Proof of the hinge loss being 1-Lipschitz
$begingroup$
Page 7 of https://web.stanford.edu/class/cs229t/scribe_notes/10_10_final.pdf
I've tried finding a proof online, but haven't been able to find it. In the notes above which are provided as part of Stanford's Statistical Learning Theory, the hinge loss is defined as:
$$
l(z,h) = max(0,1-y_ih(x_i) )
$$
where $z = (x,y)$, and $h$ is some hypothesis.
Is it possible to provide a proof that this is $1$-Lipschitz? Furthermore, in general, is there a particular method to show this for functions that may not have a gradient defined at every point?
I hope someone can correct my (wrong) intuition: If a function $f(x)$ is $K$-Lipschitz, then the magnitude of its gradient is bounded by $K$. But in this case, why is the gradient of the hinge loss necessarily bounded? Can't we choose two points $x_1$ and $x_2$ which are arbitrarily close, but correspond to different labels such that $l(z_1,h) = 0$ but $l(z_2,h) = 1$?
real-analysis machine-learning
asked Dec 3 '18 at 17:41
Sean LeeSean Lee
1749
$endgroup$
add a comment |
$begingroup$
Page 7 of https://web.stanford.edu/class/cs229t/scribe_notes/10_10_final.pdf
I've tried finding a proof online, but haven't been able to find it. In the notes above which are provided as part of Stanford's Statistical Learning Theory, the hinge loss is defined as:
$$
l(z,h) = max(0,1-y_ih(x_i) )
$$
where $z = (x,y)$, and $h$ is some hypothesis.
Is it possible to provide a proof that this is $1$-Lipschitz? Furthermore, in general, is there a particular method to show this for functions that may not have a gradient defined at every point?
I hope someone can correct my (wrong) intuition: If a function $f(x)$ is $K$-Lipschitz, then the magnitude of its gradient is bounded by $K$. But in this case, why is the gradient of the hinge loss necessarily bounded? Can't we choose two points $x_1$ and $x_2$ which are arbitrarily close, but correspond to different labels such that $l(z_1,h) = 0$ but $l(z_2,h) = 1$?
real-analysis machine-learning
asked Dec 3 '18 at 17:41
Sean LeeSean Lee
1749
$endgroup$
$begingroup$
What metric are you using on the space containing the $(z,h)$?
$endgroup$
– copper.hat
Dec 3 '18 at 17:46
$begingroup$
I'm sorry, but no metric was specified; should I just assume the Euclidean metric in this case?
$endgroup$
– Sean Lee
Dec 3 '18 at 17:50
1
$begingroup$
I don't know what you mean by Euclidean metric for the function $h$. I don't have time to read through the pdf and figure it out. By Lipschitz you need to show that $|l(x_1,h_1) -l(z_2,h_2)| le L d((z_1,h_1),(z_2,h_2))$, where $d$ is some metric.
$endgroup$
– copper.hat
Dec 3 '18 at 17:58
$begingroup$
Ah okay sure thank you I'll try to work from that angle :)
$endgroup$
– Sean Lee
Dec 3 '18 at 17:59
add a comment |
$begingroup$
Page 7 of https://web.stanford.edu/class/cs229t/scribe_notes/10_10_final.pdf
I've tried finding a proof online, but haven't been able to find it. In the notes above which are provided as part of Stanford's Statistical Learning Theory, the hinge loss is defined as:
$$
l(z,h) = max(0,1-y_ih(x_i) )
$$
where $z = (x,y)$, and $h$ is some hypothesis.
Is it possible to provide a proof that this is $1$-Lipschitz? Furthermore, in general, is there a particular method to show this for functions that may not have a gradient defined at every point?
I hope someone can correct my (wrong) intuition: If a function $f(x)$ is $K$-Lipschitz, then the magnitude of its gradient is bounded by $K$. But in this case, why is the gradient of the hinge loss necessarily bounded? Can't we choose two points $x_1$ and $x_2$ which are arbitrarily close, but correspond to different labels such that $l(z_1,h) = 0$ but $l(z_2,h) = 1$?
real-analysis machine-learning
asked Dec 3 '18 at 17:41
Sean LeeSean Lee
1749
$endgroup$
Page 7 of https://web.stanford.edu/class/cs229t/scribe_notes/10_10_final.pdf
I've tried finding a proof online, but haven't been able to find it. In the notes above which are provided as part of Stanford's Statistical Learning Theory, the hinge loss is defined as:
$$
l(z,h) = max(0,1-y_ih(x_i) )
$$
where $z = (x,y)$, and $h$ is some hypothesis.
Is it possible to provide a proof that this is $1$-Lipschitz? Furthermore, in general, is there a particular method to show this for functions that may not have a gradient defined at every point?
I hope someone can correct my (wrong) intuition: If a function $f(x)$ is $K$-Lipschitz, then the magnitude of its gradient is bounded by $K$. But in this case, why is the gradient of the hinge loss necessarily bounded? Can't we choose two points $x_1$ and $x_2$ which are arbitrarily close, but correspond to different labels such that $l(z_1,h) = 0$ but $l(z_2,h) = 1$?
real-analysis machine-learning
real-analysis machine-learning
asked Dec 3 '18 at 17:41
Sean LeeSean Lee
1749
asked Dec 3 '18 at 17:41
Sean LeeSean Lee
1749
asked Dec 3 '18 at 17:41
Sean LeeSean Lee
1749
asked Dec 3 '18 at 17:41
Sean LeeSean Lee
1749
asked Dec 3 '18 at 17:41
Sean LeeSean Lee
1749
1749
$begingroup$
What metric are you using on the space containing the $(z,h)$?
$endgroup$
– copper.hat
Dec 3 '18 at 17:46
$begingroup$
I'm sorry, but no metric was specified; should I just assume the Euclidean metric in this case?
$endgroup$
– Sean Lee
Dec 3 '18 at 17:50
1
$begingroup$
I don't know what you mean by Euclidean metric for the function $h$. I don't have time to read through the pdf and figure it out. By Lipschitz you need to show that $|l(x_1,h_1) -l(z_2,h_2)| le L d((z_1,h_1),(z_2,h_2))$, where $d$ is some metric.
$endgroup$
– copper.hat
Dec 3 '18 at 17:58
$begingroup$
Ah okay sure thank you I'll try to work from that angle :)
$endgroup$
– Sean Lee
Dec 3 '18 at 17:59
add a comment |
$begingroup$
What metric are you using on the space containing the $(z,h)$?
$endgroup$
– copper.hat
Dec 3 '18 at 17:46
$begingroup$
I'm sorry, but no metric was specified; should I just assume the Euclidean metric in this case?
$endgroup$
– Sean Lee
Dec 3 '18 at 17:50
1
$begingroup$
I don't know what you mean by Euclidean metric for the function $h$. I don't have time to read through the pdf and figure it out. By Lipschitz you need to show that $|l(x_1,h_1) -l(z_2,h_2)| le L d((z_1,h_1),(z_2,h_2))$, where $d$ is some metric.
$endgroup$
– copper.hat
Dec 3 '18 at 17:58
$begingroup$
Ah okay sure thank you I'll try to work from that angle :)
$endgroup$
– Sean Lee
Dec 3 '18 at 17:59
$begingroup$
What metric are you using on the space containing the $(z,h)$?
$endgroup$
– copper.hat
Dec 3 '18 at 17:46
$begingroup$
What metric are you using on the space containing the $(z,h)$?
$endgroup$
– copper.hat
Dec 3 '18 at 17:46
$begingroup$
I'm sorry, but no metric was specified; should I just assume the Euclidean metric in this case?
$endgroup$
– Sean Lee
Dec 3 '18 at 17:50
$begingroup$
I'm sorry, but no metric was specified; should I just assume the Euclidean metric in this case?
$endgroup$
– Sean Lee
Dec 3 '18 at 17:50
1
1
$begingroup$
I don't know what you mean by Euclidean metric for the function $h$. I don't have time to read through the pdf and figure it out. By Lipschitz you need to show that $|l(x_1,h_1) -l(z_2,h_2)| le L d((z_1,h_1),(z_2,h_2))$, where $d$ is some metric.
$endgroup$
– copper.hat
Dec 3 '18 at 17:58
$begingroup$
I don't know what you mean by Euclidean metric for the function $h$. I don't have time to read through the pdf and figure it out. By Lipschitz you need to show that $|l(x_1,h_1) -l(z_2,h_2)| le L d((z_1,h_1),(z_2,h_2))$, where $d$ is some metric.
$endgroup$
– copper.hat
Dec 3 '18 at 17:58
$begingroup$
Ah okay sure thank you I'll try to work from that angle :)
$endgroup$
– Sean Lee
Dec 3 '18 at 17:59
$begingroup$
Ah okay sure thank you I'll try to work from that angle :)
$endgroup$
– Sean Lee
Dec 3 '18 at 17:59
add a comment |
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$begingroup$
What metric are you using on the space containing the $(z,h)$?
$endgroup$
– copper.hat
Dec 3 '18 at 17:46
$begingroup$
I'm sorry, but no metric was specified; should I just assume the Euclidean metric in this case?
$endgroup$
– Sean Lee
Dec 3 '18 at 17:50
1
$begingroup$
I don't know what you mean by Euclidean metric for the function $h$. I don't have time to read through the pdf and figure it out. By Lipschitz you need to show that $|l(x_1,h_1) -l(z_2,h_2)| le L d((z_1,h_1),(z_2,h_2))$, where $d$ is some metric.
$endgroup$
– copper.hat
Dec 3 '18 at 17:58
$begingroup$
Ah okay sure thank you I'll try to work from that angle :)
$endgroup$
– Sean Lee
Dec 3 '18 at 17:59