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 for example in case of decision tree, which beyond some size will achieve zero train loss, while test loss (generalization error) will rise.

But in general, bias-variance trade-off is not applicable. 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 optimiser 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 represent in the SGD algorithm.

# 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.

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.