Why not turn momentum update equation into exponentially weighted moving average update equation? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) 2019 Moderator Election Q&A - Questionnaire 2019 Community Moderator Election ResultsAdam optimizer for projected gradient descentNesterov Momentum update equation

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Why not turn momentum update equation into exponentially weighted moving average update equation?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
2019 Moderator Election Q&A - Questionnaire
2019 Community Moderator Election ResultsAdam optimizer for projected gradient descentNesterov Momentum update equation










0












$begingroup$


In Pytorch, the update equation of SGD with (non-Nesterov) momentum is
$$ m^(i+1) = beta m^(i) + nabla L(w^(i+1)),$$
where $beta$ is the momentum coefficient, $m^(i)$ is the momentum at iteration $i$, $L$ is the loss function, $w^(i)$ is the value of weights at iteration $i$.



If we are starting with $ m^(0) = 0$, then
$$ forall i > 0 , m^(i) = sum_j=0^i-1 beta^j nabla L(w^(i-j)).$$



Now, let's write down the formulas for exponentially weighted moving average of gradients (which we'll denote as $a^(i)$) to show that one is equivalent to the other multiplicated by a constant. We will make a non-traditional assumption that $a^(0) = 0$. It doesn't matter, because as $i$ goes to infinity, the contribution of the zeroth term goes to zero.



$$ a^(i+1) = beta a^(i) + (1-beta) nabla L(w^(i+1))$$
We can rewrite it as
$$ a^(i) = (1 - beta) sum_j=0^i-1 beta^j nabla L(w^(i-j)). $$



Notice that $ forall beta in [0, 1) $ it holds that $ (1 - beta) m^(i) = a^(i) $.



It seems to me that we should change the update equation of momentum SGD to the equation of exponentially weighted moving average of gradients, i.e. add the $ 1 - beta $ coefficient to the gradient term. Here's why:



  1. It decouples learning rate from momentum coefficient. Currently, larger momentum coefficient increases the effective learning rate (i.e. by how much the weights are updated). Suppose we are in an ideal scenario, when for all iterations $i, j$ we have $nabla L(w^(i)) = nabla L(w^(j)) = nabla L$, then $lim_i to infty m^(i) = fracnabla L1 - beta$. For $beta = 0.9$ this value equals $ 10 nabla L$, for $beta = 0.99$ this value equals $ 100 nabla L$. In contrast, if we use exponentially weighted moving average formula, for all $beta$ the analagous limit would equal just $nabla L$. I concede that this is an unrealistic scenario, and in real problems gradients at steps $i, i+1, i+2, dots, i+k$ somewhat cancel each other out, but still I think it's a good point.

  2. Weighted moving average is a somewhat well known concept, while momentum isn't.

I am interested to hear, what reasons are there not to change the update formula? And if you think this is a good change, how should the authors of deep learning libraries proceed?










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$endgroup$
















    0












    $begingroup$


    In Pytorch, the update equation of SGD with (non-Nesterov) momentum is
    $$ m^(i+1) = beta m^(i) + nabla L(w^(i+1)),$$
    where $beta$ is the momentum coefficient, $m^(i)$ is the momentum at iteration $i$, $L$ is the loss function, $w^(i)$ is the value of weights at iteration $i$.



    If we are starting with $ m^(0) = 0$, then
    $$ forall i > 0 , m^(i) = sum_j=0^i-1 beta^j nabla L(w^(i-j)).$$



    Now, let's write down the formulas for exponentially weighted moving average of gradients (which we'll denote as $a^(i)$) to show that one is equivalent to the other multiplicated by a constant. We will make a non-traditional assumption that $a^(0) = 0$. It doesn't matter, because as $i$ goes to infinity, the contribution of the zeroth term goes to zero.



    $$ a^(i+1) = beta a^(i) + (1-beta) nabla L(w^(i+1))$$
    We can rewrite it as
    $$ a^(i) = (1 - beta) sum_j=0^i-1 beta^j nabla L(w^(i-j)). $$



    Notice that $ forall beta in [0, 1) $ it holds that $ (1 - beta) m^(i) = a^(i) $.



    It seems to me that we should change the update equation of momentum SGD to the equation of exponentially weighted moving average of gradients, i.e. add the $ 1 - beta $ coefficient to the gradient term. Here's why:



    1. It decouples learning rate from momentum coefficient. Currently, larger momentum coefficient increases the effective learning rate (i.e. by how much the weights are updated). Suppose we are in an ideal scenario, when for all iterations $i, j$ we have $nabla L(w^(i)) = nabla L(w^(j)) = nabla L$, then $lim_i to infty m^(i) = fracnabla L1 - beta$. For $beta = 0.9$ this value equals $ 10 nabla L$, for $beta = 0.99$ this value equals $ 100 nabla L$. In contrast, if we use exponentially weighted moving average formula, for all $beta$ the analagous limit would equal just $nabla L$. I concede that this is an unrealistic scenario, and in real problems gradients at steps $i, i+1, i+2, dots, i+k$ somewhat cancel each other out, but still I think it's a good point.

    2. Weighted moving average is a somewhat well known concept, while momentum isn't.

    I am interested to hear, what reasons are there not to change the update formula? And if you think this is a good change, how should the authors of deep learning libraries proceed?










    share|improve this question







    New contributor




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







    $endgroup$














      0












      0








      0





      $begingroup$


      In Pytorch, the update equation of SGD with (non-Nesterov) momentum is
      $$ m^(i+1) = beta m^(i) + nabla L(w^(i+1)),$$
      where $beta$ is the momentum coefficient, $m^(i)$ is the momentum at iteration $i$, $L$ is the loss function, $w^(i)$ is the value of weights at iteration $i$.



      If we are starting with $ m^(0) = 0$, then
      $$ forall i > 0 , m^(i) = sum_j=0^i-1 beta^j nabla L(w^(i-j)).$$



      Now, let's write down the formulas for exponentially weighted moving average of gradients (which we'll denote as $a^(i)$) to show that one is equivalent to the other multiplicated by a constant. We will make a non-traditional assumption that $a^(0) = 0$. It doesn't matter, because as $i$ goes to infinity, the contribution of the zeroth term goes to zero.



      $$ a^(i+1) = beta a^(i) + (1-beta) nabla L(w^(i+1))$$
      We can rewrite it as
      $$ a^(i) = (1 - beta) sum_j=0^i-1 beta^j nabla L(w^(i-j)). $$



      Notice that $ forall beta in [0, 1) $ it holds that $ (1 - beta) m^(i) = a^(i) $.



      It seems to me that we should change the update equation of momentum SGD to the equation of exponentially weighted moving average of gradients, i.e. add the $ 1 - beta $ coefficient to the gradient term. Here's why:



      1. It decouples learning rate from momentum coefficient. Currently, larger momentum coefficient increases the effective learning rate (i.e. by how much the weights are updated). Suppose we are in an ideal scenario, when for all iterations $i, j$ we have $nabla L(w^(i)) = nabla L(w^(j)) = nabla L$, then $lim_i to infty m^(i) = fracnabla L1 - beta$. For $beta = 0.9$ this value equals $ 10 nabla L$, for $beta = 0.99$ this value equals $ 100 nabla L$. In contrast, if we use exponentially weighted moving average formula, for all $beta$ the analagous limit would equal just $nabla L$. I concede that this is an unrealistic scenario, and in real problems gradients at steps $i, i+1, i+2, dots, i+k$ somewhat cancel each other out, but still I think it's a good point.

      2. Weighted moving average is a somewhat well known concept, while momentum isn't.

      I am interested to hear, what reasons are there not to change the update formula? And if you think this is a good change, how should the authors of deep learning libraries proceed?










      share|improve this question







      New contributor




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







      $endgroup$




      In Pytorch, the update equation of SGD with (non-Nesterov) momentum is
      $$ m^(i+1) = beta m^(i) + nabla L(w^(i+1)),$$
      where $beta$ is the momentum coefficient, $m^(i)$ is the momentum at iteration $i$, $L$ is the loss function, $w^(i)$ is the value of weights at iteration $i$.



      If we are starting with $ m^(0) = 0$, then
      $$ forall i > 0 , m^(i) = sum_j=0^i-1 beta^j nabla L(w^(i-j)).$$



      Now, let's write down the formulas for exponentially weighted moving average of gradients (which we'll denote as $a^(i)$) to show that one is equivalent to the other multiplicated by a constant. We will make a non-traditional assumption that $a^(0) = 0$. It doesn't matter, because as $i$ goes to infinity, the contribution of the zeroth term goes to zero.



      $$ a^(i+1) = beta a^(i) + (1-beta) nabla L(w^(i+1))$$
      We can rewrite it as
      $$ a^(i) = (1 - beta) sum_j=0^i-1 beta^j nabla L(w^(i-j)). $$



      Notice that $ forall beta in [0, 1) $ it holds that $ (1 - beta) m^(i) = a^(i) $.



      It seems to me that we should change the update equation of momentum SGD to the equation of exponentially weighted moving average of gradients, i.e. add the $ 1 - beta $ coefficient to the gradient term. Here's why:



      1. It decouples learning rate from momentum coefficient. Currently, larger momentum coefficient increases the effective learning rate (i.e. by how much the weights are updated). Suppose we are in an ideal scenario, when for all iterations $i, j$ we have $nabla L(w^(i)) = nabla L(w^(j)) = nabla L$, then $lim_i to infty m^(i) = fracnabla L1 - beta$. For $beta = 0.9$ this value equals $ 10 nabla L$, for $beta = 0.99$ this value equals $ 100 nabla L$. In contrast, if we use exponentially weighted moving average formula, for all $beta$ the analagous limit would equal just $nabla L$. I concede that this is an unrealistic scenario, and in real problems gradients at steps $i, i+1, i+2, dots, i+k$ somewhat cancel each other out, but still I think it's a good point.

      2. Weighted moving average is a somewhat well known concept, while momentum isn't.

      I am interested to hear, what reasons are there not to change the update formula? And if you think this is a good change, how should the authors of deep learning libraries proceed?







      momentum






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      Check out our Code of Conduct.











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          ValueError: Expected n_neighbors <= n_samples, but n_samples = 1, n_neighbors = 6 (SMOTE) The 2019 Stack Overflow Developer Survey Results Are InCan SMOTE be applied over sequence of words (sentences)?ValueError when doing validation with random forestsSMOTE and multi class oversamplingLogic behind SMOTE-NC?ValueError: Error when checking target: expected dense_1 to have shape (7,) but got array with shape (1,)SmoteBoost: Should SMOTE be ran individually for each iteration/tree in the boosting?solving multi-class imbalance classification using smote and OSSUsing SMOTE for Synthetic Data generation to improve performance on unbalanced dataproblem of entry format for a simple model in KerasSVM SMOTE fit_resample() function runs forever with no result