3 ms·

Genuine question, i'm no math expert: if you're using the value by multiplying it with some other value, does that not imply it qualifies as normalised? i.e. s

by tcldr 6y ago

Genuine question, i'm no math expert: if you're using the value by multiplying it with some other value, does that not imply it qualifies as normalised? i.e. say you have a width of 500 and you want to move half way across so you have this value of 0.5 to get 250. By dividing 250 by 500, aren't we in fact normalising it?

chub500 6y ago
I think generally normalized quantities maintain their original units whereas in your ratio case the unit is dropped. Units in the abstract sense I guess, such as this is a measurement along this vector or this is a ratio of any vector.
- quietbritishjim 6y ago
  I think it's the opposite way round: normalisation results in units being dropped. The typical, most common, definition of normalisation is value ÷ max possible value, giving a result in [0,1]. (More general definitions of normalisation exist e.g. if you rescale so the standard deviation is a fixed value, or even use non linear rescaling, that could count, but never mind all that.) The parent comment's example of "position along width ÷ total width" certainly fits that bill. Whenever you divide something by the max of that something, the max is going to have the same units as the original value and you're bound to end up cancelling them. Or put another way, if you rescale 10cm into 0.5, it's certainly not 0.5cm so the units are either dropped or, at least, changed e.g. you could argue you've got 0.5x where x is the unit equal to 20cm.
  - f154hfds 6y ago
    Hmm when I think of 'normalizing' I don't think of dividing by a max at all - in my experience it is more taking a quantity (perhaps in English measurement) and transferring to a more 'standard' unit (say metric). In general I don't think normalization always includes a sense of being in a bounded interval. From a mathematical perspective you could perhaps say normalization is achieved by multiplying your quantity by a 1D operator. You can't change the dimensionality this way, but are certainly changing 'units' a la mm in the x direction -> m in the x direction for example. I guess what I'm saying is that 'normalized' and the like are not the best fit for the SO question. If I could take my own shot at the SO challenge from a mathematical perspective it would perhaps be sigmoid. Where the result of a normalization function takes a 1D value and maps it to a similar 1D value, the sigmoid takes a 1D value and maps it to a similar 1D value between (0, 1). So if I want to drop the previous information and only keep the resulting map, I can say 'this is my sigmoided value' - IE it is impossible for it to be outside of that range. Unfortunately sigmoid also connotates a differentiable curve which is extraneous information... https://en.wikipedia.org/wiki/Normalization_(statistics) https://en.wikipedia.org/wiki/Normalization_(statistics) https://en.wikipedia.org/wiki/Sigmoid_function https://en.wikipedia.org/wiki/Sigmoid_function
    - quietbritishjim 6y ago
      Of course the literal meaning of "normalisation" is to make more "normal", and that can mean almost anything at all. Even the Wikipedia article you linked to starts with "normalization can have a range of meanings". If you have heard that word most used with one meaning and I have heard it most used with another meaning then that doesn't invalidate either of those definitions. The Wikipedia article you linked to gives two very broad definitions in the lead. The definition covered in the first paragraph is "adjusting values measured on different scales to a notionally common scale", which seems to be what you're talking about. The definition covered in the second paragraph is "the creation of shifted and scaled versions of statistics", in particular "some types of normalization involve only a rescaling, to arrive at values relative to some size variable". I'm not comfortable with the use of "of statistics" in that second definition: the very first example is in the article standard score [1], which is about a rescaled element of the population, not a rescaled statistic. In any case, outside of statistics, a rescaling is a common meaning for this word, and a rescaled statistic is clearly just a special case of this. I think it was clear from the context that we were talking about this meaning originally. By far the most common case of this is a linear rescaling (including translation) to [0,1] but I was already up front that this is just a special case. As for your sigmoid comment, I may have misunderstood but it sounds like you're saying that if you have a variable in the range in [0,1] then it can be described as the result of a sigmoid function. My objection to this is the same as my original objection to calling such as variable "normalised": it is a confusing variable name to use unless you actually did get it by applying a sigmoid function to something, not just because it holds a value that could hypothetically be obtained from a sigmoid function (but you didn't). [1] https://en.wikipedia.org/wiki/Standard_score https://en.wikipedia.org/wiki/Standard_score
deleted 6y ago
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quietbritishjim 6y ago
> if you're using the value by multiplying it with some other value, does that not imply it qualifies as normalised Certainly, "normalisation" can mean something more general than "rescale to the interval [0,1]". For example, "rescaling to the interval [0, 20]" might make sense in some contexts and would usually count as a type of normalisation, and would still involve muliplying by a number. But that would be multiplying by (20 / max possible value) rather than the typical case of multiplying by (1 / max possible value). So normalisation can mean something more general than the most usual common case, but it's not just any old "multiplying it with some other value". It has to specifically for the purpose of rescaling to some fixed, more useful/sensible range. > say you have a width of 500 and you want to move half way across so you have this value of 0.5 to get 250 This is a great example of multiplication that isn't normalisation! You've multiplied by 0.5 to get the midpoint, but that process isn't normalisation because you've not ended up with some more sensible range. You started with [0,infinity) (the set of all possible widths) and ended up with that same infinite range. > By dividing 250 by 500 Hang on, I'm confused about your example: are you asking about 500*0.5 (=250) or 250/500 (=0.5)? If it's the latter than that's normalisation, and not even something fancy or general but classic linear rescaling to the interval [0,1]. Yes, that's certainly normalisation. ---- Going back to my original comment: I was talking about values that were in the range 0 to 1 that hadn't reached that range by being multiplied by anything at all; they were just naturally in that range to begin with. For example, a probability would fit this bill.
- tcldr 6y ago
  > If it's the latter than that's normalisation. I'm very clumsily trying to say that if: width / maxWidth = normalWidth then: normalWidth * maxWidth = width. And so, even though we aren't necessary calculating the normal in the second equation, we can still name our variable a 'normal' as rearranging proves that is indeed what the value is. Edit: formatting

by tcldr 6y ago

chub500 6y ago

I think generally normalized quantities maintain their original units whereas in your ratio case the unit is dropped. Units in the abstract sense I guess, such as this is a measurement along this vector or this is a ratio of any vector.

quietbritishjim 6y ago
I think it's the opposite way round: normalisation results in units being dropped. The typical, most common, definition of normalisation is value ÷ max possible value, giving a result in [0,1]. (More general definitions of normalisation exist e.g. if you rescale so the standard deviation is a fixed value, or even use non linear rescaling, that could count, but never mind all that.) The parent comment's example of "position along width ÷ total width" certainly fits that bill. Whenever you divide something by the max of that something, the max is going to have the same units as the original value and you're bound to end up cancelling them. Or put another way, if you rescale 10cm into 0.5, it's certainly not 0.5cm so the units are either dropped or, at least, changed e.g. you could argue you've got 0.5x where x is the unit equal to 20cm.
- f154hfds 6y ago
  Hmm when I think of 'normalizing' I don't think of dividing by a max at all - in my experience it is more taking a quantity (perhaps in English measurement) and transferring to a more 'standard' unit (say metric). In general I don't think normalization always includes a sense of being in a bounded interval. From a mathematical perspective you could perhaps say normalization is achieved by multiplying your quantity by a 1D operator. You can't change the dimensionality this way, but are certainly changing 'units' a la mm in the x direction -> m in the x direction for example. I guess what I'm saying is that 'normalized' and the like are not the best fit for the SO question. If I could take my own shot at the SO challenge from a mathematical perspective it would perhaps be sigmoid. Where the result of a normalization function takes a 1D value and maps it to a similar 1D value, the sigmoid takes a 1D value and maps it to a similar 1D value between (0, 1). So if I want to drop the previous information and only keep the resulting map, I can say 'this is my sigmoided value' - IE it is impossible for it to be outside of that range. Unfortunately sigmoid also connotates a differentiable curve which is extraneous information... https://en.wikipedia.org/wiki/Normalization_(statistics) https://en.wikipedia.org/wiki/Normalization_(statistics) https://en.wikipedia.org/wiki/Sigmoid_function https://en.wikipedia.org/wiki/Sigmoid_function
  - quietbritishjim 6y ago
    Of course the literal meaning of "normalisation" is to make more "normal", and that can mean almost anything at all. Even the Wikipedia article you linked to starts with "normalization can have a range of meanings". If you have heard that word most used with one meaning and I have heard it most used with another meaning then that doesn't invalidate either of those definitions. The Wikipedia article you linked to gives two very broad definitions in the lead. The definition covered in the first paragraph is "adjusting values measured on different scales to a notionally common scale", which seems to be what you're talking about. The definition covered in the second paragraph is "the creation of shifted and scaled versions of statistics", in particular "some types of normalization involve only a rescaling, to arrive at values relative to some size variable". I'm not comfortable with the use of "of statistics" in that second definition: the very first example is in the article standard score [1], which is about a rescaled element of the population, not a rescaled statistic. In any case, outside of statistics, a rescaling is a common meaning for this word, and a rescaled statistic is clearly just a special case of this. I think it was clear from the context that we were talking about this meaning originally. By far the most common case of this is a linear rescaling (including translation) to [0,1] but I was already up front that this is just a special case. As for your sigmoid comment, I may have misunderstood but it sounds like you're saying that if you have a variable in the range in [0,1] then it can be described as the result of a sigmoid function. My objection to this is the same as my original objection to calling such as variable "normalised": it is a confusing variable name to use unless you actually did get it by applying a sigmoid function to something, not just because it holds a value that could hypothetically be obtained from a sigmoid function (but you didn't). [1] https://en.wikipedia.org/wiki/Standard_score https://en.wikipedia.org/wiki/Standard_score

deleted 6y ago

[deleted]

quietbritishjim 6y ago

> if you're using the value by multiplying it with some other value, does that not imply it qualifies as normalised Certainly, "normalisation" can mean something more general than "rescale to the interval [0,1]". For example, "rescaling to the interval [0, 20]" might make sense in some contexts and would usually count as a type of normalisation, and would still involve muliplying by a number. But that would be multiplying by (20 / max possible value) rather than the typical case of multiplying by (1 / max possible value). So normalisation can mean something more general than the most usual common case, but it's not just any old "multiplying it with some other value". It has to specifically for the purpose of rescaling to some fixed, more useful/sensible range. > say you have a width of 500 and you want to move half way across so you have this value of 0.5 to get 250 This is a great example of multiplication that isn't normalisation! You've multiplied by 0.5 to get the midpoint, but that process isn't normalisation because you've not ended up with some more sensible range. You started with [0,infinity) (the set of all possible widths) and ended up with that same infinite range. > By dividing 250 by 500 Hang on, I'm confused about your example: are you asking about 500*0.5 (=250) or 250/500 (=0.5)? If it's the latter than that's normalisation, and not even something fancy or general but classic linear rescaling to the interval [0,1]. Yes, that's certainly normalisation. ---- Going back to my original comment: I was talking about values that were in the range 0 to 1 that hadn't reached that range by being multiplied by anything at all; they were just naturally in that range to begin with. For example, a probability would fit this bill.

tcldr 6y ago
> If it's the latter than that's normalisation. I'm very clumsily trying to say that if: width / maxWidth = normalWidth then: normalWidth * maxWidth = width. And so, even though we aren't necessary calculating the normal in the second equation, we can still name our variable a 'normal' as rearranging proves that is indeed what the value is. Edit: formatting