Double Descent Contrary to Bias-Variance Trade-Off

Increasing parameter count leads to multiple test loss peaks and a global minima in the overparameterized regime.
Double Descent Contrary to Bias-Variance Trade-Off
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The bias-variance trade-off hypothesis implies that lowering train loss by increasing model size will lead to higher test loss. Empirically this can be observed in decision trees, which beyond some size will achieve zero train loss, while test loss (generalization error) will rise. See the bias-variance trade-off case of decision tree pruning controlled by alpha in below.

decision tree cost complexity pruning improves test accuracy until a maximum, scikit docs
decision tree cost complexity pruning improves test accuracy until a maximum, scikit docs

But in general, bias-variance trade-off is not applicable in terms of parameter count (See model norm definition below). Any kind of regularization of the optimizer, which is part of the model, will force model to look for “simple” solution (Occam’s razor), despite having capacity to fully fit the training data. Behaviour of the optimizer can have an impact on the resulting test loss (generalization error). For example early stopping is a regularization. Not having any explicit regularization, doesn’t imply that bias-variance tradeoff will be applicable. Belkin 2019 paper has an image, displayed also below, which shows that bias-variance trade-off is not applicable to fully connected NN on MNIST without regularization and early stopping. The paper refers to empirical evidence that an implicit regularization is present in the SGD algorithm.

Double Descent for fully connected NN on MNIST.
Double Descent for fully connected NN on MNIST without regularization and early stopping. Source: Belkin 2019

More general work on double descent and large datasets from OpenAI here.

Model Norm vs Parameter Count and Double Descent

But is measuring the model capacity in terms of parameter count the best? What if the model’s optimizer thanks to the regularization learns to just zero out some weights? We should be able to see this in terms of model norm, which is a norm of all model’s parameters. Machine Learning CS229 lecture notes plot a case where, the double descent disappears, when we use norm of the model’s parameters instead of their count.

double descent and weights norm model norm  Ng 2022
double descent and weights norm model norm Ng 2022


Open AI also observed double descent in Transformers, which they called “grokking”. In the grokking setup, OpenAi trained on a binary operations datasets, and observed sudden jumps in accuracy (inverse of error), as if the model suddenly found the right algorithm far in the overfitting regime despite prolonged error stagnation.

OpenAI grokking
OpenAI grokking

Additionally, OmniGrok 2022 paper connects grokking to norm of the model’s weights and the tasks dependency on good representations.

Our conclusions are: (i) grokking originates from the mismatch between training and test losses ("LU" mechanism). (ii) grokking can happen in various models for a wide range of datasets, although the grokking signature is usually most dramatic for algorithmic datasets. (iii) The dramaticness of grokking depends on how much the task relies on learning representations. This work not only reveals the mechanism of grokking, but also shows that reduced landscape analysis is a useful tool for characterizing data-model interaction and representation learning.

Bias–variance decomposition of mean squared error

In case of L2 norm, the train loss can be rewritten as a sum of bias term and variance term. However that doesn’t prove presence of any dilemma.

  • A prediction function \( f \) was fitted on the training data \( D = \lbrace (x_1, y_1), (x_2, y_2), …, (x_n, y_n) \rbrace \).
  • A single the test sample \( (x, y) \) has a noisy label \( y = Ey + \varepsilon \), but the mean label \( Ey \) is the true value.
  • Because the test label noise is independent of the training data, the fitted function has zero covariance with the test label noise: \( E \varepsilon f = 0 \).

Then the expected L2 test loss \( E(y-f)^2 \) below is conditioned on a test sample \( x \). The expectation is calculated over all draws of train samples \( D \) and label noise \( \varepsilon \).

\( E(y-f)^2 \)

\( = E(y - Ey + Ey - f)^2 \)

\( = E (Ey - f)^2 + 2 E \varepsilon (Ey - f) + \sigma^2\)

\( = E (Ey - f)^2 + \sigma^2 \)

\( = (Ey - Ef)^2 + E(Ef - f)^2 + \sigma^2 \).

Terms in above equation in given order represent:

Term Description Name
$$ (Ey - Ef)^2 $$ The model's ability to fit the true training data. bias
$$ E(Ef - f)^2 $$ the model's ability to resist the training label noise, predict mostly the same outputs on same inputs, and thus generalize variance
$$ \sigma^2 $$ the test label noise irreducible error

However, above decomposition does not explicitly prove how the individual components behave. So, it is no prove of the proposed dilemma.

What is bias variance decomposition in case of linear regression?

There is a “Deriving the final identity” section in “Linear Regression and the Bias Variance Tradeoff”, which attempts an estimate of the variance term, but I think it contains a mistake in the calculation.

The estimate is done for a linear regression problem with following simplifications:

  • training data \( D = \lbrace (x_1, y_1), …, (x_n, y_n) \rbrace \),
  • an dimension of the training data: \( p \),
  • identity matrix of dimension \( p \): \( I_p \),
  • train set \( x \) is normally distributed without covariance: \( x \in N(0, I_p \sigma_{x-train}) \),
  • train set \( y \) is normally distributed, single dimensional: \( y(x) \in N(Ey, \sigma_y) \).
  • test set \( x_t \in N(0, I_p \sigma_{x-test}) \)

I expect that variance for above simplified linear regression problem will increase if we:

  • increase the problem dimension \( p \)
  • decrease the training sample count \( n \)
  • increase the training label variance \( \sigma_y \) (this is proved in above pdf)
  • decrease the training sample variance \( \sigma_{x_{train}} \) = poor training sample
  • increase the test sample variance \( \sigma_{x_{test}} \) = unfamiliar test data

Unfortunately, I don’t have the direct proof.

What is an overparameterized model?

Usually model is called overparametrized when the number of parameters is greater than the number of training samples. It can also mean that model has capacity to achieve the test loss zero, that is to interpolate the train data. Modern ML models are over-parametrized, but use various regularization methods.

What is a generalization curve?

It is defined as the test loss as a function of number of parameters of the model.

Multiple Descent Proof for Linear Regression

In the Multiple Descent paper, the test loss as a function of number of parameters of special case of linear regression is highly controlled for in both under-parametrized and over-parametrized regimes.

Authors prove:

  • Existence of multiple descent in the generalization curve.
  • That the number of descents can be designed.

How is the multiple descent achieved?

The problem is linear regression without regularization with true linear model of zero. Where all labels contain an error \( y_i = 0 + \varepsilon_i \), which is normally distributed around zero.

However, the features have all different distribution. The authors select them, such that they can control the generalization curve. Each feature \( x_i \) is distributed either as a Gaussian mixture \( N^{mix}_{\sigma_1, 1} \) or as a standard Gaussian \( N(0, \sigma) \) . The data dimension is the feature count.

The model parameter count is increased by adding new features and corresponding labels to the data. Note that for given generated data and labels, the optimal linear model can be calculated in a single step. Authors then draw generalization curve as average test-loss normalized by the feature count (dimension) as a function of feature count.

Double Descent for fully connected NN on MNIST.
This image is wrong: \( N_{mix} \) should be first, then \( N_0 \). Multiple Descent generalization curve (dimension normalized test loss as a function of number of parameters). Source.

There is an bug in the generalization curve image above. The paper proves that the generalization curve can be pushed up if we introduce a gaussian mixture feature right after a standard gaussian. That makes sense, as gaussian mixture is easier to “mis-learn” than standard gaussian.

While this is completely artificial model, it still interesting that we can have arbitrary number of descents contrary to the bias-variance dilemma.

Created on 15 Sep 2020. Updated on: 16 Oct 2022.
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