How is the union bound applied in this proof?
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From Understanding Machine Learning: From theory to algorithms:
How explicitly is the union bound used in the proof of this theorem to get the result in the red box below?
probability-theory machine-learning
asked Jan 19 at 19:46
Oliver GOliver G
1,4971531
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add a comment |
$begingroup$
From Understanding Machine Learning: From theory to algorithms:
How explicitly is the union bound used in the proof of this theorem to get the result in the red box below?
probability-theory machine-learning
asked Jan 19 at 19:46
Oliver GOliver G
1,4971531
$endgroup$
add a comment |
$begingroup$
From Understanding Machine Learning: From theory to algorithms:
How explicitly is the union bound used in the proof of this theorem to get the result in the red box below?
probability-theory machine-learning
asked Jan 19 at 19:46
Oliver GOliver G
1,4971531
$endgroup$
From Understanding Machine Learning: From theory to algorithms:
How explicitly is the union bound used in the proof of this theorem to get the result in the red box below?
probability-theory machine-learning
probability-theory machine-learning
asked Jan 19 at 19:46
Oliver GOliver G
1,4971531
asked Jan 19 at 19:46
Oliver GOliver G
1,4971531
asked Jan 19 at 19:46
Oliver GOliver G
1,4971531
asked Jan 19 at 19:46
Oliver GOliver G
1,4971531
asked Jan 19 at 19:46
Oliver GOliver G
1,4971531
1,4971531
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let your events be $A_n = { |L_D(h) - L_S(h)| < epsilon_n$ holds for all $h in mathcal{H}_n }$.
$P(A_n^c) < delta_n $ as per the previous paragraph, where I am using $c$ to denote complement. Now, we are looking for the event that all $A_n$'s happen, i.e. for the $P(bigcap_{i=1}^n A_n) = 1 - P(bigcup_{i=1}^n A_n^c)$. Now, using union bound on that last term we get $P(bigcup_{i=1}^n A_n^c) leq sum_i P(A_i^c) leq sum_i delta_i$, so $P(bigcap_{i=1}^n A_n)$ is at least $1 - sum_i delta_i$.
Adding the elements of the set as per OPs request: the theorem says that the space we are working in is $Omega = mathcal{D}^n$, and we pick some $S$ in it. So, more formally, we have
$A_n = { S in mathcal{D}^n: |L_D(h) - L_S(h)| < epsilon_n$ holds for all $h in mathcal{H}_n }$.
answered Jan 19 at 21:10
E-AE-A
2,1121414
$endgroup$
$begingroup$
I'm a little confused on your notation of $A_n$, what is an element of that set?
$endgroup$
– Oliver G
Jan 19 at 21:39
$begingroup$
What does $S in D^n$ mean? I've never seen the notation of a set of samples being an element of a distribution. The notation $S text{~} D^n$ is what is in the book and means the $n$ elements of the set $S$ are chosen according to distribution $D$.
$endgroup$
– Oliver G
Jan 19 at 21:50
$begingroup$
And what do you mean by $Omega = D^n$? Because to me that reads: the sample space is the distribution function that maps events from a sample space to $[0,1]$.
$endgroup$
– Oliver G
Jan 19 at 22:00
1
$begingroup$
That is what I thought but it is pretty hard to try to infer the content of a book from one screenshot... Either way, then the sample space is whatever domain $S$ belongs to to the $n$. If you upload the relevant screenshots for the definitions it might be easier for people to help you.
$endgroup$
– E-A
Jan 20 at 2:02
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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$begingroup$
Let your events be $A_n = { |L_D(h) - L_S(h)| < epsilon_n$ holds for all $h in mathcal{H}_n }$.
$P(A_n^c) < delta_n $ as per the previous paragraph, where I am using $c$ to denote complement. Now, we are looking for the event that all $A_n$'s happen, i.e. for the $P(bigcap_{i=1}^n A_n) = 1 - P(bigcup_{i=1}^n A_n^c)$. Now, using union bound on that last term we get $P(bigcup_{i=1}^n A_n^c) leq sum_i P(A_i^c) leq sum_i delta_i$, so $P(bigcap_{i=1}^n A_n)$ is at least $1 - sum_i delta_i$.
Adding the elements of the set as per OPs request: the theorem says that the space we are working in is $Omega = mathcal{D}^n$, and we pick some $S$ in it. So, more formally, we have
$A_n = { S in mathcal{D}^n: |L_D(h) - L_S(h)| < epsilon_n$ holds for all $h in mathcal{H}_n }$.
answered Jan 19 at 21:10
E-AE-A
2,1121414
$endgroup$
$begingroup$
I'm a little confused on your notation of $A_n$, what is an element of that set?
$endgroup$
– Oliver G
Jan 19 at 21:39
$begingroup$
What does $S in D^n$ mean? I've never seen the notation of a set of samples being an element of a distribution. The notation $S text{~} D^n$ is what is in the book and means the $n$ elements of the set $S$ are chosen according to distribution $D$.
$endgroup$
– Oliver G
Jan 19 at 21:50
$begingroup$
And what do you mean by $Omega = D^n$? Because to me that reads: the sample space is the distribution function that maps events from a sample space to $[0,1]$.
$endgroup$
– Oliver G
Jan 19 at 22:00
1
$begingroup$
That is what I thought but it is pretty hard to try to infer the content of a book from one screenshot... Either way, then the sample space is whatever domain $S$ belongs to to the $n$. If you upload the relevant screenshots for the definitions it might be easier for people to help you.
$endgroup$
– E-A
Jan 20 at 2:02
add a comment |
$begingroup$
Let your events be $A_n = { |L_D(h) - L_S(h)| < epsilon_n$ holds for all $h in mathcal{H}_n }$.
$P(A_n^c) < delta_n $ as per the previous paragraph, where I am using $c$ to denote complement. Now, we are looking for the event that all $A_n$'s happen, i.e. for the $P(bigcap_{i=1}^n A_n) = 1 - P(bigcup_{i=1}^n A_n^c)$. Now, using union bound on that last term we get $P(bigcup_{i=1}^n A_n^c) leq sum_i P(A_i^c) leq sum_i delta_i$, so $P(bigcap_{i=1}^n A_n)$ is at least $1 - sum_i delta_i$.
Adding the elements of the set as per OPs request: the theorem says that the space we are working in is $Omega = mathcal{D}^n$, and we pick some $S$ in it. So, more formally, we have
$A_n = { S in mathcal{D}^n: |L_D(h) - L_S(h)| < epsilon_n$ holds for all $h in mathcal{H}_n }$.
answered Jan 19 at 21:10
E-AE-A
2,1121414
$endgroup$
$begingroup$
I'm a little confused on your notation of $A_n$, what is an element of that set?
$endgroup$
– Oliver G
Jan 19 at 21:39
$begingroup$
What does $S in D^n$ mean? I've never seen the notation of a set of samples being an element of a distribution. The notation $S text{~} D^n$ is what is in the book and means the $n$ elements of the set $S$ are chosen according to distribution $D$.
$endgroup$
– Oliver G
Jan 19 at 21:50
$begingroup$
And what do you mean by $Omega = D^n$? Because to me that reads: the sample space is the distribution function that maps events from a sample space to $[0,1]$.
$endgroup$
– Oliver G
Jan 19 at 22:00
1
$begingroup$
That is what I thought but it is pretty hard to try to infer the content of a book from one screenshot... Either way, then the sample space is whatever domain $S$ belongs to to the $n$. If you upload the relevant screenshots for the definitions it might be easier for people to help you.
$endgroup$
– E-A
Jan 20 at 2:02
add a comment |
$begingroup$
Let your events be $A_n = { |L_D(h) - L_S(h)| < epsilon_n$ holds for all $h in mathcal{H}_n }$.
$P(A_n^c) < delta_n $ as per the previous paragraph, where I am using $c$ to denote complement. Now, we are looking for the event that all $A_n$'s happen, i.e. for the $P(bigcap_{i=1}^n A_n) = 1 - P(bigcup_{i=1}^n A_n^c)$. Now, using union bound on that last term we get $P(bigcup_{i=1}^n A_n^c) leq sum_i P(A_i^c) leq sum_i delta_i$, so $P(bigcap_{i=1}^n A_n)$ is at least $1 - sum_i delta_i$.
Adding the elements of the set as per OPs request: the theorem says that the space we are working in is $Omega = mathcal{D}^n$, and we pick some $S$ in it. So, more formally, we have
$A_n = { S in mathcal{D}^n: |L_D(h) - L_S(h)| < epsilon_n$ holds for all $h in mathcal{H}_n }$.
answered Jan 19 at 21:10
E-AE-A
2,1121414
$endgroup$
Let your events be $A_n = { |L_D(h) - L_S(h)| < epsilon_n$ holds for all $h in mathcal{H}_n }$.
$P(A_n^c) < delta_n $ as per the previous paragraph, where I am using $c$ to denote complement. Now, we are looking for the event that all $A_n$'s happen, i.e. for the $P(bigcap_{i=1}^n A_n) = 1 - P(bigcup_{i=1}^n A_n^c)$. Now, using union bound on that last term we get $P(bigcup_{i=1}^n A_n^c) leq sum_i P(A_i^c) leq sum_i delta_i$, so $P(bigcap_{i=1}^n A_n)$ is at least $1 - sum_i delta_i$.
Adding the elements of the set as per OPs request: the theorem says that the space we are working in is $Omega = mathcal{D}^n$, and we pick some $S$ in it. So, more formally, we have
$A_n = { S in mathcal{D}^n: |L_D(h) - L_S(h)| < epsilon_n$ holds for all $h in mathcal{H}_n }$.
answered Jan 19 at 21:10
E-AE-A
2,1121414
edited Jan 19 at 21:44
answered Jan 19 at 21:10
E-AE-A
2,1121414
answered Jan 19 at 21:10
E-AE-A
2,1121414
answered Jan 19 at 21:10
E-AE-A
2,1121414
2,1121414
$begingroup$
I'm a little confused on your notation of $A_n$, what is an element of that set?
$endgroup$
– Oliver G
Jan 19 at 21:39
$begingroup$
What does $S in D^n$ mean? I've never seen the notation of a set of samples being an element of a distribution. The notation $S text{~} D^n$ is what is in the book and means the $n$ elements of the set $S$ are chosen according to distribution $D$.
$endgroup$
– Oliver G
Jan 19 at 21:50
$begingroup$
And what do you mean by $Omega = D^n$? Because to me that reads: the sample space is the distribution function that maps events from a sample space to $[0,1]$.
$endgroup$
– Oliver G
Jan 19 at 22:00
1
$begingroup$
That is what I thought but it is pretty hard to try to infer the content of a book from one screenshot... Either way, then the sample space is whatever domain $S$ belongs to to the $n$. If you upload the relevant screenshots for the definitions it might be easier for people to help you.
$endgroup$
– E-A
Jan 20 at 2:02
add a comment |
$begingroup$
I'm a little confused on your notation of $A_n$, what is an element of that set?
$endgroup$
– Oliver G
Jan 19 at 21:39
$begingroup$
What does $S in D^n$ mean? I've never seen the notation of a set of samples being an element of a distribution. The notation $S text{~} D^n$ is what is in the book and means the $n$ elements of the set $S$ are chosen according to distribution $D$.
$endgroup$
– Oliver G
Jan 19 at 21:50
$begingroup$
And what do you mean by $Omega = D^n$? Because to me that reads: the sample space is the distribution function that maps events from a sample space to $[0,1]$.
$endgroup$
– Oliver G
Jan 19 at 22:00
1
$begingroup$
That is what I thought but it is pretty hard to try to infer the content of a book from one screenshot... Either way, then the sample space is whatever domain $S$ belongs to to the $n$. If you upload the relevant screenshots for the definitions it might be easier for people to help you.
$endgroup$
– E-A
Jan 20 at 2:02
$begingroup$
I'm a little confused on your notation of $A_n$, what is an element of that set?
$endgroup$
– Oliver G
Jan 19 at 21:39
$begingroup$
I'm a little confused on your notation of $A_n$, what is an element of that set?
$endgroup$
– Oliver G
Jan 19 at 21:39
$begingroup$
What does $S in D^n$ mean? I've never seen the notation of a set of samples being an element of a distribution. The notation $S text{~} D^n$ is what is in the book and means the $n$ elements of the set $S$ are chosen according to distribution $D$.
$endgroup$
– Oliver G
Jan 19 at 21:50
$begingroup$
What does $S in D^n$ mean? I've never seen the notation of a set of samples being an element of a distribution. The notation $S text{~} D^n$ is what is in the book and means the $n$ elements of the set $S$ are chosen according to distribution $D$.
$endgroup$
– Oliver G
Jan 19 at 21:50
$begingroup$
And what do you mean by $Omega = D^n$? Because to me that reads: the sample space is the distribution function that maps events from a sample space to $[0,1]$.
$endgroup$
– Oliver G
Jan 19 at 22:00
$begingroup$
And what do you mean by $Omega = D^n$? Because to me that reads: the sample space is the distribution function that maps events from a sample space to $[0,1]$.
$endgroup$
– Oliver G
Jan 19 at 22:00
1
1
$begingroup$
That is what I thought but it is pretty hard to try to infer the content of a book from one screenshot... Either way, then the sample space is whatever domain $S$ belongs to to the $n$. If you upload the relevant screenshots for the definitions it might be easier for people to help you.
$endgroup$
– E-A
Jan 20 at 2:02
$begingroup$
That is what I thought but it is pretty hard to try to infer the content of a book from one screenshot... Either way, then the sample space is whatever domain $S$ belongs to to the $n$. If you upload the relevant screenshots for the definitions it might be easier for people to help you.
$endgroup$
– E-A
Jan 20 at 2:02
add a comment |
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