[SOLVED] IT代写 STAT340 Lecture 10: Cross Validation’

title: ‘STAT340 Lecture 10: Cross Validation’
author: ”and Wu”
date: “Fall 2021”
output: html_document

Copyright By PowCoder代写加微信 assignmentchef

“`{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)

require(ggplot2)

__Readings:__ ISLR Section 5.1

Several times this semester, we have discussed methods that required the choice of a number of components, clusters or dimensions.

__Example:__ in $k$-means, we fit $k$ clusters to our data. How do we choose $k$?

__Example:__ in PCA, how do we decide how many principal components to keep?

__Example:__ Suppose we have a data set with hundreds or even thousands of predictor variables. How do we decide which variables to include in linear regression (or any other prediction model)?

These are all examples of *model selection* problems.
That name is a bit confusing at first– it sounds like there must be a statistical model involved!

In the case of our variable selection example, there is in fact a statistical model– linear regression makes very specific assumptions about the data distribution– but we use the term “model selection” more broadly to mean situations like those above, in which we have to choose among many different *similar* ways of describing the data.

For example, do we “model” the data as coming from $3$ or $4$ clusters?

We’ll focus in this lecture specifically on selecting variables for linear regression.
Later this week, we will see ways of doing model selection for other tasks, such as clustering.

## How do we compare models?

Let’s consider a very simple example (adapted from Section 3.3.2 and Section 5.1 in ISLR).

The `mtcars` data set is a classic data set that comes packaged in R.

data(‘mtcars’);
head(mtcars);

Let’s confine our attention to the `mpg` (miles per gallon) and `hp` (horsepower) variables.
Can we predict gas mileage from the horsepower?

Let’s try fitting linear regression.

model1 <- lm(mpg ~ 1 + hp, mtcars);intercept1 <- model1$coefficients[1];slope1 <- model1$coefficients[2];# Plot the data itselfpp <- ggplot( mtcars, aes(x=hp, y=mpg)) + geom_point();pp <- pp + geom_abline(intercept=intercept1, slope=slope1, colour=’blue’ );Okay, it looks reasonable, but you might notice that the residuals have a bit of a weird behavior. Let’s plot them to see what I mean.pp <- ggplot( mtcars, aes(x=hp, y=mpg-(slope1*hp+intercept1))) + geom_point()The residuals have a kind of U-shape.This suggests that there is a non-linearity in the data that we are failing to capture.Let’s try adding another predictor: the squared horsepower.model2 <- lm(mpg ~ 1 + hp + I(hp^2), mtcars);intercept2 <- model2$coefficients[1];slope2_1 <- model2$coefficients[2];slope2_2 <- model2$coefficients[3];# Plot the data itselfpp <- ggplot( mtcars, aes(x=hp, y=mpg)) + geom_point();# As usual, there are cleaner ways to do this plot, but this is the quick and easy way to make itt.# If we were doing this more carefully, we would evaluate the cureve in the plot at more x-values than just the ones in the data to smooth things out.pp <- pp + geom_line( aes(x=hp, y=intercept2 + slope2_1*hp + slope2_2*I(hp^2) ), colour=’red’ );That looks like quite an improvement!Just for comparison:pp <- ggplot( mtcars, aes(x=hp, y=mpg)) + geom_point();pp <- pp + geom_abline(intercept=intercept1, slope=slope1, colour=’blue’ );pp <- pp + geom_line( aes(x=hp, y=intercept2 + slope2_1*hp + slope2_2*I(hp^2) ), colour=’red’ );We can also compare the squared residuals to confirm that adding the feature `hp^2` actually decreased our error:c( sum( model1$residuals^2 ), sum( model2$residuals^2 ) );Why stop there? Why not add `hp^3` as well, or even `hp^4`?Well, funny enough, that is precisely the idea behind *polynomial regression*, which you can learn more about in ISLR (Section 3.3.2; more substantial discussion in Chapter 7) or in a regression course.But that raises the question: how do we know when to stop?-You’ll find that if you add `hp^3` to the model above, that the sum of squared residuals does indeed improve.-But how do we know if that improvement is worth it?One approach to this problem would be to examine the $p$-values associated to the coefficients. See ISLR Chapter 3 for a discussion of that approach.In these notes, we will see a different, arguably more principled approach.### Overfitting and Unseen DataIf we keep adding more predictors to our model, the residuals will continue to decrease, but this will not actually mean that our model is better.Instead, what we will be doing is *over-fitting* to the data.That is, our model will really just be “memorizing” the data itself rather than learning a model.The true test of model quality is how well it does at predicting for data that we *didn’t* see.That is, if we fit our model on data $(X_i,Y_i)$ for $i=1,2,dots,n$, how well does our model do on the previously unseen $(X_{n+1},Y_{n+1})$?Specifically, in the case of regression, we want our model to minimizemathbb{E} left( hat{Y}_{n+1} – Y_{n+1} right)^2,where $hat{Y}_{n+1} = hat{beta} X_{n+1}$, with $hat{beta}$ being learned from our $n$ training observations.## Validation SetsSo rather than focusing on how well our model fits our training data, we should be trying to determine how well our model does when it gets applied to data that we haven’t seen before.Specifically, we would like to know the mean squared error (MSE),mathbb{E} left( hat{Y}_{n+1} – Y_{n+1} right)^2.__Note:__ The name is hopefully clear– it is the expectation (mean) of the squared error between our prediction and the truth.If you’ve taken an ML course, this idea should already be quite familiar. We always train (“fit” in the language of statistics) our model on a *training set*, and then assess how well the model performs on a *test set* that our model hasn’t seen before.The trouble is that in most statistical problems, we have at most a few hundred data points to work with.As a result, we can’t really afford to set aside some of our data just to use as a test set.Note that this is in contrast to many machine learning settings (e.g., training a neural net), where we often have tens or hundreds of thousands of data points to work with.Following the logic of the train/test split idea in ML, though, a natural approach is to do the following:1. Split our data into two parts, say $S_1,S_2$, such that $S_1 cup S_2 = {1,2,dots,n}$ and $S_1 cap S_2 = emptyset$.2. Obtain estimate $hat{beta}_1$ by fitting a model on the observations in $S_1$3. Evaluate the error of our fitted model on $S_2$,frac{1}{|S_2|} sum_{i in S_2} left( Y_i – hat{beta}_1 X_i right)^2.Typically, we call $S_2$, the set that we make predictions for, the *validation set*, because it is validating our model’s performance.Let’s see this in action on the `mtcars` data set.We randomly split the data set into two groups. For each model order 1, 2, 3, 4 and 5, we fit the model to the training set and then measure the sum of squared residuals of that model when applied to the validation set.One run of this experiment is summarized in `resids_onerun`.For details, refer to `mtcars_poly.R`, which is included among the supplementary files for this lecture.source(‘mtcars_poly.R’);head(resids_onerun);# Plot these resultspp <- ggplot(resids_onerun, aes(x=Order, y=log(Error) ) );pp <- pp + geom_line( size=1)Let’s pause to make sure that we understand what this plot actually shows.We split the `mtcars` dataset randomly into two sets, a “training” set and a “validation” set.For each order (1, 2, 3, 4, 5), we fit a model of that order to the training set.Then we use that model to try and predict the outcomes (`mpg`) on the validation set.So this is the performance of five different models, trained on the same data and evaluated on the same data.Looking at the plot, we see that as we add higher-order powers of `hp`, we don’t really gain much in terms of the error (i.e., sum of squared residuals) beyond order 2.Indeed, past the order-2 model, the error gets worse again!__Aside:__ this deteriorating performance is due largely to the fact that the `mtcars` data set is so small. Once we split it in half, we are fitting our model to just 32 observations.Estimating four or five coefficients from only about 30 observations is asking for trouble!This is a tell-tale sign of over-fitting of a model.This would be a good occasion for some kind of *regularization*, but we’ll come back to that.### Variance in the residualsThere’s one problem, though.That plot shows the residuals as a function of model order for one particular random set $S_1$.Let’s plot the same residuals for a few different random sets.__Note:__ the data frame `resids` contains multiple replicates of the above experiment. Once again, refer to the code in `mtcars_poly.R` for details.head(resids)pp <- ggplot(resids, aes(x=Order, y=log(Error), color=as.factor(Rep) ) );pp <- pp + geom_line( size=1)Hmm. There’s quite a lot of variance among our different estimates of the prediction error.Each of these is supposed to be estimating the errormathbb{E} left( hat{Y}_{n+1} – Y_{n+1} right)^2,but there’s so much variation among our estimates that it’s hard to know if we can trust any one of them in particular!Indeed, the variance is so high that we need to plot the error on a log scale! Once in a while, we get unlucky and pick an especially bad train/validate split, and the error is truly awful!Note that the y-axis is on a log scale, so an increase from, say, 2 to 3 is an *order of magnitude* increase in error.__Question:__ what explains the variance among the different lines in that plot?__Question:__ How might we reduce that variance?## Reducing the variance: Leave-one-out cross-validationOne source of variance in our cross-validation plots above was the fact that each replicate involved splitting the data in half and training on only one of the two halves.That means that on average, from one replicate to another, the data used to train the model changes quite a lot, and hence our estimated model changes a lot.That’s where the variance comes from!There is also the related problem that we are training on only half of the available data. As statisticians and/or machine learners, we don’t like not using all of our data!So, here’s one possible solution: instead of training on half the data and validating (i.e., evaluating) on the other half, let’s train on all of our data except for one observation, then evaluate our learned model on that one held-out data point.That is, instead of splitting our data into two halves, we1. Take one observation and set it aside (i.e., hold it out)2. Train our model on the other $n-1$ observations3. Evaluate our model on the held-out observation.# This R file implements the same experiment as we saw above, but this time# doing LOO-CV instead of source(‘mtcars_poly_loocv.R’);pp <- ggplot(resids_onerun, aes(x=Order, y=Error ) );pp <- pp + geom_line( size=1)But once again, that’s just one run. Let’s display several of them in one plot.pp <- ggplot(resids, aes(x=Order, y=log(Error), color=as.factor(Rep) ) );pp <- pp + geom_line( size=1)For each of our replicates, we are estimating our model based on $n-1$ of the observations, and then evaluating our prediction on the one held-out observation.But now we have a different kind of variance: our estimate of the error is at the mercy of the one observation that we chose to hold out.If we chose an especially “bad” observation to hold out, then our error might be especially high.Leave-one-out cross-validation (LOO-CV) tries to bridge this gap (i.e., balancing the better stability of leaving one observation out with the variability induced by evaluating on a single point) by:For each $i=1,2,dots,n$:1. Train the model on ${ (X_j, Y_j) : i
eq i }$.2. Evaluate on $(X_i, Y_i)$.3. Average the model error over all $i =1,2,dots,n$.This illustration from ISLR should give you the general idea.Let’s see that in action. As we have done many times this semester, this code is optimized for clarity and readability, not for concision or “cleverness”.There are much more “graceful” ways of doing this, and shortly we’ll see R’s built-in CV tools, which are what we would normally use for this.But here the goal is to illustrate the core ideas in a really obvious way, hence the “clumsy” code.data(‘mtcars’); # Still using mtcars data; reloading it just to remind us.nrows <- nrow(mtcars); # Number of observations in the datanorder <- 5;# For each choice of observation to hold out, we need to record the score# (i.e., squared erro) for each of the five model orders.errors <- data.frame( ‘Row’=rep(1:nrows, each=norder), ‘Order’=rep(1:norder, times=nrows), ‘Error’=rep(NA, nrows*norder));for ( i in 1:nrow(mtcars) ) {train_data <- mtcars[-c(i),]; # Leave out the i-th observationleftout <- mtcars[c(i),]; # the row containing the left-out sample.# Remember, we are fitting five different models# So that we can compare them.# Fit the linear model, then evaluate.m1 <- lm(mpg ~ 1 + hp, train_data );m1.pred <- predict( m1, leftout );idx <- (errors$Row==i & errors$Order==1); # Pick out row of the errors df.# record squared error btwn predict and trutherrors[idx,]$Error <- (m1.pred – leftout$mpg)^2 ; # Fit the quadratic model, then evaluate.m2 <- lm(mpg ~ 1 + hp + I(hp^2), train_data );m2.pred <- predict( m2, leftout );idx <- (errors$Row==i & errors$Order==2); # Pick out row of the errors df.# record squared error btwn predict and trutherrors[idx,]$Error <- (m2.pred – leftout$mpg)^2; # Fit the cubic model, then evaluate.m3 <- lm(mpg ~ 1 + hp + I(hp^2) + I(hp^3), train_data );m3.pred <- predict( m3, leftout );idx <- (errors$Row==i & errors$Order==3); # Pick out row of the errors df.# record squared error btwn predict and trutherrors[idx,]$Error <- (m3.pred – leftout$mpg)^2; # Fit the 4-th order model, then evaluate.m4 <- lm(mpg ~ 1 + hp + I(hp^2) + I(hp^3) + I(hp^4), train_data );m4.pred <- predict( m4, leftout );idx <- (errors$Row==i & errors$Order==4); # Pick out row of the errors df.# record squared error btwn predict and trutherrors[idx,]$Error <- (m4.pred – leftout$mpg)^2; # Fit the 5-th order model, then evaluate.m5 <- lm(mpg ~ 1 + hp + I(hp^2) + I(hp^3) + I(hp^4) + I(hp^5), train_data );m5.pred <- predict( m5, leftout );idx <- (errors$Row==i & errors$Order==5); # Pick out row of the errors df.# record squared error btwn predict and trutherrors[idx,]$Error <- (m5.pred – leftout$mpg)^2;Okay, so let’s make sure that we understand what is going on, here.The data frame `errors` now has `nrows*norders` rows.So for each observation in the cars data set, there are five entries in the table `errors`, recording the squared error for the models of order 1, 2, 3, 4 and 5 when that data point was held out.We said that when we do CV, we want to average across the $n$ observations, so let’s do that.We’re going to use the `aggregate` function, which is one of the ways to perform “group-by” operations in R.Group-by operations are where we pool our observations into subsets according to some criterion, and then compute a summary statistic over all of the observations in the same subset (i.e., the same “group”).Using that language, we want to group the rows of `errors` according to model order, and take the average squared error within each order.# Error ~ Order tells R to group the data according to the Order column# and that we want to summarize the Error column within observations# of the same Order.# Passing the FUN=mean argument tells R that the summary statistic we want to use# is the function mean().# We could pass other summary statistic functions in this argument.# For example, we could use median, sd, var, max, etc.,# though those would be a bit silly here.err_agg <- aggregate(Error ~ Order, data=errors, FUN=mean);head(err_agg)And we can plot that just to drive the point home.pp <- ggplot(err_agg, aes(x=Order, y=log(Error) ) );pp <- pp + geom_line( size=1)### Recap: single split versus LOO-CVSo far we have seen two different ways of estimating a model’s performance on unseen data.The first was to randomly split the data into two sets, train on one and evaluate on the other.__Pro:__ Only have to fit a model once (or just a few times, if we are going to repeat the operation and average)__Con:__ Only have half of the data available to fit the model, which leads to less accurate prediction (and thus high variance in estimated model).The second is __leave-one-out cross-validation__.__Pro:__ Use all but one observation to fit the model, so model fit is almost as good as if we had used all of the data__Con:__ Have to fit the model anew for each held-out data point, results in fitting the model $n$ different times, which can be expensive.__Con:__ Because any two training sets overlap in all but one of their elements, our fitted models are very highly correlated with one another, so we’re doing a lot of work ($n$ model fits) to get a bunch of highly correlated measurements.So the natural question is whether or not we can bridge the gap between these two extremes.### The happy medium: $K$-fold cross validationWell, there are a few different ways to bridge this gap, for example using Monte Carlo methods.Let’s discuss the most popular one here.We’ll borrow a bit from the LOO-CV idea, while lessening the correlatedness of the models fits.$K$-fold CV randomly divides the data into $K$ subsets, called *folds*.Then, one at a time, we hold out one of the folds, train our model on the $K-1$ remaining folds, and evaluate our model’s prediction error on the held-out fold.Then, we can average the errors across the $K$ folds.That is, the “recipe” for $K$-fold cross-validation is1. Randomly partition the data into $K$ (approximately) same-sized subsets, $S_1,S_2,dots,S_K$ such that $cup_k S_k = {1,2,dots,n}$ and $S_k cap S_ell = emptyset$ for all $k
eq ell$2. For each $k=1,2,dots,K$, train a model on the observations indexed by $i in cup_{ell
eq k} S_ell$ and compute the prediction errorhat{E}_k =frac{1}{|S_k|} sum_{i in S_k} (hat{y}_i – y_i)^23. Estimate the true error $mathbb{E} (hat{y}_{n+1} – y_{n+1})^2$ asfrac{1}{K} sum_{k=1}^K hat{E}_k,Schematically, this looks something like this (with $K=5$):Let’s implement this in R, just for the practice.Once again, R has built-in tools for making this easier, which we will discuss later, but this is a good opportunity to practice our R a bit.data(‘mtcars’); # We’ll continue to use the mtcars data setK <- 5; # 5-fold regularization. K between 5 and 10 is a fairly standard choice# The first thing we need to do is partition the data into K folds.# There are many different ways to do this,# including using functions from other packages# (e.g., https://www.rdocumentation.org/packages/caret/versions/6.0-90/topics/trainControl)# But here’s an approach using the R function split() that I liken <- nrow(mtcars);# sample(n,n,replace=FALSE) really just randomly permutes the data.# Then, passing that into the split function assigns these to the K different# factors defined by as.factor(1:K).# See ?split for more information.Kfolds <- split( sample(1:n, n,replace=FALSE), as.factor(1:K));程序代写 CS代考加微信: assignmentchef QQ: 1823890830 Email: [email protected]