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?



enter image description here










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    0












    $begingroup$


    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?



    enter image description here










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      0





      $begingroup$


      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?



      enter image description here










      share|cite|improve this question









      $endgroup$




      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?



      enter image description here







      probability-theory machine-learning






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      asked Jan 19 at 19:46









      Oliver GOliver G

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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 $h in mathcal{H}_n }$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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











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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 $h in mathcal{H}_n }$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          0












          $begingroup$

          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 $h in mathcal{H}_n }$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          0












          0








          0





          $begingroup$

          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 $h in mathcal{H}_n }$.






          share|cite|improve this answer











          $endgroup$



          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 $h in mathcal{H}_n }$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 19 at 21:44

























          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


















          • $begingroup$
            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


















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