‘final day points’, we can now see how close they are to which test. Which means...well, in our example, nothing...but if it were something grown-up, perhaps that there are different views on the question depending on what experiments a researcher trusts best, even if they all share the same data from the same set of experiments...
Anyway, sorry for so many words.
*I am going to have the kids track the vegetation development before the ‘more random’ stuff. Generally, the question will remain ‘what day is it?’, but the answer will be expressed as an interval, without probabilities assigned.