Timeline for How To Determine The Number Of Dimensions To A Machine Learning Problem
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 12, 2015 at 0:33 | vote | accept | user58446 | ||
| Nov 26, 2015 at 1:39 | |||||
| Apr 10, 2015 at 23:55 | comment | added | Aleksandr Blekh | @user58446: Perhaps, my wording was not perfect for that aspect. What I meant is that dimensionality, introduced by dependent variables should also be included into a model's total dimensionality, along with dimensionality, introduced by independent variables. | |
| Apr 10, 2015 at 23:25 | comment | added | user58446 | One last thing before the green check: What did you mean by counting the dependent variables seperately in your opinion. Are you saying that the dimensions contribruted by the input are completely seperate to those contributed by the output? Do they not sum to give the total dimensions for the entire problem? | |
| Apr 10, 2015 at 23:21 | vote | accept | user58446 | ||
| Apr 10, 2015 at 23:21 | |||||
| Apr 10, 2015 at 23:16 | comment | added | Aleksandr Blekh | @user58446: As far as I understand, the fact that stochastic model is based on probabilities, has nothing to do with its dimensionality. I never said that it makes sense to combine both aspects, as dimensionality and probability are two different concepts. The same applies to a range of values, which, in my opinion, has nothing to do with dimensionality of a model as well. | |
| Apr 10, 2015 at 23:09 | comment | added | user58446 | ...where did I go wrong? | |
| Apr 10, 2015 at 23:08 | comment | added | user58446 | You misunderstand: I was taking a guess at what you meant by your explanation. You mentioned that "dimensionality of a machine learning problem is equivalent to dimensionality of a statistical model", or schotastic model, which involves probability unlike a deterministic system. You also said that the dimensionality of the problem "is a sum of individual dimensions across all variables". So, I comined the range of the values (from second comment) and combined it with the stochastic model which involves probabilities for different values within the aforementioned range of possible values... | |
| Apr 10, 2015 at 23:00 | comment | added | Aleksandr Blekh | @user58446: I still don't see how those probabilities affect the dimensionality of the problem. If you think otherwise, please refer to definition of dimensionality, which includes the probability aspect. | |
| Apr 10, 2015 at 22:47 | comment | added | user58446 | The 'whole probability factor' refers to the schotastic, as opposed to deterministic, nature of the problem. However, my example in the comment may be further altered by suggesting that the range for x may be between 5 and 8, but the probability for x having specific individual values is as follows: 5 = 20%, 6 = 40%, 7 = 10%, 8 = 30%. Likewise, for y: 15 = 5%, 16 = 80%, 17 = 15%. How do these probabilities affect the dimensionality of the problem, if at all? | |
| Apr 10, 2015 at 22:36 | comment | added | Aleksandr Blekh | @user58446: I am not sure what do you mean by "whole probability factor". But, yes, I think that the total dimensionality of a problem/model/system is a sum of individual dimensions across all variables. | |
| Apr 10, 2015 at 19:29 | comment | added | user58446 | So you are saying the nuimber of dimensions is increased from the standard independent and dependent variables due to the whole 'probability factor'? Is there a new dimension for every possible value of the variable (or even worse, for every combination of all the variables) within the defined possible range of probable values? For example, if there is a probability that a variable x will be between 5 and 8, and variable y will be between 15 and 17... then we have 3 dimensions for the range of x, and 2 dimensions for the range of y, so the problem will have a total of 5 dimensions? | |
| Apr 4, 2015 at 7:44 | history | answered | Aleksandr Blekh | CC BY-SA 3.0 |