Creating thinned models in during Dropout process












4












$begingroup$



Applying dropout to a neural network amounts to sampling a “thinned” network from it. The thinned network consists of all the units that survived dropout. A neural net with n units can be seen as a collection of 2^n possible thinned neural networks.




Source:
Dropout: A Simple Way to Prevent Neural Networks fromOverfitting, pg. 1931.



How are we getting these 2^n models?






machine-learning deep-learning dropout







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    4












    $begingroup$



    Applying dropout to a neural network amounts to sampling a “thinned” network from it. The thinned network consists of all the units that survived dropout. A neural net with n units can be seen as a collection of 2^n possible thinned neural networks.




    Source:
    Dropout: A Simple Way to Prevent Neural Networks fromOverfitting, pg. 1931.



    How are we getting these 2^n models?






    machine-learning deep-learning dropout







    share|cite|improve this question









    New contributor




    ashirwad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      4












      4








      4





      $begingroup$



      Applying dropout to a neural network amounts to sampling a “thinned” network from it. The thinned network consists of all the units that survived dropout. A neural net with n units can be seen as a collection of 2^n possible thinned neural networks.




      Source:
      Dropout: A Simple Way to Prevent Neural Networks fromOverfitting, pg. 1931.



      How are we getting these 2^n models?






      machine-learning deep-learning dropout







      share|cite|improve this question









      New contributor




      ashirwad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$





      Applying dropout to a neural network amounts to sampling a “thinned” network from it. The thinned network consists of all the units that survived dropout. A neural net with n units can be seen as a collection of 2^n possible thinned neural networks.




      Source:
      Dropout: A Simple Way to Prevent Neural Networks fromOverfitting, pg. 1931.



      How are we getting these 2^n models?






      machine-learning deep-learning dropout




      machine-learning deep-learning dropout






      share|cite|improve this question









      New contributor




      ashirwad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









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      ashirwad is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









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      share|cite|improve this question








      edited 2 days ago









      Djib2011

      2,58931125




      2,58931125






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      asked 2 days ago









      ashirwadashirwad

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          2 Answers
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          active

          oldest

          votes


















          4












          $begingroup$

          The statement is a bit oversimplifying but the idea is that assuming we have $n$ nodes and each of these nodes might be "dropped", we have $2^n$ possible thinned neural networks. Obviously dropping out an entire layer would alter the whole structure of the network but the idea is straightforward: we ignore the activation/information from certain randomly selected neurons and thus encourage redundancy learning and discourage over-fitting on very specific features.



          The same idea has also been employed in Gradient Boosting Machines where instead of "ignoring neurons" we "ignore trees" at random (see Rashmi & Gilad-Bachrach (2015) DART: Dropouts meet Multiple Additive Regression Trees on that matter).



          Minor edit: I just saw Djib2011's answer. (+1) He/she specifically shows why the statement is somewhat over-simplifying. If we assume that we can drop any (or all, or none) of the neurons we have $2^n$ possible networks.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            I too haven't understood their reasoning, I always assumed it was a typo or something...



            The way I see it we if we have $n$ hidden units in a Neural Network with a single hidden layer and we apply dropout keeping $r$ of those, we'll have:



            $$
            frac{n!}{r! cdot (n-r)!}
            $$



            possible combinations (not $2^n$ as the authors state).




            Example:



            Assume a simple fully connected neural network with a single hidden layer with 4 neurons. This means the hidden layer will have 4 outputs $h_1, h_2, h_3, h_4$.



            Now, you want to apply dropout to this layer with a 0.5 probability (i.e. half of the outputs will be dropped).



            Since 2 out of the 4 outputs will be dropped, at each training iteration we'll have one of the following possibilities:




            1. $h_1, h_2$

            2. $h_1, h_3$

            3. $h_1, h_4$

            4. $h_2, h_3$

            5. $h_2, h_4$

            6. $h_3, h_4$


            or by applying the formula:



            $$
            frac{4!}{2! cdot (4-2)!} = frac{24}{2 cdot 2} = 6
            $$






            share|cite|improve this answer









            $endgroup$









            • 3




              $begingroup$
              I do not think that most implementations of dropout work by saying: If there are 100 neurons, and the probability is 0.05, I have to disable exactly 5 neurons chosen at random. Instead each neuron is disabled with a probability of 0.05, independently of what happens with the rest. Hence, the cases where all or no neurons are disabled, while unlikely, are possible.
              $endgroup$
              – Daniel López
              2 days ago






            • 1




              $begingroup$
              @DanielLópez: I think both you and Djib2011 (+1 both) are "factually correct" on this. The statement is clearly oversimplifying things. You also need to take into account that most the networks that this paper is concerned with, have thousands of neurons so certain it is kind of accepted that no layer will be totally switched off.
              $endgroup$
              – usεr11852
              2 days ago








            • 2




              $begingroup$
              Agree, but I believe the above example is transmitting the idea that exactly $n cdot text{prob}$ units are disabled with dropout, where $text{prob}$ is the dropout probability. And this is not how dropout works.
              $endgroup$
              – Daniel López
              2 days ago












            • $begingroup$
              Well... LLN is our friend. :)
              $endgroup$
              – usεr11852
              2 days ago










            • $begingroup$
              The flaw with the reasoning presented here is that dropout sets weights to 0 with some fixed probability independently. This implies that the number of zero weights at each step has a binomial distribution, because dropout has the three defining characteristics of a binomial distribution 1 dichotomous outcomes (weights are on or off) 2 fixed number of trials (number of weights in the model doesn't change) 3 probability of success is fixed & independent for each trial.
              $endgroup$
              – Sycorax
              yesterday













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            2 Answers
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            4












            $begingroup$

            The statement is a bit oversimplifying but the idea is that assuming we have $n$ nodes and each of these nodes might be "dropped", we have $2^n$ possible thinned neural networks. Obviously dropping out an entire layer would alter the whole structure of the network but the idea is straightforward: we ignore the activation/information from certain randomly selected neurons and thus encourage redundancy learning and discourage over-fitting on very specific features.



            The same idea has also been employed in Gradient Boosting Machines where instead of "ignoring neurons" we "ignore trees" at random (see Rashmi & Gilad-Bachrach (2015) DART: Dropouts meet Multiple Additive Regression Trees on that matter).



            Minor edit: I just saw Djib2011's answer. (+1) He/she specifically shows why the statement is somewhat over-simplifying. If we assume that we can drop any (or all, or none) of the neurons we have $2^n$ possible networks.






            share|cite|improve this answer











            $endgroup$


















              4












              $begingroup$

              The statement is a bit oversimplifying but the idea is that assuming we have $n$ nodes and each of these nodes might be "dropped", we have $2^n$ possible thinned neural networks. Obviously dropping out an entire layer would alter the whole structure of the network but the idea is straightforward: we ignore the activation/information from certain randomly selected neurons and thus encourage redundancy learning and discourage over-fitting on very specific features.



              The same idea has also been employed in Gradient Boosting Machines where instead of "ignoring neurons" we "ignore trees" at random (see Rashmi & Gilad-Bachrach (2015) DART: Dropouts meet Multiple Additive Regression Trees on that matter).



              Minor edit: I just saw Djib2011's answer. (+1) He/she specifically shows why the statement is somewhat over-simplifying. If we assume that we can drop any (or all, or none) of the neurons we have $2^n$ possible networks.






              share|cite|improve this answer











              $endgroup$
















                4












                4








                4





                $begingroup$

                The statement is a bit oversimplifying but the idea is that assuming we have $n$ nodes and each of these nodes might be "dropped", we have $2^n$ possible thinned neural networks. Obviously dropping out an entire layer would alter the whole structure of the network but the idea is straightforward: we ignore the activation/information from certain randomly selected neurons and thus encourage redundancy learning and discourage over-fitting on very specific features.



                The same idea has also been employed in Gradient Boosting Machines where instead of "ignoring neurons" we "ignore trees" at random (see Rashmi & Gilad-Bachrach (2015) DART: Dropouts meet Multiple Additive Regression Trees on that matter).



                Minor edit: I just saw Djib2011's answer. (+1) He/she specifically shows why the statement is somewhat over-simplifying. If we assume that we can drop any (or all, or none) of the neurons we have $2^n$ possible networks.






                share|cite|improve this answer











                $endgroup$



                The statement is a bit oversimplifying but the idea is that assuming we have $n$ nodes and each of these nodes might be "dropped", we have $2^n$ possible thinned neural networks. Obviously dropping out an entire layer would alter the whole structure of the network but the idea is straightforward: we ignore the activation/information from certain randomly selected neurons and thus encourage redundancy learning and discourage over-fitting on very specific features.



                The same idea has also been employed in Gradient Boosting Machines where instead of "ignoring neurons" we "ignore trees" at random (see Rashmi & Gilad-Bachrach (2015) DART: Dropouts meet Multiple Additive Regression Trees on that matter).



                Minor edit: I just saw Djib2011's answer. (+1) He/she specifically shows why the statement is somewhat over-simplifying. If we assume that we can drop any (or all, or none) of the neurons we have $2^n$ possible networks.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 days ago

























                answered 2 days ago









                usεr11852usεr11852

                19.3k14274




                19.3k14274

























                    0












                    $begingroup$

                    I too haven't understood their reasoning, I always assumed it was a typo or something...



                    The way I see it we if we have $n$ hidden units in a Neural Network with a single hidden layer and we apply dropout keeping $r$ of those, we'll have:



                    $$
                    frac{n!}{r! cdot (n-r)!}
                    $$



                    possible combinations (not $2^n$ as the authors state).




                    Example:



                    Assume a simple fully connected neural network with a single hidden layer with 4 neurons. This means the hidden layer will have 4 outputs $h_1, h_2, h_3, h_4$.



                    Now, you want to apply dropout to this layer with a 0.5 probability (i.e. half of the outputs will be dropped).



                    Since 2 out of the 4 outputs will be dropped, at each training iteration we'll have one of the following possibilities:




                    1. $h_1, h_2$

                    2. $h_1, h_3$

                    3. $h_1, h_4$

                    4. $h_2, h_3$

                    5. $h_2, h_4$

                    6. $h_3, h_4$


                    or by applying the formula:



                    $$
                    frac{4!}{2! cdot (4-2)!} = frac{24}{2 cdot 2} = 6
                    $$






                    share|cite|improve this answer









                    $endgroup$









                    • 3




                      $begingroup$
                      I do not think that most implementations of dropout work by saying: If there are 100 neurons, and the probability is 0.05, I have to disable exactly 5 neurons chosen at random. Instead each neuron is disabled with a probability of 0.05, independently of what happens with the rest. Hence, the cases where all or no neurons are disabled, while unlikely, are possible.
                      $endgroup$
                      – Daniel López
                      2 days ago






                    • 1




                      $begingroup$
                      @DanielLópez: I think both you and Djib2011 (+1 both) are "factually correct" on this. The statement is clearly oversimplifying things. You also need to take into account that most the networks that this paper is concerned with, have thousands of neurons so certain it is kind of accepted that no layer will be totally switched off.
                      $endgroup$
                      – usεr11852
                      2 days ago








                    • 2




                      $begingroup$
                      Agree, but I believe the above example is transmitting the idea that exactly $n cdot text{prob}$ units are disabled with dropout, where $text{prob}$ is the dropout probability. And this is not how dropout works.
                      $endgroup$
                      – Daniel López
                      2 days ago












                    • $begingroup$
                      Well... LLN is our friend. :)
                      $endgroup$
                      – usεr11852
                      2 days ago










                    • $begingroup$
                      The flaw with the reasoning presented here is that dropout sets weights to 0 with some fixed probability independently. This implies that the number of zero weights at each step has a binomial distribution, because dropout has the three defining characteristics of a binomial distribution 1 dichotomous outcomes (weights are on or off) 2 fixed number of trials (number of weights in the model doesn't change) 3 probability of success is fixed & independent for each trial.
                      $endgroup$
                      – Sycorax
                      yesterday


















                    0












                    $begingroup$

                    I too haven't understood their reasoning, I always assumed it was a typo or something...



                    The way I see it we if we have $n$ hidden units in a Neural Network with a single hidden layer and we apply dropout keeping $r$ of those, we'll have:



                    $$
                    frac{n!}{r! cdot (n-r)!}
                    $$



                    possible combinations (not $2^n$ as the authors state).




                    Example:



                    Assume a simple fully connected neural network with a single hidden layer with 4 neurons. This means the hidden layer will have 4 outputs $h_1, h_2, h_3, h_4$.



                    Now, you want to apply dropout to this layer with a 0.5 probability (i.e. half of the outputs will be dropped).



                    Since 2 out of the 4 outputs will be dropped, at each training iteration we'll have one of the following possibilities:




                    1. $h_1, h_2$

                    2. $h_1, h_3$

                    3. $h_1, h_4$

                    4. $h_2, h_3$

                    5. $h_2, h_4$

                    6. $h_3, h_4$


                    or by applying the formula:



                    $$
                    frac{4!}{2! cdot (4-2)!} = frac{24}{2 cdot 2} = 6
                    $$






                    share|cite|improve this answer









                    $endgroup$









                    • 3




                      $begingroup$
                      I do not think that most implementations of dropout work by saying: If there are 100 neurons, and the probability is 0.05, I have to disable exactly 5 neurons chosen at random. Instead each neuron is disabled with a probability of 0.05, independently of what happens with the rest. Hence, the cases where all or no neurons are disabled, while unlikely, are possible.
                      $endgroup$
                      – Daniel López
                      2 days ago






                    • 1




                      $begingroup$
                      @DanielLópez: I think both you and Djib2011 (+1 both) are "factually correct" on this. The statement is clearly oversimplifying things. You also need to take into account that most the networks that this paper is concerned with, have thousands of neurons so certain it is kind of accepted that no layer will be totally switched off.
                      $endgroup$
                      – usεr11852
                      2 days ago








                    • 2




                      $begingroup$
                      Agree, but I believe the above example is transmitting the idea that exactly $n cdot text{prob}$ units are disabled with dropout, where $text{prob}$ is the dropout probability. And this is not how dropout works.
                      $endgroup$
                      – Daniel López
                      2 days ago












                    • $begingroup$
                      Well... LLN is our friend. :)
                      $endgroup$
                      – usεr11852
                      2 days ago










                    • $begingroup$
                      The flaw with the reasoning presented here is that dropout sets weights to 0 with some fixed probability independently. This implies that the number of zero weights at each step has a binomial distribution, because dropout has the three defining characteristics of a binomial distribution 1 dichotomous outcomes (weights are on or off) 2 fixed number of trials (number of weights in the model doesn't change) 3 probability of success is fixed & independent for each trial.
                      $endgroup$
                      – Sycorax
                      yesterday
















                    0












                    0








                    0





                    $begingroup$

                    I too haven't understood their reasoning, I always assumed it was a typo or something...



                    The way I see it we if we have $n$ hidden units in a Neural Network with a single hidden layer and we apply dropout keeping $r$ of those, we'll have:



                    $$
                    frac{n!}{r! cdot (n-r)!}
                    $$



                    possible combinations (not $2^n$ as the authors state).




                    Example:



                    Assume a simple fully connected neural network with a single hidden layer with 4 neurons. This means the hidden layer will have 4 outputs $h_1, h_2, h_3, h_4$.



                    Now, you want to apply dropout to this layer with a 0.5 probability (i.e. half of the outputs will be dropped).



                    Since 2 out of the 4 outputs will be dropped, at each training iteration we'll have one of the following possibilities:




                    1. $h_1, h_2$

                    2. $h_1, h_3$

                    3. $h_1, h_4$

                    4. $h_2, h_3$

                    5. $h_2, h_4$

                    6. $h_3, h_4$


                    or by applying the formula:



                    $$
                    frac{4!}{2! cdot (4-2)!} = frac{24}{2 cdot 2} = 6
                    $$






                    share|cite|improve this answer









                    $endgroup$



                    I too haven't understood their reasoning, I always assumed it was a typo or something...



                    The way I see it we if we have $n$ hidden units in a Neural Network with a single hidden layer and we apply dropout keeping $r$ of those, we'll have:



                    $$
                    frac{n!}{r! cdot (n-r)!}
                    $$



                    possible combinations (not $2^n$ as the authors state).




                    Example:



                    Assume a simple fully connected neural network with a single hidden layer with 4 neurons. This means the hidden layer will have 4 outputs $h_1, h_2, h_3, h_4$.



                    Now, you want to apply dropout to this layer with a 0.5 probability (i.e. half of the outputs will be dropped).



                    Since 2 out of the 4 outputs will be dropped, at each training iteration we'll have one of the following possibilities:




                    1. $h_1, h_2$

                    2. $h_1, h_3$

                    3. $h_1, h_4$

                    4. $h_2, h_3$

                    5. $h_2, h_4$

                    6. $h_3, h_4$


                    or by applying the formula:



                    $$
                    frac{4!}{2! cdot (4-2)!} = frac{24}{2 cdot 2} = 6
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    Djib2011Djib2011

                    2,58931125




                    2,58931125








                    • 3




                      $begingroup$
                      I do not think that most implementations of dropout work by saying: If there are 100 neurons, and the probability is 0.05, I have to disable exactly 5 neurons chosen at random. Instead each neuron is disabled with a probability of 0.05, independently of what happens with the rest. Hence, the cases where all or no neurons are disabled, while unlikely, are possible.
                      $endgroup$
                      – Daniel López
                      2 days ago






                    • 1




                      $begingroup$
                      @DanielLópez: I think both you and Djib2011 (+1 both) are "factually correct" on this. The statement is clearly oversimplifying things. You also need to take into account that most the networks that this paper is concerned with, have thousands of neurons so certain it is kind of accepted that no layer will be totally switched off.
                      $endgroup$
                      – usεr11852
                      2 days ago








                    • 2




                      $begingroup$
                      Agree, but I believe the above example is transmitting the idea that exactly $n cdot text{prob}$ units are disabled with dropout, where $text{prob}$ is the dropout probability. And this is not how dropout works.
                      $endgroup$
                      – Daniel López
                      2 days ago












                    • $begingroup$
                      Well... LLN is our friend. :)
                      $endgroup$
                      – usεr11852
                      2 days ago










                    • $begingroup$
                      The flaw with the reasoning presented here is that dropout sets weights to 0 with some fixed probability independently. This implies that the number of zero weights at each step has a binomial distribution, because dropout has the three defining characteristics of a binomial distribution 1 dichotomous outcomes (weights are on or off) 2 fixed number of trials (number of weights in the model doesn't change) 3 probability of success is fixed & independent for each trial.
                      $endgroup$
                      – Sycorax
                      yesterday
















                    • 3




                      $begingroup$
                      I do not think that most implementations of dropout work by saying: If there are 100 neurons, and the probability is 0.05, I have to disable exactly 5 neurons chosen at random. Instead each neuron is disabled with a probability of 0.05, independently of what happens with the rest. Hence, the cases where all or no neurons are disabled, while unlikely, are possible.
                      $endgroup$
                      – Daniel López
                      2 days ago






                    • 1




                      $begingroup$
                      @DanielLópez: I think both you and Djib2011 (+1 both) are "factually correct" on this. The statement is clearly oversimplifying things. You also need to take into account that most the networks that this paper is concerned with, have thousands of neurons so certain it is kind of accepted that no layer will be totally switched off.
                      $endgroup$
                      – usεr11852
                      2 days ago








                    • 2




                      $begingroup$
                      Agree, but I believe the above example is transmitting the idea that exactly $n cdot text{prob}$ units are disabled with dropout, where $text{prob}$ is the dropout probability. And this is not how dropout works.
                      $endgroup$
                      – Daniel López
                      2 days ago












                    • $begingroup$
                      Well... LLN is our friend. :)
                      $endgroup$
                      – usεr11852
                      2 days ago










                    • $begingroup$
                      The flaw with the reasoning presented here is that dropout sets weights to 0 with some fixed probability independently. This implies that the number of zero weights at each step has a binomial distribution, because dropout has the three defining characteristics of a binomial distribution 1 dichotomous outcomes (weights are on or off) 2 fixed number of trials (number of weights in the model doesn't change) 3 probability of success is fixed & independent for each trial.
                      $endgroup$
                      – Sycorax
                      yesterday










                    3




                    3




                    $begingroup$
                    I do not think that most implementations of dropout work by saying: If there are 100 neurons, and the probability is 0.05, I have to disable exactly 5 neurons chosen at random. Instead each neuron is disabled with a probability of 0.05, independently of what happens with the rest. Hence, the cases where all or no neurons are disabled, while unlikely, are possible.
                    $endgroup$
                    – Daniel López
                    2 days ago




                    $begingroup$
                    I do not think that most implementations of dropout work by saying: If there are 100 neurons, and the probability is 0.05, I have to disable exactly 5 neurons chosen at random. Instead each neuron is disabled with a probability of 0.05, independently of what happens with the rest. Hence, the cases where all or no neurons are disabled, while unlikely, are possible.
                    $endgroup$
                    – Daniel López
                    2 days ago




                    1




                    1




                    $begingroup$
                    @DanielLópez: I think both you and Djib2011 (+1 both) are "factually correct" on this. The statement is clearly oversimplifying things. You also need to take into account that most the networks that this paper is concerned with, have thousands of neurons so certain it is kind of accepted that no layer will be totally switched off.
                    $endgroup$
                    – usεr11852
                    2 days ago






                    $begingroup$
                    @DanielLópez: I think both you and Djib2011 (+1 both) are "factually correct" on this. The statement is clearly oversimplifying things. You also need to take into account that most the networks that this paper is concerned with, have thousands of neurons so certain it is kind of accepted that no layer will be totally switched off.
                    $endgroup$
                    – usεr11852
                    2 days ago






                    2




                    2




                    $begingroup$
                    Agree, but I believe the above example is transmitting the idea that exactly $n cdot text{prob}$ units are disabled with dropout, where $text{prob}$ is the dropout probability. And this is not how dropout works.
                    $endgroup$
                    – Daniel López
                    2 days ago






                    $begingroup$
                    Agree, but I believe the above example is transmitting the idea that exactly $n cdot text{prob}$ units are disabled with dropout, where $text{prob}$ is the dropout probability. And this is not how dropout works.
                    $endgroup$
                    – Daniel López
                    2 days ago














                    $begingroup$
                    Well... LLN is our friend. :)
                    $endgroup$
                    – usεr11852
                    2 days ago




                    $begingroup$
                    Well... LLN is our friend. :)
                    $endgroup$
                    – usεr11852
                    2 days ago












                    $begingroup$
                    The flaw with the reasoning presented here is that dropout sets weights to 0 with some fixed probability independently. This implies that the number of zero weights at each step has a binomial distribution, because dropout has the three defining characteristics of a binomial distribution 1 dichotomous outcomes (weights are on or off) 2 fixed number of trials (number of weights in the model doesn't change) 3 probability of success is fixed & independent for each trial.
                    $endgroup$
                    – Sycorax
                    yesterday






                    $begingroup$
                    The flaw with the reasoning presented here is that dropout sets weights to 0 with some fixed probability independently. This implies that the number of zero weights at each step has a binomial distribution, because dropout has the three defining characteristics of a binomial distribution 1 dichotomous outcomes (weights are on or off) 2 fixed number of trials (number of weights in the model doesn't change) 3 probability of success is fixed & independent for each trial.
                    $endgroup$
                    – Sycorax
                    yesterday












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