# ML-Fundamentals - Bias Variance Tradeoff

## Introduction

If you completed the exercises simple-linear-regression, multivariate-linear-regression and logistic-linear-regression you know how to fit these models according to your training data.

This alone so far has no practical use case. The benefit of learning a model is to predict unseen data. Additionally, only with unseen data your model has not learnt from, it is possible to say if your model generalizes well or not. One way to measure this, is calculating the out of sample error $E_{out}$, which consists of the measures bias and variance.

In this notebook you will calculate two simple hypothesis for linear regression based on training data and compare them with the use of unseen validation data by calculating $E_{out}$, bias and variance.

## Requirements

### Knowledge

You should have a basic knowledge of:

• Univariate linear regression
• Out of sample error (bias variance)

Suitable sources for acquiring this knowledge are:

### Python Modules

By deep.TEACHING convention, all python modules needed to run the notebook are loaded centrally at the beginning.

import numpy as np
import matplotlib.pyplot as plt

import hashlib
def round_and_hash(value, precision=4, dtype=np.float32):
"""
Function to round and hash a scalar or numpy array of scalars.
Used to compare results with true solutions without spoiling the solution.
"""
rounded = np.array([value], dtype=dtype).round(decimals=precision)
hashed = hashlib.md5(rounded).hexdigest()
return hashed

## Exercise

Exercise inspired by lecture 8 from:

Implement the following simulation for calculating the bias and variance:

Given:

• $p(x)$ is unformly distributed in the interval of $[0, 2\pi]$
• The unknown target function is the sinus function: $t(x) = sin(x)$
• There is no noise if a $y^{i}$-value is drawn from $t(x^{(i)})$, so $y^{(i)}= t(x^{(i)})$

We consider two hypthesis sets:

• Hypthesis set $\mathcal H_1$: $h_1(x) = \theta_0 + \theta_1 x$
• Hypthesis set $\mathcal H_2$: $h_2(x) = w$ (a constant)

• Do the following 10.000 times:

• Draw two random examples $x^{(1)}$ and $x^{(2)}$ from $p(x)$ and calculate the corrsponding $y$s to get $\mathcal D = \{(x^{(1)},y^{(2)}), (x^{(1)},y^{(2)})\}$ (training data)

• Using your training data calculate the parameters $\theta_0, \theta_1$ for $\mathcal H_1$ and the parameter $w$ for $\mathcal H_2(x)$

• Numerically calculate the out of sample error $E_{out}$ for $\mathcal H_1(x)$ and $\mathcal H_2(x)$ for 100 data points uniformly distributed in the interval of $[0, 2\pi]$ (validation data)

• Now calculate the average $\theta_0, \theta_1$ and $w$ of all 10.000 experiments.

• Also calculate the average "out of sample error" $E_{out}$ for both hypothesis sets $\mathcal H_1$ and $\mathcal H_2$.

• Use the above to calculate the bias$^2$ and the variance.

• Plot the target function $sin(x)$ together with both average hypotheses $\tilde h_1(x)$ and $\tilde h_2(x)$ using the average $\theta_0, \theta_1$ and $w$

• Considering your results, which hypothesis seems to better model the target function?

Practically this explanation is all you need to solve the exercise. You are free to complete it without any further guiding or by proceeding with this notebook.

### Data Generation

Implement the function to draw two random training examples $(x^{(i)},y^{(i)})$ with:

• $x^{(i)} \in Uniform(0,2\pi)$
• $i \in \{1,2\}$
• $y^{(i)} = sin(x^{(i)})$
def train_data():
raise NotImplementedError()
x_train, y_train  = train_data()
print(x_train, y_train)
# If your implementation is correct, these tests should not throw an exception

assert len(x_train) == 2
assert len(y_train) == 2
np.testing.assert_array_equal(np.sin(x_train), y_train)
for i in range(1000):
x_tmp, _ = train_data()
assert x_tmp.min() >= 0.0
assert x_tmp.max() <= 2*np.pi

### Hypothesis

For our training data we will now model two different hypothesis sets:

$\mathcal H_1: h_1(x) = \theta_0 + \theta_1 x$

and

$\mathcal H_2: h_2(x) = w$

Implement the functions to calculate the parameters $\theta_0, \theta_1$ for $h_1$ and $w$ for $h_2$ using the two drawn examples.

For later purpose (passing functions as argument) it is important that both functions accept the same amount of parameters and also return the same amount. Therefore we also pass $x$ to get_w, although we do not need it. And for the same reason get_thetas should return a list of two values instead of two seperate values.

def get_thetas(x, y):
raise NotImplementedError()

def get_w(x, y):
raise NotImplementedError()
thetas = get_thetas(x_train, y_train)
w = get_w(x_train, y_train)
print(thetas[0], thetas[1])
print(w)
# If your implementation is correct, these tests should not throw an exception

x_train_temp = np.array([0,1])
y_train_temp = np.array([np.sin(x_i) for x_i in x_train_temp])
thetas_test = get_thetas(x_train_temp, y_train_temp)
w_test = get_w(x_train_temp, y_train_temp)

np.testing.assert_almost_equal(thetas_test[0], 0.0)
np.testing.assert_almost_equal(thetas_test[1], 0.8414709848078965)
np.testing.assert_almost_equal(w_test, 0.42073549240394825)

Implement the hypothesis $h_1(x)$ and $h_2(x)$. Your function should return a function.

def get_hypothesis_1(thetas):
raise NotImplementedError()

def get_hypothesis_2(w):
raise NotImplementedError()
# we want to compute numerically the expectation w.r.t. x
x_grid = np.linspace(0, 2*np.pi, 100)
y_grid = np.sin(x_grid)
# If your implementation is correct, these tests should not throw an exception

h1_test = get_hypothesis_1(thetas_test)
h2_test = get_hypothesis_2(w_test)
np.testing.assert_almost_equal(h1_test(x_grid)[10], 0.5340523361780719)
np.testing.assert_almost_equal(h2_test(x_grid)[10], 0.42073549240394825)

### Plot

Following the original exercise it is not yet necessary to plot anything. But it also does not hurt to do so, since we need to implement code for the plot anyways.

Write the function to plot:

• the two examples $(x^{(1)},y^{(2)})$ and $(x^{(2)},y^{(2)})$
• the true target function $sin(x)$ in the interval $[0, 2 \pi]$.
• the hypothesis $h_1(x)$ in the interval $[0, 2 \pi]$
• the hypothesis $h_2(x)$ in the interval $[0, 2 \pi]$

Your plot should look similar to this one:

def plot_true_target_function_x_y_h1_h2(x, y, hypothesis1, hypothesis2):
raise NotImplementedError()
thetas = get_thetas(x_train, y_train)
w = get_w(x_train, y_train)
plot_true_target_function_x_y_h1_h2(x_train, y_train, get_hypothesis_1(thetas), get_hypothesis_2(w))

### Out of Sample Error

The out of sample error $E_{out}(h)$ is the expected error on new unseen data.

$E_{out}(h) = \mathbb E_{x,y}[loss(h(x), y)] = \int_{\mathcal X \times \mathcal Y} loss(h(x), y) p(x,y) dx dy$

In our example we don't need to take the expectation w.r.t. $y$ because we have have no noise:

$E_{out}(h) = \mathbb E_{x}[loss(h(x), t(x))] = \int_{\mathcal X } loss(h(x), t(x)) p(x) dx$

Here we will compute the out of sample error $E_{out}(h)$ numerically.

We already have discretized the $x$-axis (x_grid) for $p(x)\neq 0$.
So to compute the expectation we just need to average over x_grid (remember $p(x)$ is uniform).

$E_{out}(h) = \mathbb E_{x}[loss(h(x),t(x))] \approx \frac{1}{m} \sum_{j=1}^{m} loss(h(x^{(j)}), y^{(j)})$

with

• $m$: number of elements in x_grid
• $x^{(j)}$ is the $j$-element of x_grid

Implement the function to numerically calculate the out of sample error $E_{out}$ with the mean squared error as loss function.

def out_of_sample_error(y_preds, y):
raise NotImplementedError()
# If your implementation is correct, these tests should not throw an exception

e_out_h1_test = out_of_sample_error(h1_test(x_grid), y_grid)
np.testing.assert_almost_equal(e_out_h1_test, 11.525485917588728)

### Repeat

Now instead of drawing two examples (one training data set), draw now 10.000 times different training sets with two examples.
Calculate $E_{out}$ for the differnt $h_1$ and $h_2$ numerically.

For each run, keep track of the following parameters and return them at the end of the function:

• $\{x^{(1)},x^{(2)}\}$
• $\{y^{(1)},y^{(2)}\}$
• $\theta_0$
• $\theta_1$
• $w$
• $E_{out}$
def run_experiment(m, x_val, y_val):
raise NotImplementedError()
xs, ys, t0s, t1s, ws, e_out_h1s, e_out_h2s = run_experiment(
10000, x_grid, y_grid)

### Average and Plot

Now we can calculate the average of $\theta_0, \theta_1$, $w$ and $E_{out}$ and already plot the resulting averaged $\tilde h_1(x)$ and $\tilde h_2(x)$ together with the target function $sin(x)$.

Your plot should look similar to the one below:

t0_avg = t0s.mean()
t1_avg = t1s.mean()
thetas_avg = [t0_avg, t1_avg]
w_avg = ws.mean()
h1_avg = get_hypothesis_1(thetas_avg)
h2_avg = get_hypothesis_2(w_avg)
print(thetas_avg)
plot_true_target_function_x_y_h1_h2([], [], h1_avg, h2_avg)
expectation_Eout_1 = e_out_h1s.mean()
print ("expectation of E_out of model 1:", expectation_Eout_1)
expectation_Eout_2 = e_out_h2s.mean()
print ("expectation of E_out of model 2:", expectation_Eout_2)

### Bias

The bias for the mean-squared error is:

$bias^2 = \mathbb E_{x,y} \left[(\tilde h(x) - y)^2\right]$

The expectation w.r.t. $y$ vanishes if we have no noise:

$bias^2 = \mathbb E_x \left[(\tilde h(x) - t(x))^2\right]$

with:

• the average hypothesis $\tilde h(x)$

Implement the function to calculate the expecation of the bias numerically.

def bias(y_true, y_predicted):
raise NotImplementedError()
bias_1 = bias(y_grid,  h1_avg(x_grid))
print ("Bias of model 1:", bias_1)
bias_2 = bias(y_grid,  h2_avg(x_grid))
print ("Bias of model 2:", bias_2)

### Variance

Variance for the mean-squared-error:

$variance = \mathbb E_x \left[ \mathbb E_D \left[\left(h^D(x) - \tilde h(x)\right)^2 \right] \right]$

with:

• the average hypothesis $\tilde h(x)$
• the learnt hypothesis $h^D(x)$ for training data set $D$.

Implement the function to calculate the variances for each of the 10.000 experiements and return them as list or array.

Now we benefit from our implementation of get_w, get_thetas, respectively get_hypothesis1,get_hypothesis2, which accept and return the same amount of parameters, so we can write a generalized function.

thetas_avg = [0.17, 0.68]
def variances(hypothesis_func, param_func, xs, ys, x_val, y_preds):
return NotImplementedError()
vars_1 = variances(get_hypothesis_1,
get_thetas,
xs, ys,
x_grid,
h1_avg(x_grid))
var_1_avg = vars_1.mean()
print(var_1_avg)
vars_2 = variances(get_hypothesis_2,
get_w,
xs, ys,
x_grid,
h2_avg(x_grid)).mean()
var_2_avg = vars_2.mean()
print(var_2_avg)
print("model 1: E_out ≈ bias^2 + variance:  %f ≈ %f + %f" % (expectation_Eout_1, bias_1, var_1_avg))
print("model 2: E_out ≈ bias^2 + variance:  %f ≈ %f + %f" % (expectation_Eout_2, bias_2, var_2_avg))

## Summary and Outlook

[TODO]

### Notebook License (CC-BY-SA 4.0)

The following license applies to the complete notebook, including code cells. It does however not apply to any referenced external media (e.g., images).

Exercise: Bias Variance Tradeoff
by Christian Herta, Klaus Strohmenger