What advantage does Guassian kernel have than any other kernels, such as linear kernel, polynomial kernel and...
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Guassian kernel is so important in SVM as we know. The parameter gamma is designed for this kind of kernel. My question is what makes Guassian kernel so unique? What advantage does it have over other kernels?
machine-learning svm kernel
asked Oct 31 '18 at 14:37
Jason ShuJason Shu
1
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This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
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$begingroup$
Guassian kernel is so important in SVM as we know. The parameter gamma is designed for this kind of kernel. My question is what makes Guassian kernel so unique? What advantage does it have over other kernels?
machine-learning svm kernel
asked Oct 31 '18 at 14:37
Jason ShuJason Shu
1
$endgroup$
bumped to the homepage by Community♦ 4 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
$begingroup$
Guassian kernel is so important in SVM as we know. The parameter gamma is designed for this kind of kernel. My question is what makes Guassian kernel so unique? What advantage does it have over other kernels?
machine-learning svm kernel
asked Oct 31 '18 at 14:37
Jason ShuJason Shu
1
$endgroup$
Guassian kernel is so important in SVM as we know. The parameter gamma is designed for this kind of kernel. My question is what makes Guassian kernel so unique? What advantage does it have over other kernels?
machine-learning svm kernel
machine-learning svm kernel
asked Oct 31 '18 at 14:37
Jason ShuJason Shu
1
asked Oct 31 '18 at 14:37
Jason ShuJason Shu
1
asked Oct 31 '18 at 14:37
Jason ShuJason Shu
1
asked Oct 31 '18 at 14:37
Jason ShuJason Shu
1
asked Oct 31 '18 at 14:37
Jason ShuJason Shu
1
1
bumped to the homepage by Community♦ 4 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 4 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
add a comment |
2 Answers
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The advantange is that it is related to euclidean distance, and if your problem benefits from this type of feature, e.g. you have a highly non-linear problem where values nearby some vector point tend to be of a different class, it will be able to find relationships that other kernels would not, and create highly complex decision boundaries.
answered Oct 31 '18 at 17:25
anymous.askeranymous.asker
55118
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add a comment |
$begingroup$
Mathematically speaking, one can show that an RBF is simply a low-band pass filter. To put it simply, RBF induces a smooth solution. Sometimes that is an important characteristic that matters. However, it isn't always important or useful. For example, it's somewhat well known that a smoothing prior does not outperform a linear/polynomial kernel for problems like Text Classification.
There's also other reasons though why people tend to stick with an RBF kernel instead of trying to design their own or use a polynomial. One main one is that an RBF is non-parametric -- meaning that it's complexity grows with data. While if one uses a polynomial kernel, then you have a parametric model which has a finite and fix size. There essentially comes a point when your polynomial kernel becomes saturated no matter how much data you throw at it. Hence, it's generally seen as a good benchmark to start with an RBF unless you have reason to believe otherwise.
There's other properties that make it mathematically useful, like it's invariant, stationary contain less parameters than polynomial kernels.
TL;DR generally works well.
answered Jan 1 at 2:50
TophatTophat
1,332212
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2 Answers
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2 Answers
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$begingroup$
The advantange is that it is related to euclidean distance, and if your problem benefits from this type of feature, e.g. you have a highly non-linear problem where values nearby some vector point tend to be of a different class, it will be able to find relationships that other kernels would not, and create highly complex decision boundaries.
answered Oct 31 '18 at 17:25
anymous.askeranymous.asker
55118
$endgroup$
add a comment |
$begingroup$
The advantange is that it is related to euclidean distance, and if your problem benefits from this type of feature, e.g. you have a highly non-linear problem where values nearby some vector point tend to be of a different class, it will be able to find relationships that other kernels would not, and create highly complex decision boundaries.
answered Oct 31 '18 at 17:25
anymous.askeranymous.asker
55118
$endgroup$
add a comment |
$begingroup$
The advantange is that it is related to euclidean distance, and if your problem benefits from this type of feature, e.g. you have a highly non-linear problem where values nearby some vector point tend to be of a different class, it will be able to find relationships that other kernels would not, and create highly complex decision boundaries.
answered Oct 31 '18 at 17:25
anymous.askeranymous.asker
55118
$endgroup$
The advantange is that it is related to euclidean distance, and if your problem benefits from this type of feature, e.g. you have a highly non-linear problem where values nearby some vector point tend to be of a different class, it will be able to find relationships that other kernels would not, and create highly complex decision boundaries.
answered Oct 31 '18 at 17:25
anymous.askeranymous.asker
55118
answered Oct 31 '18 at 17:25
anymous.askeranymous.asker
55118
answered Oct 31 '18 at 17:25
anymous.askeranymous.asker
55118
answered Oct 31 '18 at 17:25
anymous.askeranymous.asker
55118
55118
add a comment |
add a comment |
$begingroup$
Mathematically speaking, one can show that an RBF is simply a low-band pass filter. To put it simply, RBF induces a smooth solution. Sometimes that is an important characteristic that matters. However, it isn't always important or useful. For example, it's somewhat well known that a smoothing prior does not outperform a linear/polynomial kernel for problems like Text Classification.
There's also other reasons though why people tend to stick with an RBF kernel instead of trying to design their own or use a polynomial. One main one is that an RBF is non-parametric -- meaning that it's complexity grows with data. While if one uses a polynomial kernel, then you have a parametric model which has a finite and fix size. There essentially comes a point when your polynomial kernel becomes saturated no matter how much data you throw at it. Hence, it's generally seen as a good benchmark to start with an RBF unless you have reason to believe otherwise.
There's other properties that make it mathematically useful, like it's invariant, stationary contain less parameters than polynomial kernels.
TL;DR generally works well.
answered Jan 1 at 2:50
TophatTophat
1,332212
$endgroup$
add a comment |
$begingroup$
Mathematically speaking, one can show that an RBF is simply a low-band pass filter. To put it simply, RBF induces a smooth solution. Sometimes that is an important characteristic that matters. However, it isn't always important or useful. For example, it's somewhat well known that a smoothing prior does not outperform a linear/polynomial kernel for problems like Text Classification.
There's also other reasons though why people tend to stick with an RBF kernel instead of trying to design their own or use a polynomial. One main one is that an RBF is non-parametric -- meaning that it's complexity grows with data. While if one uses a polynomial kernel, then you have a parametric model which has a finite and fix size. There essentially comes a point when your polynomial kernel becomes saturated no matter how much data you throw at it. Hence, it's generally seen as a good benchmark to start with an RBF unless you have reason to believe otherwise.
There's other properties that make it mathematically useful, like it's invariant, stationary contain less parameters than polynomial kernels.
TL;DR generally works well.
answered Jan 1 at 2:50
TophatTophat
1,332212
$endgroup$
add a comment |
$begingroup$
Mathematically speaking, one can show that an RBF is simply a low-band pass filter. To put it simply, RBF induces a smooth solution. Sometimes that is an important characteristic that matters. However, it isn't always important or useful. For example, it's somewhat well known that a smoothing prior does not outperform a linear/polynomial kernel for problems like Text Classification.
There's also other reasons though why people tend to stick with an RBF kernel instead of trying to design their own or use a polynomial. One main one is that an RBF is non-parametric -- meaning that it's complexity grows with data. While if one uses a polynomial kernel, then you have a parametric model which has a finite and fix size. There essentially comes a point when your polynomial kernel becomes saturated no matter how much data you throw at it. Hence, it's generally seen as a good benchmark to start with an RBF unless you have reason to believe otherwise.
There's other properties that make it mathematically useful, like it's invariant, stationary contain less parameters than polynomial kernels.
TL;DR generally works well.
answered Jan 1 at 2:50
TophatTophat
1,332212
$endgroup$
Mathematically speaking, one can show that an RBF is simply a low-band pass filter. To put it simply, RBF induces a smooth solution. Sometimes that is an important characteristic that matters. However, it isn't always important or useful. For example, it's somewhat well known that a smoothing prior does not outperform a linear/polynomial kernel for problems like Text Classification.
There's also other reasons though why people tend to stick with an RBF kernel instead of trying to design their own or use a polynomial. One main one is that an RBF is non-parametric -- meaning that it's complexity grows with data. While if one uses a polynomial kernel, then you have a parametric model which has a finite and fix size. There essentially comes a point when your polynomial kernel becomes saturated no matter how much data you throw at it. Hence, it's generally seen as a good benchmark to start with an RBF unless you have reason to believe otherwise.
There's other properties that make it mathematically useful, like it's invariant, stationary contain less parameters than polynomial kernels.
TL;DR generally works well.
answered Jan 1 at 2:50
TophatTophat
1,332212
answered Jan 1 at 2:50
TophatTophat
1,332212
answered Jan 1 at 2:50
TophatTophat
1,332212
answered Jan 1 at 2:50
TophatTophat
1,332212
1,332212
add a comment |
add a comment |
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