2023-03-22 Beyond Linear Models¶

Last time¶

• Discussion
• Bias-variance tradeoff
• Linear models

Today¶

• Assumptions of linear models
• Look at your data!
• Partial derivatives
• Loss functions

Bias-variance tradeoff¶

The expected error in our approximation $\hat f(x)$ of noisy data $y = f(x) + \epsilon$ (with $\epsilon \sim \mathcal N(0, \sigma)$), can be decomposed as $$ E[(\hat f(x) - y)^2] = \sigma^2 + \big(\underbrace{E[\hat f(x)] - f(x)}_{\text{Bias}}\big)^2 + \underbrace{E[\hat f(x)^2] - E[\hat f(x)]^2}_{\text{Variance}} . $$ The $\sigma^2$ term is irreducible error (purely due to observation noise), but bias and variance can be controlled by model selection. More complex models are more capable of expressing the underlying function $f(x)$, thus are capable of reducing bias. However, they are also more affected by noise, thereby increasing variance.

Stacking many realizations¶

Interpretation¶

• Re-run the cell above for different values of degree. (Set it back to a number around 7 to 10 before moving on.)
• Low-degree polynomials are not rich enough to capture the peak of the function.
• As we increase degree, we are able to resolve the peak better, but see more eratic behavior near the ends of the interval. This erratic behavior is overfitting, which we'll quantify as variance.
• This tradeoff is fundamental: richer function spaces are more capable of approximating the functions we want, but they are more easily distracted by noise.

Mean and variance over the realizations¶

Why do we call it a linear model?¶

We are currently working with algorithms that express the regression as a linear function of the model parameters. That is, we search for coefficients $c = [c_1, c_2, \dotsc]^T$ such that

$$ V(x) c \approx y $$

where the left hand side is linear in $c$. In different notation, we are searching for a predictive model

$$ f(x_i, c) \approx y_i \text{ for all $(x_i, y_i)$} $$

that is linear in $c$.

Standard assumptions for regression¶

(the way we've been doing it so far)¶

1. The independent variables $x$ are error-free
2. The prediction (or "response") $f(x,c)$ is linear in $c$
3. The noise in the measurements $y$ is independent (uncorrelated)
4. The noise in the measurements $y$ has constant variance

There are reasons why all of these assumptions may be undesirable in practice, thus leading to more complicated methods.

Anscombe's quartet¶

Loss functions¶

The error in a single prediction $f(x_i,c)$ of an observation $(x_i, y_i)$ is often measured as $$ \frac 1 2 \big( f(x_i, c) - y_i \big)^2, $$ which turns out to have a statistical interpretation when the noise is normally distributed. It is natural to define the error over the entire data set as \begin{align} L(c; x, y) &= \sum_i \frac 1 2 \big( f(x_i, c) - y_i \big)^2 \\ &= \frac 1 2 \lVert f(x, c) - y \rVert^2 \end{align} where I've used the notation $f(x,c)$ to mean the vector resulting from gathering all of the outputs $f(x_i, c)$. The function $L$ is called the "loss function" and is the key to relaxing the above assumptions.

Gradient of scalar-valued function¶

Let's step back from optimization and consider how to differentiate a function of several variables. Let $f(\boldsymbol x)$ be a function of a vector $\boldsymbol x$. For example,

$$ f(\boldsymbol x) = x_1^2 + \sin(x_2) e^{3x_3} . $$

We can evaluate the partial derivative by differentiating with respect to each component $x_i$ separately (holding the others constant), and collect the result in a vector,

\begin{align} \frac{\partial f}{\partial \boldsymbol x} &= \begin{bmatrix} \frac{\partial f}{\partial x_1} & \frac{\partial f}{\partial x_2} & \frac{\partial f}{\partial x_3} \end{bmatrix} \\ &= \begin{bmatrix} 2 x_1 & \cos(x_2) e^{3 x_3} & 3 \sin(x_2) e^{3 x_3} \end{bmatrix}. \end{align}

Gradient of vector-valued functions¶

Now let's consider a vector-valued function $\boldsymbol f(\boldsymbol x)$, e.g.,

$$ \boldsymbol f(\boldsymbol x) = \begin{bmatrix} x_1^2 + \sin(x_2) e^{3x_3} \\ x_1 x_2^2 / x_3 \end{bmatrix} . $$

and write the derivative as a matrix,

\begin{align} \frac{\partial \boldsymbol f}{\partial \boldsymbol x} &= \begin{bmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \frac{\partial f_1}{\partial x_3} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \frac{\partial f_2}{\partial x_3} \\ \end{bmatrix} \\ &= \begin{bmatrix} 2 x_1 & \cos(x_2) e^{3 x_3} & 3 \sin(x_2) e^{3 x_3} \\ x_2^2 / x_3 & 2 x_1 x_2 / x_3 & -x_1 x_2^2 / x_3^2 \end{bmatrix}. \end{align}

Geometry of partial derivatives¶

• Handy resource on partial derivatives for matrices and vectors: https://explained.ai/matrix-calculus/index.html#sec3

Derivative of a dot product¶

Let $f(\boldsymbol x) = \boldsymbol y^T \boldsymbol x = \sum_i y_i x_i$ and compute the derivative

$$ \frac{\partial f}{\partial \boldsymbol x} = \begin{bmatrix} y_1 & y_2 & \dotsb \end{bmatrix} = \boldsymbol y^T . $$

Note that $\boldsymbol y^T \boldsymbol x = \boldsymbol x^T \boldsymbol y$ and we have the product rule,

$$ \frac{\partial \lVert \boldsymbol x \rVert^2}{\partial \boldsymbol x} = \frac{\partial \boldsymbol x^T \boldsymbol x}{\partial \boldsymbol x} = 2 \boldsymbol x^T . $$

Also, $$ \frac{\partial \lVert \boldsymbol x - \boldsymbol y \rVert^2}{\partial \boldsymbol x} = \frac{\partial (\boldsymbol x - \boldsymbol y)^T (\boldsymbol x - \boldsymbol y)}{\partial \boldsymbol x} = 2 (\boldsymbol x - \boldsymbol y)^T .$$

Variational notation¶

It's convenient to express derivatives in terms of how they act on an infinitessimal perturbation. So we might write

$$ \delta f = \frac{\partial f}{\partial x} \delta x .$$

(It's common to use $\delta x$ or $dx$ for these infinitesimals.) This makes inner products look like a normal product rule

$$ \delta(\mathbf x^T \mathbf y) = (\delta \mathbf x)^T \mathbf y + \mathbf x^T (\delta \mathbf y). $$

A powerful example of variational notation is differentiating a matrix inverse

$$ 0 = \delta I = \delta(A^{-1} A) = (\delta A^{-1}) A + A^{-1} (\delta A) $$

and thus $$ \delta A^{-1} = - A^{-1} (\delta A) A^{-1} $$

Practice¶

1. Differentiate $f(x) = A x$ with respect to $x$
2. Differentiate $f(A) = A x$ with respect to $A$

Optimization¶

Given data $(x,y)$ and loss function $L(c; x,y)$, we wish to find the coefficients $c$ that minimize the loss, thus yielding the "best predictor" (in a sense that can be made statistically precise). I.e., $$ \bar c = \arg\min_c L(c; x,y) . $$

It is usually desirable to design models such that the loss function is differentiable with respect to the coefficients $c$, because this allows the use of more efficient optimization methods. Recall that our forward model is given in terms of the Vandermonde matrix,

$$ f(x, c) = V(x) c $$

and thus

$$ \frac{\partial f}{\partial c} = V(x) . $$

Derivative of loss function¶

We can now differentiate our loss function $$ L(c; x, y) = \frac 1 2 \lVert f(x, c) - y \rVert^2 = \frac 1 2 \sum_i (f(x_i,c) - y_i)^2 $$ term-by-term as \begin{align} \nabla_c L(c; x,y) = \frac{\partial L(c; x,y)}{\partial c} &= \sum_i \big( f(x_i, c) - y_i \big) \frac{\partial f(x_i, c)}{\partial c} \\ &= \sum_i \big( f(x_i, c) - y_i \big) V(x_i) \end{align} where $V(x_i)$ is the $i$th row of $V(x)$.

Alternative derivative¶

Alternatively, we can take a more linear algebraic approach to write the same expression is \begin{align} \nabla_c L(c; x,y) &= \big( f(x,c) - y \big)^T V(x) \\ &= \big(V(x) c - y \big)^T V(x) \\ &= V(x)^T \big( V(x) c - y \big) . \end{align} A necessary condition for the loss function to be minimized is that $\nabla_c L(c; x,y) = 0$.

• Is the condition sufficient for general $f(x, c)$?
• Is the condition sufficient for the linear model $f(x,c) = V(x) c$?
• Have we seen this sort of equation before?