Question about the proof that ${h_a: h_a(x) = 1 text{ if } x < a, a in Bbb R }$ is learnable.
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From Understanding Machine Learning :
My three questions about the proof in the below image are:
(1.) In the red box below, why is $(b_0, b_1)$ a valid interval?
(2.) How does $b_s$ being a threshold corresponding to an ERM hypothesis $h_S$ imply that $b_s in (b_0,
b_1)$?
(3.) In the blue box, what makes this condition sufficient?
Assume $S={(x, y)}_{sampled}$ is some training set. Order all the $2$-tuples by their $x$ component and just write the $y$ values to get $S=(y_{x_0}, y_{x_1}, dots, y_{x_n}).$ As an example suppose we have $S=(1,0,1,0)$. Then we have $large S = (1, 0_{b_1}, 1_{b_0}, 0)$. But then we have improper interval notation.
What am I misunderstanding in the proof of this lemma?
probability proof-explanation machine-learning
asked Jan 15 at 15:45
Oliver GOliver G
1,5791530
$endgroup$
add a comment |
$begingroup$
From Understanding Machine Learning :
My three questions about the proof in the below image are:
(1.) In the red box below, why is $(b_0, b_1)$ a valid interval?
(2.) How does $b_s$ being a threshold corresponding to an ERM hypothesis $h_S$ imply that $b_s in (b_0,
b_1)$?
(3.) In the blue box, what makes this condition sufficient?
Assume $S={(x, y)}_{sampled}$ is some training set. Order all the $2$-tuples by their $x$ component and just write the $y$ values to get $S=(y_{x_0}, y_{x_1}, dots, y_{x_n}).$ As an example suppose we have $S=(1,0,1,0)$. Then we have $large S = (1, 0_{b_1}, 1_{b_0}, 0)$. But then we have improper interval notation.
What am I misunderstanding in the proof of this lemma?
probability proof-explanation machine-learning
asked Jan 15 at 15:45
Oliver GOliver G
1,5791530
$endgroup$
add a comment |
$begingroup$
From Understanding Machine Learning :
My three questions about the proof in the below image are:
(1.) In the red box below, why is $(b_0, b_1)$ a valid interval?
(2.) How does $b_s$ being a threshold corresponding to an ERM hypothesis $h_S$ imply that $b_s in (b_0,
b_1)$?
(3.) In the blue box, what makes this condition sufficient?
Assume $S={(x, y)}_{sampled}$ is some training set. Order all the $2$-tuples by their $x$ component and just write the $y$ values to get $S=(y_{x_0}, y_{x_1}, dots, y_{x_n}).$ As an example suppose we have $S=(1,0,1,0)$. Then we have $large S = (1, 0_{b_1}, 1_{b_0}, 0)$. But then we have improper interval notation.
What am I misunderstanding in the proof of this lemma?
probability proof-explanation machine-learning
asked Jan 15 at 15:45
Oliver GOliver G
1,5791530
$endgroup$
From Understanding Machine Learning :
My three questions about the proof in the below image are:
(1.) In the red box below, why is $(b_0, b_1)$ a valid interval?
(2.) How does $b_s$ being a threshold corresponding to an ERM hypothesis $h_S$ imply that $b_s in (b_0,
b_1)$?
(3.) In the blue box, what makes this condition sufficient?
Assume $S={(x, y)}_{sampled}$ is some training set. Order all the $2$-tuples by their $x$ component and just write the $y$ values to get $S=(y_{x_0}, y_{x_1}, dots, y_{x_n}).$ As an example suppose we have $S=(1,0,1,0)$. Then we have $large S = (1, 0_{b_1}, 1_{b_0}, 0)$. But then we have improper interval notation.
What am I misunderstanding in the proof of this lemma?
probability proof-explanation machine-learning
probability proof-explanation machine-learning
asked Jan 15 at 15:45
Oliver GOliver G
1,5791530
asked Jan 15 at 15:45
Oliver GOliver G
1,5791530
edited Jan 15 at 16:01
Oliver G
asked Jan 15 at 15:45
Oliver GOliver G
1,5791530
asked Jan 15 at 15:45
Oliver GOliver G
1,5791530
asked Jan 15 at 15:45
Oliver GOliver G
1,5791530
1,5791530
add a comment |
add a comment |
1 Answer
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$S=(1,0,1,0)$ is not in the function class. If $h_a(x_1)=1$ and $h_a(x_2)=0$, then $x_1<x_2$. So $b_0$ is always smaller than $b_1$.
answered Jan 15 at 20:06
Micky MesserMicky Messer
287
$endgroup$
$begingroup$
That answers $1$ and $2$. Do you have any idea about $3$? I can't see why those conditions are sufficient.
$endgroup$
– Oliver G
Jan 16 at 15:34
$begingroup$
Recall that at $a^*$ the loss is zero. If you move the threshold to $a_0$ or $a_1$, the loss will be $epsilon$. Take into account that the loss funtion is monotonous (in this case, not in general) and you see that it is sufficient
$endgroup$
– Micky Messer
Jan 19 at 8:51
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
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oldest
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$begingroup$
$S=(1,0,1,0)$ is not in the function class. If $h_a(x_1)=1$ and $h_a(x_2)=0$, then $x_1<x_2$. So $b_0$ is always smaller than $b_1$.
answered Jan 15 at 20:06
Micky MesserMicky Messer
287
$endgroup$
$begingroup$
That answers $1$ and $2$. Do you have any idea about $3$? I can't see why those conditions are sufficient.
$endgroup$
– Oliver G
Jan 16 at 15:34
$begingroup$
Recall that at $a^*$ the loss is zero. If you move the threshold to $a_0$ or $a_1$, the loss will be $epsilon$. Take into account that the loss funtion is monotonous (in this case, not in general) and you see that it is sufficient
$endgroup$
– Micky Messer
Jan 19 at 8:51
add a comment |
$begingroup$
$S=(1,0,1,0)$ is not in the function class. If $h_a(x_1)=1$ and $h_a(x_2)=0$, then $x_1<x_2$. So $b_0$ is always smaller than $b_1$.
answered Jan 15 at 20:06
Micky MesserMicky Messer
287
$endgroup$
$begingroup$
That answers $1$ and $2$. Do you have any idea about $3$? I can't see why those conditions are sufficient.
$endgroup$
– Oliver G
Jan 16 at 15:34
$begingroup$
Recall that at $a^*$ the loss is zero. If you move the threshold to $a_0$ or $a_1$, the loss will be $epsilon$. Take into account that the loss funtion is monotonous (in this case, not in general) and you see that it is sufficient
$endgroup$
– Micky Messer
Jan 19 at 8:51
add a comment |
$begingroup$
$S=(1,0,1,0)$ is not in the function class. If $h_a(x_1)=1$ and $h_a(x_2)=0$, then $x_1<x_2$. So $b_0$ is always smaller than $b_1$.
answered Jan 15 at 20:06
Micky MesserMicky Messer
287
$endgroup$
$S=(1,0,1,0)$ is not in the function class. If $h_a(x_1)=1$ and $h_a(x_2)=0$, then $x_1<x_2$. So $b_0$ is always smaller than $b_1$.
answered Jan 15 at 20:06
Micky MesserMicky Messer
287
answered Jan 15 at 20:06
Micky MesserMicky Messer
287
answered Jan 15 at 20:06
Micky MesserMicky Messer
287
answered Jan 15 at 20:06
Micky MesserMicky Messer
287
287
$begingroup$
That answers $1$ and $2$. Do you have any idea about $3$? I can't see why those conditions are sufficient.
$endgroup$
– Oliver G
Jan 16 at 15:34
$begingroup$
Recall that at $a^*$ the loss is zero. If you move the threshold to $a_0$ or $a_1$, the loss will be $epsilon$. Take into account that the loss funtion is monotonous (in this case, not in general) and you see that it is sufficient
$endgroup$
– Micky Messer
Jan 19 at 8:51
add a comment |
$begingroup$
That answers $1$ and $2$. Do you have any idea about $3$? I can't see why those conditions are sufficient.
$endgroup$
– Oliver G
Jan 16 at 15:34
$begingroup$
Recall that at $a^*$ the loss is zero. If you move the threshold to $a_0$ or $a_1$, the loss will be $epsilon$. Take into account that the loss funtion is monotonous (in this case, not in general) and you see that it is sufficient
$endgroup$
– Micky Messer
Jan 19 at 8:51
$begingroup$
That answers $1$ and $2$. Do you have any idea about $3$? I can't see why those conditions are sufficient.
$endgroup$
– Oliver G
Jan 16 at 15:34
$begingroup$
That answers $1$ and $2$. Do you have any idea about $3$? I can't see why those conditions are sufficient.
$endgroup$
– Oliver G
Jan 16 at 15:34
$begingroup$
Recall that at $a^*$ the loss is zero. If you move the threshold to $a_0$ or $a_1$, the loss will be $epsilon$. Take into account that the loss funtion is monotonous (in this case, not in general) and you see that it is sufficient
$endgroup$
– Micky Messer
Jan 19 at 8:51
$begingroup$
Recall that at $a^*$ the loss is zero. If you move the threshold to $a_0$ or $a_1$, the loss will be $epsilon$. Take into account that the loss funtion is monotonous (in this case, not in general) and you see that it is sufficient
$endgroup$
– Micky Messer
Jan 19 at 8:51
add a comment |
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