EE 541 - Unit 3
Dr. Brandon Franzke
Fall 2026
Real systems operate with incomplete information:
Objective: Given measurements (\(X\)), estimate unknowns (\(Y\))
Uncertainty is inherent, not a nuisance
Many estimates of \(Y\) are possible - best needs a definition
Problem:
Predictor function: \(g: \mathbb{R} \to \mathbb{R}\)
Example: Room temperature estimation
Among all possible functions \(g\), which one is “best”?
Need:
Loss function \(\ell(y, \hat{y})\) measures prediction quality
Common choices:
| Loss | Formula | Properties |
|---|---|---|
| Squared | \((y - \hat{y})^2\) | Differentiable, convex, emphasizes large errors |
| Absolute | \(|y - \hat{y}|\) | Robust to outliers, non-differentiable at 0 |
| 0-1 | \(\mathbb{1}[y \neq \hat{y}]\) | Classification, discontinuous |
| Huber | \(\begin{cases} \frac{1}{2}(y-\hat{y})^2 & |y-\hat{y}| \leq \delta \\ \delta|y-\hat{y}| - \frac{\delta^2}{2} & \text{otherwise} \end{cases}\) | Robust + differentiable |
Risk: Expected loss over joint distribution
Goal: Find \(g^* = \arg\min_g R(g)\)
Squared error leads to the expected value (mean), absolute error the median
Mathematical advantages:
Differentiable:
Smooth objective
Unique minimum: Strictly convex \(\Rightarrow\) no local minima
Closed-form solutions: Often analytically tractable
Decomposition:
Separates systematic error from random scatter
Statistical connection:
Under Gaussian noise \(N \sim \mathcal{N}(0, \sigma^2)\):
Maximum likelihood estimation:
MSE \(\leftrightarrow\) Gaussian MLE
Definition: for estimator \(g\)
Expanding the expectation:
Empirical MSE (from data):
Properties:
Other forms:
Hierarchy of function classes:
Constant: \(\hat{Y} = c\)
Linear (affine): \(\hat{Y} = aX + b\)
Polynomial: \(\hat{Y} = \sum_{k=0}^p a_k X^k\)
Unrestricted: \(\hat{Y} = g(X)\)
Each step up fits more but requires knowing more
Definition: A linear estimator has the form
where \(a, b \in \mathbb{R}\) are constants.
Parameters:
Special cases:
Affine, strictly: superposition holds only for \(b = 0\)
Vector case: \(\hat{\mathbf{y}} = \mathbf{A}\mathbf{x} + \mathbf{b}\)
Mean \(\mathbb{E}[Y]\): baseline prediction
Variance \(\text{Var}(Y)\): the error to beat
Covariance \(\text{Cov}(X,Y)\): linear link
Correlation \(\rho \in [-1, 1]\): scale-free link
These moments completely determine the optimal linear estimator
Additive noise model:
Signal-to-Noise Ratio (SNR):
For zero-mean:
In dB: \(\text{SNR}_{\text{dB}} = 10\log_{10}(\text{SNR})\)
Linear estimation floor:
Sample space \(\Omega\): Set of all possible outcomes
Random variable \(X: \Omega \to \mathbb{R}\)
Maps each outcome to a real number.
Notation convention:
Discrete example: Die roll
Continuous example: Measurement noise
Joint distribution \(p(x, y)\):
Probability (density) that \(X = x\) AND \(Y = y\) simultaneously.
For discrete: \(P(X = x, Y = y)\)
For continuous: \(p(x,y)\) where
What the joint determines:
Independence means:
Joint factors into product of marginals.
Marginal distribution: Integrate out the unused variable
Conditional distribution: Slice and renormalize
Bayes’ rule is just rearranging:
Conditioning is just:
Discrete:
Continuous:
Interpretation: Center of mass of the distribution
Expectation of a function:
Example: Die roll
\(\mathbb{E}[X] = 1 \cdot \frac{1}{6} + 2 \cdot \frac{1}{6} + \cdots + 6 \cdot \frac{1}{6} = 3.5\)
3.5 is not a possible outcome - it is the long-run average
Linearity:
This holds even if \(X\) and \(Y\) are dependent.
Why? Just rearranging sums/integrals:
Example: Sum of two dice
\(\mathbb{E}[X_1 + X_2] = \mathbb{E}[X_1] + \mathbb{E}[X_2] = 3.5 + 3.5 = 7\)
i.e., Don’t need to enumerate all 36 outcomes.
Definition:
where \(\mu = \mathbb{E}[X]\)
Equivalent form:
When to use which:
Standard deviation: \(\sigma = \sqrt{\text{Var}(X)}\)
Same units as \(X\) (variance has squared units)
Properties:
Definition:
Equivalent form:
Interpretation:
Correlation (normalized covariance):
\(\text{Cov}(X,Y) = 0\) does NOT imply independence.
\(X \sim \mathcal{N}(0,1)\) with \(Y = X^2\): clearly dependent, but \(\text{Cov}(X, Y) = 0\).
Exception: jointly Gaussian \((X,Y)\) - zero covariance implies independence
For two random variables:
If uncorrelated (\(\text{Cov}(X,Y) = 0\)):
Compare with (linear) expectation:
General form:
Appears in portfolio diversification, error propagation, averaged measurements
Joint distribution \((X, Y) \sim p(x,y)\) contains all information.
Bayes’ rule:
Once we observe \(X = x\), the distribution of \(Y\) changes from \(p(y)\) to \(p(y|x)\).
Example: Temperature sensor, \(X = Y + N\)
Conditioning reduces uncertainty on average: \(\mathbb{E}[\text{Var}(Y|X)] \leq \text{Var}(Y)\) - a particular observation can widen the spread
Discrete case:
Definition: The conditional expectation of \(Y\) given \(X = x\) is:
Discrete case:
Continuous case:
Facts:
Example: \(Y = X + N\), \(N \sim \mathcal{N}(0,1)\), independent \(X\) and \(N\)
Example: Binary communication channel
Joint distribution:
Conditional distributions:
Conditional expectations:
MSE for different predictors:
Result: \(\mathbb{E}[X|Y]\) achieves minimum MSE
Joint Gaussian: \((X, Y) \sim \mathcal{N}(\mu, K)\)
Parameters:
Conditional mean:
Conditional variance:
Linear in \(x\), spread independent of \(x\)
Figure: \(\mu = (1, 2)\), \(\sigma_X = 2\), \(\sigma_Y = 3\); at \(\rho = 0.8\) the slope is \(1.2\) and the spread \(1.8\), at \(\rho = 0\) the slope is \(0\) and the spread the full \(3\)
Computing \(\mathbb{E}[Y|X=x]\) requires \(p(y|x)\)
From Bayes:
When tractable:
When intractable (often no closed form):
Example: Nonlinear sensor
where \(N \sim \mathcal{N}(0, 0.1)\)
Computing \(\mathbb{E}[Y|X=x]\) requires:
Law of Total Expectation:
Proof:
where \(g(x) = \mathbb{E}[Y|X=x]\)
Intuition: Average of conditional averages = unconditional average
Example: Class grades
Property: For any function \(h(X)\):
Proof: Fix \(X = x\). Then \(h(X) = h(x)\) is a constant:
Special cases:
Example: Signal processing
Only \(\mathbb{E}[Y|X]\) is needed, not \(\mathbb{E}[XY|X]\)
But functions of \(Y\) do not factor out:
Definition:
Computing formula:
Measures remaining uncertainty after observing \(X\)
Law of Total Variance:
Interpretation:
Minimum MSE: \(\mathbb{E}[\text{Var}(Y|X)]\), and no predictor does better
Goal: Find the best predictor of \(Y\) given \(X\)
Function space:
Optimization problem:
Infinite-dimensional optimization
Strategy: Decompose the problem by conditioning on \(X\)
Decomposition:
Why this helps:
When \(X = x\) is fixed:
This transforms the problem:
Solution approach:
Step 1: Express MSE using conditional expectation
Step 2: For fixed \(X = x\), \(g(X) = g(x)\) is a constant
where \(a = g(x)\)
Step 3: Minimize over \(a\)
Solution: \(a^* = \mathbb{E}[Y|X=x]\)
Step 4: Since this holds for all \(x\):
Therefore:
Orthogonality principle: Optimal error orthogonal to any function of \(X\):
for any measurable \(h\).
Uniqueness: Only one function satisfies this \(\Rightarrow\) unique MMSE
Optimality test: If \(\mathbb{E}[\epsilon \cdot h(X)] \neq 0\), can reduce MSE by adjusting estimate in direction of \(h(X)\)
Computational foundation:
Geometric meaning: MMSE = projection in Hilbert space
Proof: \(\mathbb{E}[(Y - \mathbb{E}[Y|X])|X] = 0\) by definition, multiplying by \(h(X)\) preserves this.
Cannot extract more information from \(X\)
We solved pointwise: For each \(x\), found \(g^*(x) = \mathbb{E}[Y|X=x]\)
Question: Is this globally optimal over all functions?
Two concerns:
MSE decomposes additively
Each term \(\mathbb{E}[(Y - g(x))^2|X=x]\) depends only on \(g(x)\), not on \(g(x')\) for \(x' \neq x\).
Therefore:
Idea: Prediction at \(x\) doesn’t help prediction at \(x'\)
Claim: No function can beat \(\mathbb{E}[Y|X]\)
Proof: For any function \(g\):
Add and subtract \(\mathbb{E}[Y|X]\):
Expand the square:
The cross term vanishes:
Because \(\mathbb{E}[Y|X] - g(X)\) is a function of \(X\) and:
for any \(h(X)\) (orthogonality).
Therefore:
Minimum achievable MSE:
Result:
Proof:
For fixed \(X = x\):
Therefore:
Irreducible: no predictor can remove conditional variance
Special cases:
A single outlier can dominate the sample MSE:
If \(y_k\) is an outlier: \((y_k - \hat{y}_k)^2 \gg (y_i - \hat{y}_i)^2\) for \(i \neq k\)
Effect on the estimate:
Example: Sensor failure
The mean absorbs the spike; typical-case error grows.
Alternative loss functions:
When outliers are present:
Median regression (LAD):
Solution: \(g^*(x) = \text{median}(Y|X=x)\)
Huber regression:
where \(\ell_\delta(e) = \begin{cases} \frac{1}{2}e^2 & |e| \leq \delta \\ \delta|e| - \frac{\delta^2}{2} & |e| > \delta \end{cases}\)
Trimmed mean: Remove top/bottom α% before computing \(\mathbb{E}[Y|X]\)
Definition: For vectors \(\mathbf{x}, \mathbf{y} \in \mathbb{R}^n\)
Geometric interpretation:
where \(\theta\) is the angle between vectors.
Special cases:
Norm from inner product:
Problem: Given \(\mathbf{y}\) and subspace \(\mathcal{S}\), find \(\hat{\mathbf{y}} \in \mathcal{S}\) that minimizes
Solution: \(\hat{\mathbf{y}} = \text{Proj}_{\mathcal{S}}(\mathbf{y})\)
Projection onto a line (1D subspace):
Given direction \(\mathbf{u}\) with \(\|\mathbf{u}\| = 1\):
Projection onto span of \(\{\mathbf{u}_1, \mathbf{u}_2\}\):
If \(\mathbf{u}_1 \perp \mathbf{u}_2\) and both unit vectors:
The error is what remains:
Theorem: \(\hat{\mathbf{y}}\) minimizes \(\|\mathbf{y} - \hat{\mathbf{y}}\|^2\) over \(\mathcal{S}\)
if and only if
Why orthogonality implies optimality:
Suppose error \(\mathbf{e} = \mathbf{y} - \hat{\mathbf{y}}\) is not orthogonal to \(\mathcal{S}\).
Then \(\mathbf{e}\) has a component along \(\mathcal{S}\).
Moving \(\hat{\mathbf{y}}\) in that direction reduces \(\|\mathbf{e}\|\).
Contradiction: \(\hat{\mathbf{y}}\) wasn’t optimal.
Orthogonality conditions:
For \(\mathcal{S} = \text{span}\{\mathbf{v}_1, ..., \mathbf{v}_k\}\):
This gives \(k\) equations to solve for \(\hat{\mathbf{y}}\).
Random variable inner product:
Define \(\langle X, Y \rangle = \mathbb{E}[XY]\)
For zero-mean variables:
Orthogonal random variables:
(For zero-mean: \(\text{Cov}(X,Y) = 0\))
The connection:
| Vectors | Random Variables |
|---|---|
| \(\langle \mathbf{x}, \mathbf{y} \rangle = \mathbf{x}^T\mathbf{y}\) | \(\langle X, Y \rangle = \mathbb{E}[XY]\) |
| \(\mathbf{x} \perp \mathbf{y}\) | \(X \perp Y\) |
| Projection minimizes \(|\mathbf{y} - \hat{\mathbf{y}}|^2\) | LMMSE minimizes \(\mathbb{E}[(Y - \hat{Y})^2]\) |
| Error \(\perp\) subspace | Error \(\perp\) predictors |
This is why linear algebra solves estimation problems.
Curse of dimensionality:
To compute \(\mathbb{E}[Y|X]\), need \(p(y|x)\)
Density estimation convergence:
Consequence:
Linear class advantages:
MMSE requires \(p(y|x)\) to compute \(\mathbb{E}[Y|X]\)
Reality: We rarely know conditional distributions
What we often have:
Practical solution: Restrict to linear predictors
Function class:
New optimization problem:
From infinite to finite:
Expanding the MSE:
MSE is quadratic in \((a, b)\)
Step 1: Find optimal \(b\) for fixed \(a\)
Take derivative with respect to \(b\):
Solving for \(b\):
Interpretation:
Substituting back:
Step 2: Find optimal \(a\) using \(b^* = \mathbb{E}[Y] - a\mathbb{E}[X]\)
Substitute into MSE:
Define centered variables:
Then: \(\text{MSE}(a) = \mathbb{E}[(\tilde{Y} - a\tilde{X})^2]\)
Take derivative:
Solution:
Note: \(\mathbb{E}[\tilde{X}\tilde{Y}] = \text{Cov}(X,Y)\) and \(\mathbb{E}[\tilde{X}^2] = \text{Var}(X)\)
Alternative approach: Require error orthogonal to predictors
From MMSE theory: optimal error satisfies
for functions \(h\) in the predictor class.
For linear predictors \(\hat{Y} = aX + b\):
Predictor class = span\(\{1, X\}\)
Orthogonality conditions:
\(\mathbb{E}[(Y - aX - b) \cdot 1] = 0\)
\(\mathbb{E}[(Y - aX - b) \cdot X] = 0\)
Substitute \(b\) and solve:
Same answer, different route
Orthogonality generalizes to vectors without derivatives.
Complete solution:
Alternative forms:
Using correlation \(\rho = \frac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y}\):
Standardized form: If \(\mathbb{E}[X] = \mathbb{E}[Y] = 0\) and \(\sigma_X = \sigma_Y = 1\):
Minimum MSE:
Explained variance:
Correlation measures linear predictability, not dependence
Properties:
Restriction can only lose:
Performance gap:
Measures nonlinearity of \(\mathbb{E}[Y|X]\).
When are they equal?
if and only if \(\mathbb{E}[Y|X]\) is linear in \(X\).
For jointly Gaussian \((X,Y)\):
Uniform-Quadratic example:
MMSE estimator:
Nonlinear in \(X\)
Linear MMSE (suboptimal): slope \(\text{Cov}(X,Y)/\text{Var}(X) = \mathbb{E}[X^3]/\text{Var}(X) = 0\)
Performance gap:
The gap is Var(\(X^2\)) - the part of \(\mathbb{E}[Y|X]\) no line can follow
MMSE (unrestricted):
LMMSE (restricted):
LMMSE error may still correlate with \(X^2, X^3, \sin(X), ...\)
This correlation is unexploited information—structure in \((X, Y)\) that linear predictors cannot capture.
When LMMSE achieves MMSE:
If \(\mathbb{E}[Y|X]\) is already linear in \(X\):
Bivariate Gaussian \((X, Y) \sim \mathcal{N}(\mu, K)\)
Parameters:
where \(k_{XY} = \rho\sigma_X\sigma_Y\)
Joint density:
where \(z_x = \frac{x-\mu_X}{\sigma_X}\) and \(z_y = \frac{y-\mu_Y}{\sigma_Y}\) are standardized variables
Fully determined by first and second moments
Result: \(Y|X=x\) is Gaussian
Conditional mean:
Conditional variance (constant):
Therefore:
Observations:
Endpoints:
Regression line:
passes through \((\mu_X, \mu_Y)\) with slope \(\rho\sigma_Y/\sigma_X\)
Method 1: Complete the square
Start with joint density, condition on \(X = x\):
Method 2: Linear projection
Find \(a, b\) minimizing \(\mathbb{E}[(Y - aX - b)^2]\):
Method 3: Schur complement
Using block matrix operations on \(K\):
(all methods give same result)
For jointly Gaussian \((X, Y)\):
Why equal?
Minimum MSE:
Implications:
Maximum entropy property: Among all distributions with given mean and covariance, Gaussian has maximum entropy
Central Limit Theorem: Sum of many independent effects \(\to\) Gaussian
Preserved under linear operations: If \((X,Y)\) Gaussian and \(Z = aX + bY + c\):
Sufficient statistics: For Gaussian, sample mean and covariance are sufficient
Problem: the statistics are unknown
LMMSE needs: \(\text{Cov}(X,Y)\) and \(\text{Var}(X)\)
Solution: Learn from streaming data
Population gradient:
Widrow-Hoff: replace the expectation with its one-sample estimate
\(x_i y_i\), \(x_i^2\): rank-one instantaneous estimates of \(\mathbb{E}[XY]\), \(\mathbb{E}[X^2]\)
LMS update rule:
where \(e_i = y_i - \hat{y}_i\) is prediction error and \(\eta\) the step size (sometimes write \(\mu\))
Convergence: \(a_i \to a^*_{\text{LMMSE}}\) as \(i \to \infty\)
Convergence condition (in the mean):
Too large \(\eta\) \(\Rightarrow\) divergence
Steady-state error:
Misadjustment (small \(\eta\)): \(M = \frac{\eta \mathbb{E}[X^2]}{2}\)
Trade-off:
Time constant:
iterations to converge
Advantages:
Relation: LMS = stochastic gradient descent for MSE
Theory (Population)
Given joint distribution \(p(x,y)\):
Linear MMSE solution:
Practice (Sample)
Given data \(\{(x_i, y_i)\}_{i=1}^n\):
Sample estimates:
Convergence: \(\hat{a} \to a^*\) as \(n \to \infty\) (law of large numbers)
This is linear regression
Population risk (unknown):
Empirical risk (computable):
Optimization:
Sample optimization gives sample LMMSE
Linear model limitations:
Can only model linear relationships
Feature transformation:
where \(\boldsymbol{\phi}(x) = [\phi_0(x), \phi_1(x), ..., \phi_K(x)]^T\)
Still linear MMSE in feature space:
Example: Polynomial features
Now can fit parabolas while still using linear methods
Batch gradient descent:
Gradients:
Update:
One sample per step recovers LMS: the single-sample gradient step is exactly the LMS update; minibatches interpolate between the two
LMMSE \(\neq\) MMSE when \(\mathbb{E}[Y|X]\) is nonlinear.
Single neuron:
where \(\sigma\) is activation (ReLU, sigmoid, etc.)
Without activation: Linear regression
With activation: Nonlinear transform
Deep network: Composition of layers
Objective: \(\frac{1}{n}\sum_i (y_i - \hat{y}_i)^2\) (same MSE)
Backpropagation: Chain rule for gradients
Random vector: \(\mathbf{X} = [X_1, X_2, ..., X_p]^T\)
Mean vector:
Covariance matrix:
Element structure:
Properties:
Definition: \(\mathbf{A}\) is positive definite if
Equivalent conditions:
Geometric interpretation:
The set \(\{\mathbf{x} : \mathbf{x}^T \mathbf{A} \mathbf{x} = 1\}\) is an ellipsoid.
Covariance ellipse: for Gaussian data, \((\mathbf{x} - \boldsymbol{\mu})^T \mathbf{K}_{XX}^{-1} (\mathbf{x} - \boldsymbol{\mu}) = c^2\) contains about 95% of the data at \(c \approx 2.45\) - the shape of a data cloud is its covariance
Scalar case:
Given: \(X\) (single predictor), predict \(Y\)
Vector case:
Given: \(\mathbf{X} = [X_1, X_2, ..., X_p]^T\) (multiple predictors)
Example: Predict temperature \(Y\) from:
Matrix notation:
where \(\mathbf{a} = [a_1, a_2, ..., a_p]^T\) are weights
Minimize MSE over linear predictors:
Taking derivatives:
Substituting \(b\) and simplifying:
Define:
Reminder (vector derivatives, \(\mathbf{A}\) symmetric):
Normal equations:
Solution (if \(\mathbf{K}_{XX}\) invertible):
LMMSE predictor:
Minimum MSE:
Variance reduction = \(\mathbf{k}_{XY}^T \mathbf{K}_{XX}^{-1} \mathbf{k}_{XY} \geq 0\)
Column space perspective:
Data matrix columns: \(\mathbf{x}_1, \mathbf{x}_2, ..., \mathbf{x}_p\)
LMMSE = Projection:
Orthogonality:
Equivalently: \(\mathbf{X}^T(\mathbf{y} - \mathbf{X}\mathbf{a}) = \mathbf{0}\) - the normal equations in sample form, \(\mathbf{X}^T\mathbf{X}\mathbf{a} = \mathbf{X}^T\mathbf{y}\)
Projection matrix:
Properties:
Joint Gaussian distribution:
\(\begin{bmatrix} \mathbf{X} \\ \mathbf{Y} \end{bmatrix} \sim \mathcal{N}\left(\begin{bmatrix} \boldsymbol{\mu}_X \\ \boldsymbol{\mu}_Y \end{bmatrix}, \begin{bmatrix} \mathbf{K}_{XX} & \mathbf{K}_{XY} \\ \mathbf{K}_{YX} & \mathbf{K}_{YY} \end{bmatrix}\right)\)
Probability density:
where \(\mathbf{z} = \begin{bmatrix} \mathbf{x} - \boldsymbol{\mu}_X \\ \mathbf{y} - \boldsymbol{\mu}_Y \end{bmatrix}\)
Linear operations preserve Gaussianity
If \(\mathbf{Z} = \mathbf{A}\mathbf{X} + \mathbf{b}\), then:
Marginals are Gaussian:
Given joint Gaussian, conditional is also Gaussian:
Conditional mean (MMSE estimator):
Conditional covariance (error covariance):
The subtracted term is the Schur complement - the uncertainty removed by observing \(\mathbf{X}\)
Observations:
Direct inversion issues:
Ill-conditioning: When predictors highly correlated
Rank deficiency: perfect collinearity, or a sample estimate \(\hat{\mathbf{K}}_{XX}\) with \(p > n\)
Solutions:
QR decomposition: \(\mathbf{X} = \mathbf{Q}\mathbf{R}\)
SVD: \(\mathbf{X} = \mathbf{U}\boldsymbol{\Sigma}\mathbf{V}^T\)
Regularization:
Multiple responses:
Linear model:
where \(\mathbf{W} \in \mathbb{R}^{p \times m}\) is the weight matrix
MSE matrix:
Minimize trace (total MSE):
Solution:
where \(\mathbf{K}_{XY} = \mathbb{E}[\mathbf{X}\mathbf{Y}^T] - \mathbb{E}[\mathbf{X}]\mathbb{E}[\mathbf{Y}]^T\)
Solves as \(m\) separate LMMSE problems - one per output
Computational cost:
When \(\mathbf{K}_{XX}\) has off-diagonal terms:
Eigendecomposition approach:
Any symmetric positive semi-definite matrix:
where:
Eigenvectors are orthogonal
This orthogonality simplifies coordinate transformation.
Coordinate transformation:
Covariance in new coordinates:
Result: Diagonal covariance matrix
Components of \(\mathbf{Z}\) are uncorrelated:
Inverse transform (reconstruction):
Since \(\mathbf{U}\) is orthonormal: \(\mathbf{U}^T\mathbf{U} = \mathbf{I}\)
Given covariance:
Find eigenvalues: Solve \(\det(\mathbf{K} - \lambda\mathbf{I}) = 0\)
Block structure simplifies:
Result: \(\lambda_1 = 10, \lambda_2 = 3, \lambda_3 = 2\)
Find eigenvectors: Solve \((\mathbf{K} - \lambda_i\mathbf{I})\mathbf{v} = 0\)
Result: \(\mathbf{K}_{ZZ} = \text{diag}(10, 3, 2)\)
Coupled problem:
Every component of \(\mathbf{a}\) depends on every entry of \(\mathbf{K}_{XX}\)
In the KL basis \(\mathbf{Z} = \mathbf{U}^T\mathbf{X}\):
\(\mathbf{K}_{ZZ} = \boldsymbol{\Lambda}\), so the normal equations diagonalize:
Each coefficient solved independently
Cost: eigendecomposition is \(O(p^3)\), like inversion - the gain is structure, not speed
Not all dimensions equally informative
Eigenvalues measure variance (information) per dimension.
Truncated representation: Keep only first \(k < p\) components:
Selection criterion: Retain fraction \(\alpha\) of total variance:
Common choice: \(\alpha = 0.95\) or \(0.99\)
Reconstruction error:
Error equals sum of discarded eigenvalues.
Applications: