SVM - why does scaling the parameters w and b result in nothing meaningful?
The functional margin tells us how confident the SVM is in it's classification, for big values it's better than for small ones. Now the question that bugs me is as to why we can't make w
and b
arbitrarily big? My book suggests that the reason for that is that in the end the SVM might classify the training set well, but it cannot generalize, therefore we have to regularize the expression, dividing it by the norm of w
. That explanation I did not quite understand, that is I get that:
x + 1 and 2x + 2 are the same vector, only scaled differently, but how is that significant here? One thing that comes to mind is that we should have the same w
and b
for the same vector, since it doesn't make sense for them to be different values and maybe in the test data set we might have the same vector that is scaled differently, but that should not hold meaning.
But how and why will it generalize badly?
machine-learning svm data-science
add a comment |
The functional margin tells us how confident the SVM is in it's classification, for big values it's better than for small ones. Now the question that bugs me is as to why we can't make w
and b
arbitrarily big? My book suggests that the reason for that is that in the end the SVM might classify the training set well, but it cannot generalize, therefore we have to regularize the expression, dividing it by the norm of w
. That explanation I did not quite understand, that is I get that:
x + 1 and 2x + 2 are the same vector, only scaled differently, but how is that significant here? One thing that comes to mind is that we should have the same w
and b
for the same vector, since it doesn't make sense for them to be different values and maybe in the test data set we might have the same vector that is scaled differently, but that should not hold meaning.
But how and why will it generalize badly?
machine-learning svm data-science
add a comment |
The functional margin tells us how confident the SVM is in it's classification, for big values it's better than for small ones. Now the question that bugs me is as to why we can't make w
and b
arbitrarily big? My book suggests that the reason for that is that in the end the SVM might classify the training set well, but it cannot generalize, therefore we have to regularize the expression, dividing it by the norm of w
. That explanation I did not quite understand, that is I get that:
x + 1 and 2x + 2 are the same vector, only scaled differently, but how is that significant here? One thing that comes to mind is that we should have the same w
and b
for the same vector, since it doesn't make sense for them to be different values and maybe in the test data set we might have the same vector that is scaled differently, but that should not hold meaning.
But how and why will it generalize badly?
machine-learning svm data-science
The functional margin tells us how confident the SVM is in it's classification, for big values it's better than for small ones. Now the question that bugs me is as to why we can't make w
and b
arbitrarily big? My book suggests that the reason for that is that in the end the SVM might classify the training set well, but it cannot generalize, therefore we have to regularize the expression, dividing it by the norm of w
. That explanation I did not quite understand, that is I get that:
x + 1 and 2x + 2 are the same vector, only scaled differently, but how is that significant here? One thing that comes to mind is that we should have the same w
and b
for the same vector, since it doesn't make sense for them to be different values and maybe in the test data set we might have the same vector that is scaled differently, but that should not hold meaning.
But how and why will it generalize badly?
machine-learning svm data-science
machine-learning svm data-science
asked Nov 12 at 7:45
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