ML Regression
Notes taken during ML Regression by Coursera.
Linear Algebra Refresher
Matrices and Vectors
 Denoted by
n x m
(n rows, m columns).
Special Matrices
 Square matrix: a matrix whose dimensions are
n x n
.  Zero matrix: denoted by
0n x m
, a matrix where all the values are 0. (acts like 0 in scalar land)  Identity matrix: main diagonals are 1 and others 0. When multipling with another matrix, the result is the other matrix. (acts like 1 in scalar land)
 Column matrix:
n x 1
and row matrix:1 x n
. Aka vector.
Arithmetic
 Adding or subtract matrices: add or subtract each field. Must be the same size. (see Vector Addition
 Scalar multiplication: multiply each value in matrix by scalar (see Vector Scaling).
 Matrix multiplication:
A x B
 A must have same number of columns as B does rows eg
A(2 x 3) * B(3 x 2)
is valid. The resulting size will be2 x 2
.
 A must have same number of columns as B does rows eg

Example:
A = [1, 2, 4] B = [4, 2] [4, 3, 5] [5, 8] [7, 1] (2 x 3) (3 x 2) Outcome = [(1 * 4) + (2 * 5) + (4 * 7)] [(1 * 2) + (2 * 8) + (4 * 1)] [(4 * 4) + (3 * 5) + (5 * 8)] [(4 * 2) + (3 * 8) + (5 * 1)]

Commutative property does not apply in matrix multiplication: order matters.
Determinant
 Function that takes a square matric and convert it to a number. Formula looks like:
[a c]
[b d] == a*d  c*b
Matrix Inverse
 Given a square matrix
A
of sizen x n
we want to find another matrix such that:
AB = BA = I n
Matrix calculus
 Each of the points in a matrix can be function, can determine the derivative of each point.
Week 1  Simple Regression
 Course Outline:
 Module 1: Simple regression
 Low number of inputs
 Fit a simple line through the data
 Need to define "goodnessoffit" metric for each possible line.
 Gradient descent algorithm:
 Get estimated parameters
 Interpret
 Use to form predictions
 Module 2: Multiple relationships
 More complicated relationship than just a line.
 Incorporate more inputs when training the model.
 Module 3: Assessing performance
 Determine when "overfitting" the data.
 "Biasvariance tradeoff"
 Simple models are well behave but can be too simple to describe a behaviour accurately.
 Complex models can have odd behaviour.
 Module 4: Ridge Regression
Ridge total cost = measure of fit + measure of model complexity
 Cross validation (??)
 Module 5: Feature Selection & Lasso Regression
 What are the most important features for model?
Lasso total cost = measure of fit + (different) measure of model complexity
 Will learn: coordinate descent algorithm.
 What are the most important features for model?
 Module 6: Nearest neighbor & kernel regression
 Methods useful when you have a lot of data:
 Nearest neighbor: find closest piece of data to what you're looking at and use for predictions
 Kernel regression
 Methods useful when you have a lot of data:
 Module 1: Simple regression
 Assumed background
 Basic calculus (LOL)
 Basic linear algebra:
 Vectors
 Matrices
 Matrix multiply
Regression fundamentals
 Housing prices
 Look at recent sales in the neighbour hood to help inform your house price.
 Data:
 "For each house that sold recently, record square feet the house had and what the price was."
x1 = sq_ft, y2 = price````x2 = sq_ft, y2 = price
,x3 = sq_ft, y3 = price
etc
 Regression model:
y[i] = f(x[i]) + E[i]
 Outcome is the function applied to the input plus some error.E(e[i]) = 0
 expected value of error is 0  will be equally likely to have positive or negative values.
 Tasks in regression:
 Task 1: Determining options for data model.
 Task 2: Estimate the specific fit from the data.
 Block diagram for regression (diagram from memory)
ML Model
 Equation of a line: "intercept + slope * variable of interest" (
f(x) = W₀ + W₁ * Xᵢ
)  Single point includes the error (the distance from point back to line): (
f(x) = W₀ + W₁ * Xᵢ + εᵢ
).  Parameters of model =
W₀
(intercept) andW₁
(slope) Aka: regression coefficients.
Quality metric
 "Cost" of using a given line.

Can be calculated with "Residual sum of squares (RSS)"
 Take each error (
εᵢ
)  calculated as(actual_value  predicted value)²
 Add them for each point.
 In pseudo:
RSS(W₀, W₁) = sum([(y  (W₀ + W₁ * Xᵢ))² for y in actual_prices])
 Take each error (

Goal is to search over all possible lines to find the lowest RSS.
Using fitted line
 Estimated parameters:
Ŵ₀
 Intercept: value of
y
whenx == 0
 In the context of square feet features, it'd be houses that are just land.
 Usually not very meaningful (aka intercept is negative, does that mean land only sales mean the seller gets money??)
 Intercept: value of
Ŵ₁
 Estimated slope
 "Per unit change in input"  "when you go up 1 square foot, how much does the price go up?"
Finding the best fit (minimizing the cost)
min([RSS(W₀, W₁) for (W₀, W₁) in ALL_POSSIBLE_PARAMS])
 Convex / concave functions:
 Concave
 Looks like arc of cave.
 Line between 2 points lie below curve of function
 Max is where derivative = 0

Convex
 Opposite of concave
 Any two points will be above the curve of the function
 Min is where derivative = 0

Other functions:
 Can be multiple solutions to derivative = 0 or no solutions
 Finding the max or min analytically:
 Get derivate of function eg (
2w + 20
).  Set derivate to 0 then solve for
w
: (2w + 20 = 0 == w = 10
):  Finding the max via hill climbing:

Idea:
 Keep increasing
w
via some "step size" until "converged" eg: derivate = 0.  Divide space:
 On one side, the derivative will be more than 0, on the other less.
 In pseudo:
``` # g = func # w = value
def has_converged(g, w): return derivate(g, w) ~ 0
while not has_converged(g, w): w = w + STEP_SIZE ```
 While not converged:
 Take previous w
 Move in direction specified by derivated via some step size
 Choosing stepsize and convergence criteria:
 Fixed step size = set to constant size.
 Can result in jumping over optimal, resulting in slow convergence.
 Decrease the stepsize as you get closer to convergence.
 Convergence will never get exactly to 0, but you set a threshold for convergence success.
 Threshold depends on dataset.
 Commons choices:
 Algorithms that decrease with number of iterations:
η[t] = α / t
(step size at position t is some alpha over t)η[t] = α / sqrt(t)
(step size at position t is some alpha over square root of t)
 Keep increasing
Multiple dimensions: gradients
 When you have functions in higher dimensions, don't talk about "derivates" talk about "gradients".
 Notation example:
 Gradient definition:
 A vector (array) with partial derivatives.
 Partial derivatives:
 Same as derivate but involves other derivates. Treats them like constants.
 Worked example:
 Gradient descent == hill descent but compute the gradient instead of derivate for each iteration.
Asymmetric cost functions
 "Prefer to under estimate than over"
 Eg: don't want to predict my house price as too high.
Week 2  Multiple Regression
 Polynomial regression:
 Even with only a single input variable, a linear equation model may not represent the relationship between the output variable.
 Could use a higherorder function (quadratic, polynomial etc):
yᵢ = W₀ + W₁Xᵢ + W₂Xᵢ² ... + WpXᵢ^p + εi
 Treat different powers of
x
as "features":  feature 1 =
1 (constant)
 feature 2 =
x
 ..
 feature p + 1 =
x^p
 Different coefficients = parameters.
 More inputs added to regression model lets you factor in other relationships between your output variable.
 Question What letter will be used to represent number of observations?
N
 Question What letter will be used to represent no of inputs? (
x[i]
)d
 Question What letter will be used to represent no of features? (
hⱼ(x)
)D
 Commonly used equation:
 Interpreting coefficients of fitted function
 Coefficient's should be considered in the context of the entire model.
 Eg: Number of bedrooms might have a negative coefficient if the square feet of the house is low.
 If in a situation where you can't "fix" an input (eg if all features are a power of one input), then you can't interpret the coefficient.
 Linear Algebra Review
 Stages for computing the least squares fit:

Step 1: Rewrite regression model using linear algebra
 Rewrite the regression model for a single observation in matrix notation:

Multiple two vectors:
 A row vector (aka transposed column vector) of parameters / coefficients
 A column vector of features.

Then add the error term.
 Rewrite model for all observations:
* Rows of middle matrix (matrix ``H``) = vector of features from previous section.
 Step 2: Compute the cost
 Algorithm = search all different fits to find the smallest cost (RSS).
 RSS in matrix notation:
y  Hw
is a residual vector. Using vector multiplication of that vector times the transpose, you end up with a scalar of the residual sum of squares.
 Step 3: Take the gradient of the RSS:
2 * H_vector_transposed * (y_vector  H_vector * w_vector)

Step 4: Approach 1, closedform solution: solve for W.
 Set result to 0 and solve.
 Matrix inverse * matrix = identity matrix (??)
Week 3  Assessing Performance
 Goal: figure out how much you are "losing" using your model compare to perfection.
 Example: low predictions causing house to be listed too cheap.
 Loss can be measured with a loss function: L(y, f_\hat{w}(\mathbf{x}))
 Examples:
 Absolute error: L(y, f_\hat{w}(\mathbf{x})) = y  f_\hat{w}(\mathbf{x})
 Squared error: L(y, f_\hat{w}(\mathbf{x})) = (y  f_\hat{w}(\mathbf{x}))^2
 Root meansquared error  L2 Loss squared.
 Can have a very high cost if difference is large, compared to absolute error.
 Root meansquared error  L2 Loss squared.
 Examples:
 Compute training error:
 Define some loss function (as above).
 Computing training error. * Example: Average loss on training set using squared error: * = 1/N \sum\limits_{i=1}^{N} L(y, f_\hat{w}(\mathbf{x}))

average of loss function
* = $$1/N \sum\limits_{i=1}^{N} (y  f_\hat{w}(\mathbf{x}))^2$$

average of squared error
* = $$\sqrt{1/N \sum\limits_{i=1}^{N} (y  f _\hat{w}(\mathbf{x}))^2}$$
(convert to root mean squared error (RMSE))
* provides more intuitive format (dollars, instead of squared dollars)
* Training error vs model complexity:
* Training error obviously is lowers as complexity of model increases (can get almost perfect fit with high complexity models):
* Doesn't necessarily mean that predictions will be good, in fact they can get extremely bad with overfit models.
* Summary: low training error != good predictions.
* Generalisation error: theoretic idea for figuring out ideal model.
* Weight house price pairs (house, price)
by how they are to occur in dataset and use to evaluate predictions, use to evaluate predictions.
* For a given square footage, how likely is it to occur in the dataset?
* For houses with a given square footage, what house prices are we likely to see?
* Theoretical idea: impossible to compute generation error: requires every possible dataset in existence.
* Test error: like generalisation error but actually computable.
* Basically, use test error* to roughly approximate generation error.
* Average loss on houses in test set: 1/Ntest \sum_\limits{i=1}^{Ntest} L(y, f_\hat{w}(\mathbf{x}))
* Note: f_\hat{w}(\mathbf{x})
was fit with training data.
 Defining overfitting:
 Overfitting if there exists a model with estimated params $w'$
such that:
1. Training error ($$\hat{w}$$
) < Training error ($$w'$$
).
2. True error ($$\hat{w}$$
) > True error ($$w'$$
) * In other words: overfit if training error is low compared to another model but "true error" is high. Better to have high training error but low "true error". * Training/test split: how to think about dividing data between training and test split. * General rule of thumb: just enough points in test set to approximate generalisation error well. * To do: figure out how to approximate generalisation error. * If you don't have enough for training data, then need to use other methods like "cross validation". * 3 sources of error: 1. Noise 2. Bias 3. Variance
 Noise:
 Data is inherently noisy; there are heaps of variables you can't factor into model.
 Variance of noise term: spread of $\epsilon_i$
 Can't really control noise, there will be factors that influence observations outside of feature set.
 Data is inherently noisy; there are heaps of variables you can't factor into model.
 Bias:
 Difference between fit and "true function".
 "Is our model flexible enough to capture true relationship between data and model."
 Bias(x) = f_{w(true)}(\mathbf{x})  f_\hat{w}(\mathbf{x})
 Bias is basically how good our fit is; more parameters generally means less bias.
 Difference between fit and "true function".
 Variance:
 How much variance does model have?
 High complexity models == high variance.
 Variance is basically how wild our model is; more parameters means higher variance.
 Biasvariance tradeoff:
 As error decreases, bias decreases and variance increases.
 $MSE = bias^2 + variance$
(meansquared error) * Want to find sweet spot in model where bias and variance are together as low as possible. * We can't compute "true function" but there are ways to optimise an approximation a practical way. * Error vs amount of data: * True error decreases as data points in training set increase to limit of bias + noise. * Training error goes up as data points increase to same limit. * Validation set: * Choosing tuning parameters, $\lambda$
(eg degree of polynomial):
* Naive approach: For each model complexity $$\lambda $$
: 1. Estimate params on training data. 2. Assess performance on test data.
3. Choose $$\lambda $$
with lowest test
* Problem with this the test data was already used for selecting $$ \lambda $$
 will be overfit.
* Better approach: split data 3 ways: training, validation and test set
* Select $$\lambda $$
that minimizes error on validation set.
* Get approximate generalisation error of $$ \hat{w}_{\lambda\star} $$
using test set. * Typical splits: * 80% / 10% / 10% * 50% / 25% / 25%
Week 4  Ridge Regression
i* Symptom of overfitting: * When model is extremely overfit, magnitude of coefficients can be extremely large. * Overfitting not unique to polynomial regression: can also occur if you have lots of inputs. * Number of observations can influence overfitting: * Few observations (N small) can cause model to be quickly overfit as complexity grows. * Many observations (N very large) can be harder to overfit (but harder to find datasets).
![Number of Observations](./images/numberofobservations.png)
 The larger the inputs, the more change of data not including inputs for all data points causing overfitting.
 Balancing fits and magnitude of coefficients:
 Can improve quality metric by factoring in coefficient magnitude to avoid complex models.
Total cost = measure of fit (RSS) + measure of coefficient magnitudes
. Want to find the ideal balance between the two.
 Ways to measure coefficient magnitude:
 Add coefficients: $\hat{w}_0 + \hat{w}_1 ... \hat{w}_d$
 the negative coefficients would cancel out the others.
 Add the abs value of coefficients: $\sum\limits_{i=0}^{D}  \hat{w}_i $
 Aka "L1 norm"
 Sum squared values of coefficients: $\sum\limits_{i=0}^{D} (\hat{w}_i)^2$
 Add coefficients: $\hat{w}_0 + \hat{w}_1 ... \hat{w}_d$
(sum of squared values of coefficients) * Aka "L2 norm" * Resulting ridge objective and its extreme solution. * Can use an extra parameter to tune how much weight is applied to the measure of coefficient magnitude: $RSS(\mathbf{w}) + \lambda \mathbf{w}_2^2$ * Set the param to 0 = only RSS would weight in quality metric. * Set the param to $\infty$, $\mathbf{w} = 0$
would be the solution.
* Need to find a balance between the two: enter "ridge regression" (aka $$L_2 $$
regularisation.
 How ridge regression balances bias and variance.
 High $\lambda$
= high bias, low variance.
* As $$\lambda $$
get closer to infinitely, coefficients get smaller and less general.
* Low $$\lambda $$
= low bias, high variance.
* As $$\lambda $$
get closer to 0, coefficients get larger (RSS takes over) and shit gets crazy.
 Ridge regression demo
 "Leave one out" (LOO) cross validation can show approximate average mean squared error (MSE) and is a useful technique for choosing $\lambda$
.
 The ridge coefficient path:

Graphical representation of how the value of lambda affects the coefficient:
Again, at 0 it's just the RSS ($$\hat{w}_{LS} $$
), as it gets larger, they approach 0.
 Some sweet spot between smallish coefficients and a well fit model.
 Computing the gradient of the ridge objective:
 Can rewrite the L2 norm ($$ \mathbf{w} _2^2 $) in vector notation as follows:$w^tw $$
.
 Then, the entire ridge regression total cost can be rewritten like: $y  (\mathbf{H}\mathbf{w})^2(\mathbf{H}\mathbf{w}) +  \mathbf{w} _2^2$
, which is just the RSS + the L2 norm.
 Can take the gradient of both terms and you get: $2\mathbf{H}^T( y  (\mathbf{H}\mathbf{w})) + 2\mathbf{w}$
.
 The gradient of the L2 norm is analygous to the 1d case: derivate of $w^2$
= $2w$
 Approach 1: closedform solution:
 Summary: set the gradient = 0 and solve for $\hat{w}$
. * Steps:
1. Multiple the $$\mathbf{w} $$
vector by the identity matrix to make the derivation easier.
$$\triangle cost(\mathbf{w})) = 2\mathbf{H}^T(\mathbf{y}  \mathbf{H}\mathbf{w}) + 2\lambda\mathbf{I}\mathbf{w} = 0 $$
2. Divide both sides by 2.
$$ =\mathbf{H}^T(\mathbf{y}  \mathbf{H}\mathbf{w}) + \lambda\mathbf{I}\mathbf{w} = 0 $$
3. Multiple out.
$$\mathbf{H}^T\mathbf{y} + \mathbf{H}^T \mathbf{H}\mathbf{w} + \lambda\mathbf{I}\mathbf{w} = 0 $$
4. Add $$\mathbf{H}^T\mathbf{y} $$
to both sides.
$$\mathbf{H}^T \mathbf{H}\mathbf{w} + \lambda\mathbf{I}\mathbf{w} = \mathbf{H}^T\mathbf{y} $$
5. Since the $$ \hat{w} $$
appears in both expressions, can factor it out.
$$(\mathbf{H}^T \mathbf{H} + \lambda\mathbf{I})\mathbf{w} = \mathbf{H}^T\mathbf{y} $$
6. Multiple both sides by the inverse of $$\mathbf{H}^T \mathbf{H} + \lambda\mathbf{I} $$
.
$$\mathbf{w} = (\mathbf{H}^T \mathbf{H} + \lambda\mathbf{I})^{1}\mathbf{H}^T\mathbf{y} $$

Discussing the closedform solution
* Can prove the closedform solution is congruent with the above notes, by setting $$\lambda = 0 $$
and seeing that results are equal to the least squares closed form solution:
$$\hat{w}^{ridge} = (\mathbf{H}^T\mathbf{H})^{1}\mathbf{H}^T\mathbf{y} = \hat{w}^{LS} $$results

Setting lambda to infinity results in 0 because the inverse of infinity matrix is like dividing by infinity.

Recall previous solution for $\hat{w}^{LS} = (\mathbf{H}^T\mathbf{H})^{1}\mathbf{H}^T\mathbf{y}$ :

Invertible if number of linear independant observations is more than number of features ($$ N > D $$ )  an added bonus of using ridge regression.

Complexity of the inverse: $O(D^3)$


Properties for ridge regression solution:
 Inverible always if $\lambda > 0$even if $N < D$ .

Complexity of inverse is the same. May be prohibitive to do this solution with lots of features.

Called "regularised" because adding the $\lambda\mathbf{I}$is "regularising" the $\mathbf{H}^T \mathbf{H}$ , making it invertable ever with lots of features.


Approach 2: gradient descent
 Same as RSS gradient descent but factors in the L2 norm cost component gradient.
 Selecting tuning parameters via cross validation.
 If you don't have enough data to build a validation set, for testing the performance of $\lambda$
, you can grab the validation set as a subset of the training data, but instead of grabbing a single slice, can grab all slices and average the results.
 Kfold cross validation.
 Algorithm:
 For each lambda you're considering:
 For each data slice of size $N/K$
 For each lambda you're considering:
 Algorithm:
: * Go through each validation block on training data compute predictions.
* Compute error for predictions using the $$\lambda $$
value.
* Calculate average error: $$CV(\lambda) = 1/K \sum\limits_{k=1}^K error_K(\lambda) $$
* Choosing ideal value of K:
* Best approximation occurs for validation sets of size 1 (K = N) aka "leaveoneout" (LOO) cross validation.
* Computationally intensive: requires computing N fits of model per $$\lambda $$
.
* Common to use $$K=5 $$
(5fold CV) or $K=10$
(10fold CV) if $K=N$
is computational infeasible to run.
 How to handle intercept term.
 The goal of the ridge regression is to lower the coefficients values $ but that doesn't necessarily make sense for $\hat{w}_0$
. * Solutions:
* Option 1: leave out the $$\hat{w}_0 $$
coefficient when computing the optimal fit. * In closedform solution, can use a modified identity matrix with the first position set to 0.
* In gradient descent, can just skip the ridge component when computing the j value for $$\hat{w}_0 $$
. * Option 2: transform data so the expected intercept value is around 0 (called "centring").
Week 5  Feature Selection & Lasso
 Feature selection task motivation
 Efficiency
 when faced with large feature sets, predictions can get expensive.
 When $\hat{w}$
 Efficiency
is sparse (eg has a lot of 0s in the dataset) can eliminate all 0 features. * Interpretability * which features are relevant to housing predictions? * All subsets algorithm * Search over every combination of features computing the RSS: * Iterate through every N = 1 sized combination, then N = 2 and so on up to feature D. * Slow to compute with a lot of features: complexity = $2^{D}$ * With 15 features $2^{15} = 32768$ * Greedy algorithms * "Forward stepwise algorithm" * Find best N = 1 feature subset (maybe based on training error) * Next, find best N = 2 subsets that includes previous feature. * Use validation set or cross validation to choose when to stop greedy procedure  training error alone is insufficient. * Quicker to compute than all subsets, but may not find optimal combination. * Complexity: * 1st step: D models, 2nd step: D  1 models, 3rd step: D  2 models etc * Worst case complexity: $D^2$ * Other greedy algorithms * Backward stepwise: start with full model and work backwards removing one. * Combining forward and backward steps: add steps to remove features not considered relevant. * Regularised regression for feature selection * Ridge regression (L2 penalty) encourages small weights but not exactly 0. * Core idea: can we use a similar idea but actually get weights to 0 and effectively remove them from the model? * Potential method 1: Thresholding. * Remove features that falls below some threshold for $w$
. * Problem: redundant features * If you have bathroom and number of showers, the weights would be distributed over both features. If you remove one, the coefficient should double on the other. * This means you could potentially end up discarding all redundant features. * Potential method 2: use L1 norm for regularized regression aka "Lasso regression" aka "L1 regularised regression".
* $$ RSS(\mathbf{w}) + \lambda * \mathbf{w}_1 $$
* When tuning param, $$\lambda $$is 0, $$\hat{w}^{lasso} = \hat{w}^{LS} $$
(ie result is just least squares.
* When $$\lambda = \infty $$
, $\hat{w}^{lasso} = 0$
(ie coefficients shrink to 0). * Coefficient path for Lasso:
![Coefficient path for Lasso](coefficientpathforlasso.png)
* For certain lambda values, features jump out of the model and other fall to 0. Individual impact of features becomes clearer.
 Optimising the lasso objective
 The derivate of $w_j$can't be calculated when $w_j = 0$
. Therefore, can't calculate the derivate. * Use "subgradients" over gradients. * Coordinate descent * Goal: minimise a function $g(w_0, w_1, ..., W_D)$
, a function with multivariables. * Hard to find minimum for all coordinates, be easy for a single coordinate with all the other fixed.. * Algorithm: * While not converged: * Pick a coordinate (w coefficient) and fix the other. * Find the minimum of that function. * No stepsize required * Converges for lasso objective * Converges to optimum in some cases. * Can pick next coordinate at random, round robin or something smarter... * Normalising features * Idea: put all features into same numeric range (eg number of bathrooms, house square feet) * Apply to a column of the feature matrix. * Need to apply to training and test data. * Formula:
$$\underline{h}_j(\mathbf{x}_k) = \frac{h_j(\mathbf{x}_k)}{ \sqrt{\sum\limits_{i=1}^{N} h_j(\mathbf{x}_i)^2}} $$
 Coordinate descent for least squares regression
 For each j feature, fix all other coordinates $ \mathbf{w}_{j} $and take the partial with respect to $\mathbf{w}_j$
 Gradient derived to $2P_j + 2w_j$($$P_j $$
== residual without jth feature)
* Setting to 0 and solving for $$\hat{w}_j $$
results in $\hat{w}_j = P_j$
* Intuition: if the residual between the predictions without j and the actual prediction is large, then the weight of j will be large and vice versa (to do: check this as your understanding improves).
 Coordinate descent for lasso (for normalised features)
 Same as cord descent for least squares, except we set $\hat{w}_j$
using a "thresholding" function (in code):
```
def set_w_j(p_j, lambda):
if p_j < lambda / 2:
return p_j + lambda / 2
if lambda / 2 < p_j < lambda / 2:
return 0
if p_j > lambda / 2:
return p_j  lambda / 2
```
* General idea here is "soft thresholding", aiming to get values to 0 that fit within some range:
![Soft thresholding]("./images/corddescentsoftthresholding.png")
 Coordinate descent for lasso (for unnormalised features)
 Normalisation factor is used during the set $w_j$
portion of the regression.
 Choosing the penalty strength and other practical issues with lasso.
 Same as ridge regression:
 If enough data, validation set.
 Compute average error
 Same as ridge regression:
 Summary
 Searching for best features
 All subsets
 Greedy algorithms
 Lasso regularised regression approach
 Contrast greedy and optimal algorithms
 Describe geometrically why L1 penalty leads to sparsity
 Estimate lasso regression parameters using an iterative coordinate descent algorithm
 Implement Kfold cross validation to select lasso tuning parameter $\lambda$
 Searching for best features
 Note: be careful about interpreting features, need to consider in the context of the entire model.
Week 6  Nearest Neighbors & Kernel Regression
 Limitations of parametric regression
 A polynomial fit might work well in certain regions of input space and not well in others.
 Eg cubic fit might work well for higher square feet houses, but not so well for lower.
 Ideal method:
 Flexible enough to support "local structure" ie different fit at certain input space regions.
 Doesn't require us to infer "structure breaks" ie the places where we want a different fit.
 Nearest neighbour regression approach
 For some input, find the closest observation. That's it.
 What people do naturally when predicting house prices.

Formally:
 For some input, find closest $\mathbf{x}_i$
using some distance metric.
2. Predict: $$\hat{y}_q = y_{nn} $$
 Distance metrics
 1D feature sets: Euclidian distance
 $distance(x_j, x_q) = x_j  x_q$
 Multi dimensions:
 Can use a weight on each feature to determine importance in computing distance ie sqft closeness is important, but year renovated may not be.
 $ distance(\mathbf{x_j}, \mathbf{x_q}) = \sqrt{a_1(x_i[1]  x_q[1])^2) + .. a_D(x_i[D]  x_q[D])^2)} $
 Note: $a_D$
 1D feature sets: Euclidian distance
is the "weight" in this equation.

1Nearest Neighbour Algorithm
 Need to define your distance function.

In pseudo:
``` closest_distance = float('inf')
for i in all_data_points: dist = calculate_distance(i, input) if dist < closest_distance: closest_distance = dist
return closest_distance
```

Notes:
 Sensitive to regions with missing data; need heaps of data for it to be good.
 Sensitive to noisy datasets.
 kNearest neighbours regression
 Same as 1Nearest Neighbour but average over values of "k"nearest neighbours.
 kNearest neighbours in practise
 Usually has a much better fit than 1nearest.
 Issues in regions where there's sparse data and at boundaries.
 Weighted knearest neighbours
 Idea: weight neighbours by how close or far they are to data point.
 Formula (where $C_{qNN1}$
refers to some weight for NN1:
$$\hat{y}_q = \dfrac{C_{qNN1}Y_{qNN1} + C_{qNN2}Y_{qNN2} .. C_{qNNK}Y_{qNNK}}{\sum\limits_{j=1}^{K}C_{qNNj}} $$
* Weighting data points:
* Could just use inverse of distance: $$C_{qNN1} = \dfrac{1}{distance(\mathbf{X_j}, \mathbf{X_q})} $$
* Can use "kernel" functions:
* Gaussian kernel
* Uniform kernel etc

Kernel regression

Instead of weighing nneighbours, weigh all points with some "kernel":
$\hat{y}_q \frac{\sum\limits_{y=1}^{N} C_{qi}Y_{qi}}{\sum\limits_{y=1}^{N} C_{qi}} = \frac{\sum\limits_{y=1}^{N} kernel_{\lambda}(distance(\mathbf{x_i},\mathbf{x_q})) * y_i}{kernel_{\lambda}(distance(\mathbf{x_i},\mathbf{x_q}))}$

In stats called "NadayaWatson" kernel weighted averages.
 Need to choice a good "bandwidth" (lambda value)
 Too high = over smoothing; low variance, high bias.
 Too low = over fitting; high variance, high bias.
 Need to choose kernel but bandwidth more important.
 Use validation set (if enough data) or crossvalidation to choose $\lambda$

value.
 Global fits of parametric models vs local fits of kernel regression
 If you were to predict datapoint by averaging all observations, you'd end up with a constant fit.
 Kernel gives constant fit at a single point; a "locally constant fit".