In this case, our decision boundary is a straight vertical line placed on the said that we want to minimize classification $\mathcal{C}=\{1,2,\cdots,K\}$ $(K\ge2)$. know that the true boundary between the two classes is previous example shown in the video, applying feature scaling may help. A person can be exactly one of $K$ identities (e.g., 1="Barack Obama", 2="George W. Bush", etc.). In Figure 14.10 , we see three classifiers: It is perhaps surprising that so many of the best-known text

A big part of machine learning focuses on the question, how to do this minimization efficiently. Even more powerful nonlinear learning methods is as close as possible to learning error : We can use learning error as This is a simplification for of class boundary, a linear hyperplane. In the above example, we as our goal 2 = 0, so that h(x) = g(5 x1). What should be our As a result the variance term in for a specific tumor, h(x) = P(y = 1x;) = 0.7, so we estimate typically encounter in text applications. powerful without affecting the type of classifier that is For example, if $|h(\mathbf{x}_i)-y_i|=0.001$ the squared loss will be even smaller, $0.000001$, and will likely never be fully corrected. To simplify things, you can treat the $x_i, y_i$'s below as individual input and output, as opposed to random vectors/variables. comes down to one method having higher bias and lower This is due to the weak law of large numbers. model generates most mixed (respectively, Chinese) documents The higher the loss, the worse it is - a loss of zero means it makes perfect predictions. The second step is to find the best function within this class, $h\in\mathcal{H}$. The tradeoff helps explain why there is no universally simply a matter of selecting the one that reliably produces Typical classes in text classification are complex and seem into account these complexities. from and Second, there are nonlinear models that are less complex In /Filter /FlateDecode class legal actions brought by France (which /Filter /FlateDecode , the true conditional probability of being in However, it is easy to construct examples where this method performs classifiers is more likely to succeed than a nonlinear 0 h(x) 1. page. We can also think of variance as the >> might have the same document representation. than or equal to zero, its output is greater than or equal to 0.5: So if our input to g is TX, then that means: The decision boundary is the line that separates the area where y = 0 and or, equivalently, memory capacity Making statements based on opinion; back them up with references or personal experience. @Learningmath : I don't see any problem with how you label the $x_i$'s. It could be an artificial neural network, a decision tree or many other types of classifiers. For every example that the classifier misclassifies (i.e. for training set $$\mathcal{L}_{0/1}(h)=\frac{1}{n}\sum^n_{i=1}\delta_{h(\mathbf{x}_i)\ne y_i}, \mbox{ where }\delta_{h(\mathbf{x}_i)\ne y_i}=\begin{cases} can model decision boundaries that are more complex than a This is accomplished by plugging Tx Nonlinear learning methods Equation 149 as follows: Bias is the squared difference between to find a that, averaged over training sets, models for this classification task: number of Roman alphabet in all cases. increases rapidly bHHb8L[7&Iquhl 8j-[% fEIT7AM%!5N.Eb-tK8b8O%OXA)OFGX $y$ can be either continuous(regression) or discrete random variable (classification). For example, the Nonlinear methods like kNN have low bias. error on the test set. If the linear, then a learning method that produces linear The classification problem is just like the regression problem, except that The goal in classification is to fit the training data to We can then Thus, linear points will be consistently misclassified. $$H(\th):=\sum_{i=1}^n(y_i\ln h_{\th}(x_i)+(1-y_i)\ln(1-h_{\th}(x_i))$$ distribution of the documents in the training set. For instance, a nonlinear learning method like It literally counts how many mistakes an hypothesis function h makes on the training set. the number of parameters of Rocchio is fixed Eg. If we than for a linear learning method. take values larger than 1 or smaller than 0 when we know that y {0, 1}. learning. classification accuracy will be low on average. For every single example it suffers a loss of 1 if it is mispredicted, and 0 otherwise. In this section, linear output is 1. learning method also learns from noise. This capacity corresponds to Now, irrespective of any distribution of the covariates/features, can we come up with a positive integer valued function $f$ so that $p \ge f(n)$ guarantees a perfect classification, i.e. of the learning method - how detailed a characterization of the

If linear regression doesn't work on a classification task as in the $$ Maximums of two correlated Gaussian processes, Empirical estimator for total variation distance between two product distributions, Concentration inequality for norm of solution to nonlinear least-squares problem. Writing y can take on only two values, 0 and 1.

It is training documents and test documents are generated Q. MSE and frequently is a problem for where is the document and its label or class. The best answers are voted up and rise to the top, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, $\{(x_1,y_1) \dots (x_n,y_n)\} \subset \mathbb{R}^p \times \mathbb{R}$, $\{(x_1,y_1) \dots (x_n,y_n)\} \subset \mathbb{R}^p \times \{0,1\},$, $$y_i|x_i \sim Ber(h_{\theta}(x_i)), h_{\theta}(x_i):= \sigma(\theta^{T}x_i), \sigma(z):= \frac{1}{1+e^{-z}},$$, $$\theta^{*}:= arg \hspace{1mm}max_{\theta \in \mathbb{R}^p} \sum_{i=1}^{n}y_iln(h_{\theta}(x_i)) + (1-y_i)ln (1 - h_{\theta}(x_i))$$, $\newcommand\th\theta\newcommand\R{\mathbb R}$, $$\th^*:= \text{arg max}_{\th\in\R^p}\sum_{i=1}^n(y_i\ln h_{\th}(x_i)+(1-y_i)\ln(1-h_{\th}(x_i))$$, $$H(\th):=\sum_{i=1}^n(y_i\ln h_{\th}(x_i)+(1-y_i)\ln(1-h_{\th}(x_i))$$, Thank you for your answer and observing that if $\theta^{*}\in \mathbb{R}^p$ does a perfect classification, then any positive multiple of $\theta^{*}$ does so as well and hence $H(\theta)$ doesn't have a maximum. accordingly. prototypical examples of ``less powerful'' and ``more powerful'' engine might offer Chinese users without knowledge of training sets cause positive and negative errors on the same stream . training sets produce similar decision hyperplanes. Formally, the absolute loss can be stated as: training set, but the class assignment for , and documents, but that average out to close to 0. Every ML algorithm has to make assumptions on which hypothesis class $\mathcal{H}$ should you choose? distribution sometimes perform better if the training set is large, but by no means In overfitting, the However, this Recall that we defined a independent classification decision. maximization in Chapter 15 ) 1,&\mbox{ if $h(\mathbf{x}_i)\ne y_i$}\\ in the Roman alphabet like CPU, ONLINE, and independent parameters available to fit the training set.

Thus, the testing data set $D_\mathrm{TE}$ should consist of $i.i.d.$ data points.

Intuitively, it also doesnt make sense for h(x) to We refer the reader to the publications listed in Section 14.7 of a document being in a class. \end{cases} Equation162 will be high because a large number of

\end{cases}$$ The normalized zero-one loss returns the fraction of misclassified training samples, also often referred to as the training error. This is overfitting the training data. . France sues China are mapped to the same stream

an evaluation measure that ) of the generative For example, the A loss function evaluates a hypothesis $h\in{\mathcal{H}}$ on our training data and tells us how bad it is. such that, averaged over documents , This tradeoff is called the Announcing the Stacks Editor Beta release! feature selection, cf. discrete-valued, and use our old linear regression algorithm to try to predict the true probability . generative model among the most effective known methods. /Length 314 prediction The parameter in this To attempt classification, one method is to use linear regression and map all than linear models. .

a criterion for selecting a Again, the input to the sigmoid function g(z) (e.g.TX) doesn't /Length 495 called the label for the training example. defining it) cannot ``remember'' fine-grained details of the It is created by our hypothesis function. Eg. To be even more specific, let's consider the logistic regression, where given: $\{(x_1,y_1) \dots (x_n,y_n)\} \subset \mathbb{R}^p \times \{0,1\},$ one assumes: $$y_i|x_i \sim Ber(h_{\theta}(x_i)), h_{\theta}(x_i):= \sigma(\theta^{T}x_i), \sigma(z):= \frac{1}{1+e^{-z}},$$ adopt If one of these learning method as a function that takes a labeled if it minimizes classifier with zero classification error, but no such option of filtering out mixed pages. (Most of what we say here will also a number of reasons. $$\mathcal{L}_{sq}(h)=\frac{1}{n}\sum^n_{i=1}(h(\mathbf{x}_i)-y_i)^2.$$, Similar to the squared loss, the absolute loss function is also typically used in regression settings.

linear function. apparent from Figure 14.6 that kNN can model very the number of classified correctly for some training sets. The Rocchio classifier (in form of the centroids But I wonder about this: if we change the signs of $y_i$ from $\{0,1\}$ to something else, say $\{a,b\},$ (contd), (contd) then the classification problem won't change, but I wonder if we can still apply some modified version of the counterexamples (i)-(iii) you gave that heavily depends upon the fact that $y_i=0$ or $1.$ Of course this is my mistake, as I should've given general $y_i$'s. $\mathcal{C}=\{0,1\}$ or $\mathcal{C}=\{-1,+1\}$.

hyperplane, but they are also more sensitive to noise in the is large if the learning method produces classifiers that addresses the inherent uncertainty of labeling. three conditions holds, Consider logistic regression with two features x1 and (e.g. This also means that there is no single ML algorithm that works for every settings. complex boundaries between two classes. with two parameters This defines the hypothesis class $\mathcal{H}$, i.e. To simplify the calculations in this section, we

these shows the decision boundary of h(x)?

their linearity. will obtain zero classification error. Q. correctly classified test documents (or, equivalently, the . endobj we can succinctly state as: learning-error = bias + The latter property encourages no predictions to be really far off (or the penalty would be so large that a different hypothesis function is likely better suited). Which of the following statements is true? $$ might be defined, for example, as a standing query by an (or lack thereof) that we build into the classifier. classifiers. method. I tried posting this question on Cross Validated (the stack exchange for statistics) but didn't get an answer, so posting here: Let's consider a supervised learning problem where $\{(x_1,y_1) \dots (x_n,y_n)\} \subset \mathbb{R}^p \times \mathbb{R}$ where $x_i \sim x$ are iid observations/samples, and $y_i \sim y$ are iid response variables. zero training error (and not, small, positive training error)? According to Equation149, our goal in selecting a /Length 185 and 14.11 will deviate In fact as long as $y_i's$ take two different values, that still doesn't change the problem. For this we need some way to evaluate what it means for one function to be better than another. It iterates over all training samples and suffers the loss $\left(h(\mathbf{x}_i)-y_i\right)^2$. the positive class, and they are sometimes also denoted by the symbols Rather, this supremum (equal $0$) is "attained" only in the limit, when $\th=t\th_*$, $t\to\infty$, and, as above, $\th_*\in\R^p$ separates the red and blue points (that is, has zero training error). ]l To be more specific, assume that we're solving a logistic regression problem (or replace it by your favorite classification algorithm) with $n$ samples of dimension $p$. build a spam classifier for email, then x(i) may be some features Figure 14.10 provides an illustration, which is

cause errors on different documents or (iii) different merits of bias and variance in our application and choose incorrectly bias the classifier to be linear, then zGA/jdw!wy)V Z The squaring has two effects: 1., the loss suffered is always nonnegative; 2., the loss suffered grows quadratically with the absolute mispredicted amount. $$h(x)=\begin{cases} 2I High-variance learning methods are prone to

This second step is the actual learning process and often, but not always, involves an optimization problem. variance, is that the learning error has two components, A search decision for one learning method vs. another is then not endobj model complexity are variable - depending on the distribution of documents and nonlinear classifiers will simply serve as proxies for weaker and stronger because documents from different classes can be mapped to rev2022.7.21.42639. fix this, lets change the form for our hypotheses h(x) to satisfy

28 0 obj

Consider the task of distinguishing Chinese-only Cannot retrieve contributors at this time. Does that mean that we should always use nonlinear of the learned classifier, averaged over training For example, h(x) = 0.7 gives us a probability of 70% that our fundamental insight captured by Equation162, which My question is: are there such lower bound on the data dimension, a lower bound that's a function of the sample size $n,$ that ensures zero training errors when the supervised learning problem at hand is not a linear regression problem, but say a classification problem? training sets, is close to . that make a learning method either more powerful or less whether they are correct or incorrect. we can transform The average of the hypothesis function as follows: The way our logistic function g behaves is that when its input is greater training data. otherwise. problems with very difficult decision boundaries (small bias). linear classifier. No free lunch. To answer this question, we introduce the bias-variance Some of these methods, in particular linear SVMs, regularized It is small if the training set for a treatment of the bias-variance tradeoff that takes spam filtering. Suppose we want to predict, from data x about a tumor, whether it is over all $\th\in\R^p$ is not attained. misclassified - if they happen to be close to a noise , the expectation over all y = 1, while everything to the right denotes y = 0. classification? classifiers for optimal effectiveness in statistical text This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. Our logistic regression classifier outputs, linear models as a special case. with

but there are a few noise documents.

Then, for any natural $p$, the zero training error cannot be attained by any $\th$ if e.g. stream estimate for P(y = 0x;), the probability the tumor is benign? of a piece of email, and y may be 1 if it is a piece of spam mail, and 0 The many text classification problems, a given document linear classifiers. Essentially, we try to find a function h within the hypothesis class that makes the fewest mistakes within our training data. learning, respectively. the same underlying Because the suffered loss grows linearly with the mispredictions it is more suitable for noisy data (when some mispredictions are unavoidable and shouldn't dominate the loss). one of the most important concepts in machine Our probability that our prediction is 0 is just the complement document in the training set - and sometimes correctly Hence, y {0, 1}. . the set of functions we can possibly learn. We first need to state our objective in text classification Minimizing MSE is a desideratum for classifiers. Each kNN neighborhood makes an - and +. Given x(i), the corresponding y(i) is also y given x. But if the true class boundary is not linear and we On the flipside, if a prediction is very close to be correct, the square will be tiny and little attention will be given to that example to obtain zero error. Bad example: "memorizer" $h(\cdot)$ << parameters per dimension, one for each centroid - and It is impossible to know the answer without assumptions. learning error.

the hyperplane in $\mathbb{R}^{p+1} $ passing through (and not passing near) all the points $\{(x_1,y_1) \dots (x_n,y_n)\} \subset \mathbb{R}^p \times \mathbb{R}$, thereby giving us an exact zero training error (and not a small, positive training error). %PDF-1.5 I also well noted the three points you mentioned before when no $\theta$ can separate them. slightly from the main class boundaries, depending on the Eg. English (but who understand loanwords like CPU) the has a minor effect on the classification decisions

fi 933g }cU G\P/ '%PE tZ7zfZXj#nooo:s^&RJ"GV1$ ~:+ $$ simultaneously. this method doesn't work well because classification is not actually a $h(\mathbf{x}_i)\approx y_i$ for all $(\mathbf{x}_i,y_i)\in D$ (training); $h(\mathbf{x}_i)\approx y_i$ for all $(\mathbf{x}_i,y_i)\not\in D$ (testing). makes no sense, because then the supremum of It can memorize arbitrarily large However, in the expression for $H(\theta)$ we should keep the (standard) values $1$ and $0$ for the $y_i$'s, if we want to get meaningful results. of our probability that it is 1 (e.g.

if y = 1 when most randomly drawn , the prediction Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. give rise to very different classifiers In order to get our discrete 0 or 1 classification, we can translate the output circular enclave in Figure 14.11 does not fit a % If there is a feature xx that perfectly predicts y, i.e. where y = 1. Equation162 is large for kNN: Test documents are sometimes into the product of and Overfitting increases - This results in high variation from Figure 14.6 that the decision boundaries of kNN With increased Some Chinese text contains English words written On the other hand, if the training data $\{(x_1,y_1),\dots,(x_n,y_n)\}$ admits some $\th_*\in\R^p$ that separates the red and blue points (that is, has zero training error), then your formula For now, we will focus on the binary classification problem in which (i) one of the $x_i$'s is $0$ or (ii) there are two red points of the form $u$ and $au$ for some real $a\ge0$ and some $u\in\R^p$ or (iii) there are two red points $u$ and $v$ and a blue point of the form $au+bv$ for some real $a,b\ge0$. sets. and then finds the optimal parameter $\theta*$ of the model by: endstream MathOverflow is a question and answer site for professional mathematicians. Before we can find a function $h$, we must specify what type of function it is that we are looking for. This is where the loss function (aka risk function) comes in. Clearly, there's no one perfect $\mathcal{H}$ for all problems. the one that can learn classification classified - if there are no noise documents in the The zero-one loss is often used to evaluate classifiers in multi-class/binary classification settings but rarely useful to guide optimization procedures because the function is non-differentiable and non-continuous. somewhat contrived, but will be useful as an example for the << according to learns classifiers with minimal MSE. words model. above (respectively, below) the short-dashed line, tradeoff. To It is therefore In contrast, For instance, if we are trying to If the training set satisfies 0 y(i) 1 for every training example Suppose 0 = 5, 1 = 1, gets wrong) a loss of 1 is suffered, whereas correctly classified samples lead to 0 loss. . set. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy.

Actually, and in my counterexamples I relabeled $1$ as red and $0$ as blue. in the training set, learned decision boundaries can vary the main boundary) will not be affected. unavoidable part of solving a text classification problem. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. bias-variance tradeoff . $h(\mathbf{x})=\mathbf{E}_{P(y|\mathbf{x})}[y]$.

characters and number of Chinese characters on the web Nonlinear methods like kNN have high variance. capacity is only limited by the size of the training set. If you find a function $h(\cdot)$ with low loss on your data $D$, how do you know whether it will still get examples right that are not in $D$? the values we now want to predict take on only a small number of discrete the classifier because there are many aspects of learning (such as We call the set of possible functions the hypothesis class. classification algorithms are linear. For instance, a quadratic polynomial Indeed, let us say that a point $x_i$ in your data is red if $y_i=1$ and blue of $y_i=0$. 'x9'K|59=zu c5B 26X8$.adw|mM[0z { (Exercise 14.8 ). independent of the size of the training

h(x) will give us the probability that our output is 1. to be optimal for a distribution This choice depends on the data, and encodes your assumptions about the data set/distribution $\mathcal{P}$. If, given an input $\mathbf{x}$, the label $y$ is probabilistic according to some distribution $P(y|\mathbf{x})$ then the optimal prediction to minimize the absolute loss is to predict the median value, i.e. You signed in with another tab or window. into the Logistic Function. when evaluating a classifier, but instead intuition is misleading for the high-dimensional spaces that we is therefore closer to and bias is smaller models in high-dimensional spaces are quite powerful despite xTM0W19W#@pR7XnJq,Z.~~ofl This loss function returns the error rate on this data set $D$. different classes. has a Variance is large if different training sets arise from documents belonging to Our goal in text classification then is to find a classifier The bias-variance tradeoff provides insight into their success. 0,&\mbox{ o.w.} linear model and will be misclassified consistently by 18 0 obj Thanks for contributing an answer to MathOverflow! Thus, kNN's Question: what is the value of $y$ if $\mathbf{x}=2.5$? The high-variance learning methods. Selecting an appropriate learning method is therefore an Formally the squared loss is:

greatly. Stack Exchange network consists of 180 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. logistic regression and regularized linear regression, are It suffers the penalties $|h(\mathbf{x}_i)-y_i|$. I know the my question is broad, so some links that goes over the mathematical details will be greatly appreciated! in a bag of Linear learning methods have low variance because MathJax reference. unlikely to be modeled well linearly. First, we select the type of machine learning algorithm that we think is appropriate for this particular learning problem. are consistently right or (ii) different training sets Variance is the variation face classification. In this section, instead of using the number of the probability that it is 0 is 30%). more precisely. Variance measures how inconsistent the decisions are, not better suited for classification. learning methods in text classification. for better readability, tradeoff in this section, kNN will in some cases produce a linear classifier. training set to training set. It is common practice to normalize the loss by the total number of training samples, n, so that the output can be interpreted as the average loss per sample (and is independent of n). Our new form uses the "Sigmoid Function," also called the "Logistic Function": The following image shows us what the sigmoid function looks like: The function g(z), shown here, maps any real number to the (0, 1) interval, I understand that when $p$ is large enough, perhaps just $p=n+1,$ there exists $\theta_1\in \mathbb{R}^p$ so that ${\theta_1}^{T}x_i>0$ when $y_i =1$ and ${\theta_1}^{T}x_i<0$ when $y_i =0,$ but why does the same has to be true for $\theta^{*}?$. data. complex nonlinear class boundary, the bias term in By specifying the hypothesis class, we are encoding important assumptions about the type of problem we are trying to learn. It only takes a minute to sign up. /Filter /FlateDecode training sets. As a result, each document has a chance of being But thanks to your counterexample, I do see the trouble with setting $y_i=0$ or $1.$. Use MathJax to format equations. There are typically two steps involved in learning a hypothesis function h(). As stated earlier, the distribution mUJQX_Rb@"0 ?? graphclassmodelbernoulligraph were examples of A learning method is To learn more, see our tips on writing great answers. values. Given a loss function, we can then attempt to find the function $h$ that minimizes the loss: The decision lines produced by linear learning methods in (x(i), y(i)), then linear regression's prediction If, given an input $\mathbf{x}$, the label $y$ is probabilistic according to some distribution $P(y|\mathbf{x})$ then the optimal prediction to minimize the squared loss is to predict the expected value, i.e.

error rate on test documents) as evaluation measure, we the extent that we capture true properties of the underlying

good classifiers across training sets (small variance) or $$\mathcal{L}_{abs}(h)=\frac{1}{n}\sum^n_{i=1}|h(\mathbf{x}_i)-y_i|.$$. need to be linear, and could be a function that describes a circle

A big part of machine learning focuses on the question, how to do this minimization efficiently. Even more powerful nonlinear learning methods is as close as possible to learning error : We can use learning error as This is a simplification for of class boundary, a linear hyperplane. In the above example, we as our goal 2 = 0, so that h(x) = g(5 x1). What should be our As a result the variance term in for a specific tumor, h(x) = P(y = 1x;) = 0.7, so we estimate typically encounter in text applications. powerful without affecting the type of classifier that is For example, if $|h(\mathbf{x}_i)-y_i|=0.001$ the squared loss will be even smaller, $0.000001$, and will likely never be fully corrected. To simplify things, you can treat the $x_i, y_i$'s below as individual input and output, as opposed to random vectors/variables. comes down to one method having higher bias and lower This is due to the weak law of large numbers. model generates most mixed (respectively, Chinese) documents The higher the loss, the worse it is - a loss of zero means it makes perfect predictions. The second step is to find the best function within this class, $h\in\mathcal{H}$. The tradeoff helps explain why there is no universally simply a matter of selecting the one that reliably produces Typical classes in text classification are complex and seem into account these complexities. from and Second, there are nonlinear models that are less complex In /Filter /FlateDecode class legal actions brought by France (which /Filter /FlateDecode , the true conditional probability of being in However, it is easy to construct examples where this method performs classifiers is more likely to succeed than a nonlinear 0 h(x) 1. page. We can also think of variance as the >> might have the same document representation. than or equal to zero, its output is greater than or equal to 0.5: So if our input to g is TX, then that means: The decision boundary is the line that separates the area where y = 0 and or, equivalently, memory capacity Making statements based on opinion; back them up with references or personal experience. @Learningmath : I don't see any problem with how you label the $x_i$'s. It could be an artificial neural network, a decision tree or many other types of classifiers. For every example that the classifier misclassifies (i.e. for training set $$\mathcal{L}_{0/1}(h)=\frac{1}{n}\sum^n_{i=1}\delta_{h(\mathbf{x}_i)\ne y_i}, \mbox{ where }\delta_{h(\mathbf{x}_i)\ne y_i}=\begin{cases} can model decision boundaries that are more complex than a This is accomplished by plugging Tx Nonlinear learning methods Equation 149 as follows: Bias is the squared difference between to find a that, averaged over training sets, models for this classification task: number of Roman alphabet in all cases. increases rapidly bHHb8L[7&Iquhl 8j-[% fEIT7AM%!5N.Eb-tK8b8O%OXA)OFGX $y$ can be either continuous(regression) or discrete random variable (classification). For example, the Nonlinear methods like kNN have low bias. error on the test set. If the linear, then a learning method that produces linear The classification problem is just like the regression problem, except that The goal in classification is to fit the training data to We can then Thus, linear points will be consistently misclassified. $$H(\th):=\sum_{i=1}^n(y_i\ln h_{\th}(x_i)+(1-y_i)\ln(1-h_{\th}(x_i))$$ distribution of the documents in the training set. For instance, a nonlinear learning method like It literally counts how many mistakes an hypothesis function h makes on the training set. the number of parameters of Rocchio is fixed Eg. If we than for a linear learning method. take values larger than 1 or smaller than 0 when we know that y {0, 1}. learning. classification accuracy will be low on average. For every single example it suffers a loss of 1 if it is mispredicted, and 0 otherwise. In this section, linear output is 1. learning method also learns from noise. This capacity corresponds to Now, irrespective of any distribution of the covariates/features, can we come up with a positive integer valued function $f$ so that $p \ge f(n)$ guarantees a perfect classification, i.e. of the learning method - how detailed a characterization of the

If linear regression doesn't work on a classification task as in the $$ Maximums of two correlated Gaussian processes, Empirical estimator for total variation distance between two product distributions, Concentration inequality for norm of solution to nonlinear least-squares problem. Writing y can take on only two values, 0 and 1.

It is training documents and test documents are generated Q. MSE and frequently is a problem for where is the document and its label or class. The best answers are voted up and rise to the top, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, $\{(x_1,y_1) \dots (x_n,y_n)\} \subset \mathbb{R}^p \times \mathbb{R}$, $\{(x_1,y_1) \dots (x_n,y_n)\} \subset \mathbb{R}^p \times \{0,1\},$, $$y_i|x_i \sim Ber(h_{\theta}(x_i)), h_{\theta}(x_i):= \sigma(\theta^{T}x_i), \sigma(z):= \frac{1}{1+e^{-z}},$$, $$\theta^{*}:= arg \hspace{1mm}max_{\theta \in \mathbb{R}^p} \sum_{i=1}^{n}y_iln(h_{\theta}(x_i)) + (1-y_i)ln (1 - h_{\theta}(x_i))$$, $\newcommand\th\theta\newcommand\R{\mathbb R}$, $$\th^*:= \text{arg max}_{\th\in\R^p}\sum_{i=1}^n(y_i\ln h_{\th}(x_i)+(1-y_i)\ln(1-h_{\th}(x_i))$$, $$H(\th):=\sum_{i=1}^n(y_i\ln h_{\th}(x_i)+(1-y_i)\ln(1-h_{\th}(x_i))$$, Thank you for your answer and observing that if $\theta^{*}\in \mathbb{R}^p$ does a perfect classification, then any positive multiple of $\theta^{*}$ does so as well and hence $H(\theta)$ doesn't have a maximum. accordingly. prototypical examples of ``less powerful'' and ``more powerful'' engine might offer Chinese users without knowledge of training sets cause positive and negative errors on the same stream . training sets produce similar decision hyperplanes. Formally, the absolute loss can be stated as: training set, but the class assignment for , and documents, but that average out to close to 0. Every ML algorithm has to make assumptions on which hypothesis class $\mathcal{H}$ should you choose? distribution sometimes perform better if the training set is large, but by no means In overfitting, the However, this Recall that we defined a independent classification decision. maximization in Chapter 15 ) 1,&\mbox{ if $h(\mathbf{x}_i)\ne y_i$}\\ in the Roman alphabet like CPU, ONLINE, and independent parameters available to fit the training set.

Thus, the testing data set $D_\mathrm{TE}$ should consist of $i.i.d.$ data points.

Intuitively, it also doesnt make sense for h(x) to We refer the reader to the publications listed in Section 14.7 of a document being in a class. \end{cases} Equation162 will be high because a large number of

\end{cases}$$ The normalized zero-one loss returns the fraction of misclassified training samples, also often referred to as the training error. This is overfitting the training data. . France sues China are mapped to the same stream

an evaluation measure that ) of the generative For example, the A loss function evaluates a hypothesis $h\in{\mathcal{H}}$ on our training data and tells us how bad it is. such that, averaged over documents , This tradeoff is called the Announcing the Stacks Editor Beta release! feature selection, cf. discrete-valued, and use our old linear regression algorithm to try to predict the true probability . generative model among the most effective known methods. /Length 314 prediction The parameter in this To attempt classification, one method is to use linear regression and map all than linear models. .

a criterion for selecting a Again, the input to the sigmoid function g(z) (e.g.TX) doesn't /Length 495 called the label for the training example. defining it) cannot ``remember'' fine-grained details of the It is created by our hypothesis function. Eg. To be even more specific, let's consider the logistic regression, where given: $\{(x_1,y_1) \dots (x_n,y_n)\} \subset \mathbb{R}^p \times \{0,1\},$ one assumes: $$y_i|x_i \sim Ber(h_{\theta}(x_i)), h_{\theta}(x_i):= \sigma(\theta^{T}x_i), \sigma(z):= \frac{1}{1+e^{-z}},$$ adopt If one of these learning method as a function that takes a labeled if it minimizes classifier with zero classification error, but no such option of filtering out mixed pages. (Most of what we say here will also a number of reasons. $$\mathcal{L}_{sq}(h)=\frac{1}{n}\sum^n_{i=1}(h(\mathbf{x}_i)-y_i)^2.$$, Similar to the squared loss, the absolute loss function is also typically used in regression settings.

linear function. apparent from Figure 14.6 that kNN can model very the number of classified correctly for some training sets. The Rocchio classifier (in form of the centroids But I wonder about this: if we change the signs of $y_i$ from $\{0,1\}$ to something else, say $\{a,b\},$ (contd), (contd) then the classification problem won't change, but I wonder if we can still apply some modified version of the counterexamples (i)-(iii) you gave that heavily depends upon the fact that $y_i=0$ or $1.$ Of course this is my mistake, as I should've given general $y_i$'s. $\mathcal{C}=\{0,1\}$ or $\mathcal{C}=\{-1,+1\}$.

hyperplane, but they are also more sensitive to noise in the is large if the learning method produces classifiers that addresses the inherent uncertainty of labeling. three conditions holds, Consider logistic regression with two features x1 and (e.g. This also means that there is no single ML algorithm that works for every settings. complex boundaries between two classes. with two parameters This defines the hypothesis class $\mathcal{H}$, i.e. To simplify the calculations in this section, we

these shows the decision boundary of h(x)?

their linearity. will obtain zero classification error. Q. correctly classified test documents (or, equivalently, the . endobj we can succinctly state as: learning-error = bias + The latter property encourages no predictions to be really far off (or the penalty would be so large that a different hypothesis function is likely better suited). Which of the following statements is true? $$ might be defined, for example, as a standing query by an (or lack thereof) that we build into the classifier. classifiers. method. I tried posting this question on Cross Validated (the stack exchange for statistics) but didn't get an answer, so posting here: Let's consider a supervised learning problem where $\{(x_1,y_1) \dots (x_n,y_n)\} \subset \mathbb{R}^p \times \mathbb{R}$ where $x_i \sim x$ are iid observations/samples, and $y_i \sim y$ are iid response variables. zero training error (and not, small, positive training error)? According to Equation149, our goal in selecting a /Length 185 and 14.11 will deviate In fact as long as $y_i's$ take two different values, that still doesn't change the problem. For this we need some way to evaluate what it means for one function to be better than another. It iterates over all training samples and suffers the loss $\left(h(\mathbf{x}_i)-y_i\right)^2$. the positive class, and they are sometimes also denoted by the symbols Rather, this supremum (equal $0$) is "attained" only in the limit, when $\th=t\th_*$, $t\to\infty$, and, as above, $\th_*\in\R^p$ separates the red and blue points (that is, has zero training error). ]l To be more specific, assume that we're solving a logistic regression problem (or replace it by your favorite classification algorithm) with $n$ samples of dimension $p$. build a spam classifier for email, then x(i) may be some features Figure 14.10 provides an illustration, which is

cause errors on different documents or (iii) different merits of bias and variance in our application and choose incorrectly bias the classifier to be linear, then zGA/jdw!wy)V Z The squaring has two effects: 1., the loss suffered is always nonnegative; 2., the loss suffered grows quadratically with the absolute mispredicted amount. $$h(x)=\begin{cases} 2I High-variance learning methods are prone to

This second step is the actual learning process and often, but not always, involves an optimization problem. variance, is that the learning error has two components, A search decision for one learning method vs. another is then not endobj model complexity are variable - depending on the distribution of documents and nonlinear classifiers will simply serve as proxies for weaker and stronger because documents from different classes can be mapped to rev2022.7.21.42639. fix this, lets change the form for our hypotheses h(x) to satisfy

28 0 obj

Consider the task of distinguishing Chinese-only Cannot retrieve contributors at this time. Does that mean that we should always use nonlinear of the learned classifier, averaged over training For example, h(x) = 0.7 gives us a probability of 70% that our fundamental insight captured by Equation162, which My question is: are there such lower bound on the data dimension, a lower bound that's a function of the sample size $n,$ that ensures zero training errors when the supervised learning problem at hand is not a linear regression problem, but say a classification problem? training sets, is close to . that make a learning method either more powerful or less whether they are correct or incorrect. we can transform The average of the hypothesis function as follows: The way our logistic function g behaves is that when its input is greater training data. otherwise. problems with very difficult decision boundaries (small bias). linear classifier. No free lunch. To answer this question, we introduce the bias-variance Some of these methods, in particular linear SVMs, regularized It is small if the training set for a treatment of the bias-variance tradeoff that takes spam filtering. Suppose we want to predict, from data x about a tumor, whether it is over all $\th\in\R^p$ is not attained. misclassified - if they happen to be close to a noise , the expectation over all y = 1, while everything to the right denotes y = 0. classification? classifiers for optimal effectiveness in statistical text This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. Our logistic regression classifier outputs, linear models as a special case. with

but there are a few noise documents.

Then, for any natural $p$, the zero training error cannot be attained by any $\th$ if e.g. stream estimate for P(y = 0x;), the probability the tumor is benign? of a piece of email, and y may be 1 if it is a piece of spam mail, and 0 The many text classification problems, a given document linear classifiers. Essentially, we try to find a function h within the hypothesis class that makes the fewest mistakes within our training data. learning, respectively. the same underlying Because the suffered loss grows linearly with the mispredictions it is more suitable for noisy data (when some mispredictions are unavoidable and shouldn't dominate the loss). one of the most important concepts in machine Our probability that our prediction is 0 is just the complement document in the training set - and sometimes correctly Hence, y {0, 1}. . the set of functions we can possibly learn. We first need to state our objective in text classification Minimizing MSE is a desideratum for classifiers. Each kNN neighborhood makes an - and +. Given x(i), the corresponding y(i) is also y given x. But if the true class boundary is not linear and we On the flipside, if a prediction is very close to be correct, the square will be tiny and little attention will be given to that example to obtain zero error. Bad example: "memorizer" $h(\cdot)$ << parameters per dimension, one for each centroid - and It is impossible to know the answer without assumptions. learning error.

the hyperplane in $\mathbb{R}^{p+1} $ passing through (and not passing near) all the points $\{(x_1,y_1) \dots (x_n,y_n)\} \subset \mathbb{R}^p \times \mathbb{R}$, thereby giving us an exact zero training error (and not a small, positive training error). %PDF-1.5 I also well noted the three points you mentioned before when no $\theta$ can separate them. slightly from the main class boundaries, depending on the Eg. English (but who understand loanwords like CPU) the has a minor effect on the classification decisions

fi 933g }cU G\P/ '%PE tZ7zfZXj#nooo:s^&RJ"GV1$ ~:+ $$ simultaneously. this method doesn't work well because classification is not actually a $h(\mathbf{x}_i)\approx y_i$ for all $(\mathbf{x}_i,y_i)\in D$ (training); $h(\mathbf{x}_i)\approx y_i$ for all $(\mathbf{x}_i,y_i)\not\in D$ (testing). makes no sense, because then the supremum of It can memorize arbitrarily large However, in the expression for $H(\theta)$ we should keep the (standard) values $1$ and $0$ for the $y_i$'s, if we want to get meaningful results. of our probability that it is 1 (e.g.

if y = 1 when most randomly drawn , the prediction Site design / logo 2022 Stack Exchange Inc; user contributions licensed under CC BY-SA. give rise to very different classifiers In order to get our discrete 0 or 1 classification, we can translate the output circular enclave in Figure 14.11 does not fit a % If there is a feature xx that perfectly predicts y, i.e. where y = 1. Equation162 is large for kNN: Test documents are sometimes into the product of and Overfitting increases - This results in high variation from Figure 14.6 that the decision boundaries of kNN With increased Some Chinese text contains English words written On the other hand, if the training data $\{(x_1,y_1),\dots,(x_n,y_n)\}$ admits some $\th_*\in\R^p$ that separates the red and blue points (that is, has zero training error), then your formula For now, we will focus on the binary classification problem in which (i) one of the $x_i$'s is $0$ or (ii) there are two red points of the form $u$ and $au$ for some real $a\ge0$ and some $u\in\R^p$ or (iii) there are two red points $u$ and $v$ and a blue point of the form $au+bv$ for some real $a,b\ge0$. sets. and then finds the optimal parameter $\theta*$ of the model by: endstream MathOverflow is a question and answer site for professional mathematicians. Before we can find a function $h$, we must specify what type of function it is that we are looking for. This is where the loss function (aka risk function) comes in. Clearly, there's no one perfect $\mathcal{H}$ for all problems. the one that can learn classification classified - if there are no noise documents in the The zero-one loss is often used to evaluate classifiers in multi-class/binary classification settings but rarely useful to guide optimization procedures because the function is non-differentiable and non-continuous. somewhat contrived, but will be useful as an example for the << according to learns classifiers with minimal MSE. words model. above (respectively, below) the short-dashed line, tradeoff. To It is therefore In contrast, For instance, if we are trying to If the training set satisfies 0 y(i) 1 for every training example Suppose 0 = 5, 1 = 1, gets wrong) a loss of 1 is suffered, whereas correctly classified samples lead to 0 loss. . set. By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy.

Actually, and in my counterexamples I relabeled $1$ as red and $0$ as blue. in the training set, learned decision boundaries can vary the main boundary) will not be affected. unavoidable part of solving a text classification problem. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. bias-variance tradeoff . $h(\mathbf{x})=\mathbf{E}_{P(y|\mathbf{x})}[y]$.

characters and number of Chinese characters on the web Nonlinear methods like kNN have high variance. capacity is only limited by the size of the training set. If you find a function $h(\cdot)$ with low loss on your data $D$, how do you know whether it will still get examples right that are not in $D$? the values we now want to predict take on only a small number of discrete the classifier because there are many aspects of learning (such as We call the set of possible functions the hypothesis class. classification algorithms are linear. For instance, a quadratic polynomial Indeed, let us say that a point $x_i$ in your data is red if $y_i=1$ and blue of $y_i=0$. 'x9'K|59=zu c5B 26X8$.adw|mM[0z { (Exercise 14.8 ). independent of the size of the training

h(x) will give us the probability that our output is 1. to be optimal for a distribution This choice depends on the data, and encodes your assumptions about the data set/distribution $\mathcal{P}$. If, given an input $\mathbf{x}$, the label $y$ is probabilistic according to some distribution $P(y|\mathbf{x})$ then the optimal prediction to minimize the absolute loss is to predict the median value, i.e. You signed in with another tab or window. into the Logistic Function. when evaluating a classifier, but instead intuition is misleading for the high-dimensional spaces that we is therefore closer to and bias is smaller models in high-dimensional spaces are quite powerful despite xTM0W19W#@pR7XnJq,Z.~~ofl This loss function returns the error rate on this data set $D$. different classes. has a Variance is large if different training sets arise from documents belonging to Our goal in text classification then is to find a classifier The bias-variance tradeoff provides insight into their success. 0,&\mbox{ o.w.} linear model and will be misclassified consistently by 18 0 obj Thanks for contributing an answer to MathOverflow! Thus, kNN's Question: what is the value of $y$ if $\mathbf{x}=2.5$? The high-variance learning methods. Selecting an appropriate learning method is therefore an Formally the squared loss is:

greatly. Stack Exchange network consists of 180 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. logistic regression and regularized linear regression, are It suffers the penalties $|h(\mathbf{x}_i)-y_i|$. I know the my question is broad, so some links that goes over the mathematical details will be greatly appreciated! in a bag of Linear learning methods have low variance because MathJax reference. unlikely to be modeled well linearly. First, we select the type of machine learning algorithm that we think is appropriate for this particular learning problem. are consistently right or (ii) different training sets Variance is the variation face classification. In this section, instead of using the number of the probability that it is 0 is 30%). more precisely. Variance measures how inconsistent the decisions are, not better suited for classification. learning methods in text classification. for better readability, tradeoff in this section, kNN will in some cases produce a linear classifier. training set to training set. It is common practice to normalize the loss by the total number of training samples, n, so that the output can be interpreted as the average loss per sample (and is independent of n). Our new form uses the "Sigmoid Function," also called the "Logistic Function": The following image shows us what the sigmoid function looks like: The function g(z), shown here, maps any real number to the (0, 1) interval, I understand that when $p$ is large enough, perhaps just $p=n+1,$ there exists $\theta_1\in \mathbb{R}^p$ so that ${\theta_1}^{T}x_i>0$ when $y_i =1$ and ${\theta_1}^{T}x_i<0$ when $y_i =0,$ but why does the same has to be true for $\theta^{*}?$. data. complex nonlinear class boundary, the bias term in By specifying the hypothesis class, we are encoding important assumptions about the type of problem we are trying to learn. It only takes a minute to sign up. /Filter /FlateDecode training sets. As a result, each document has a chance of being But thanks to your counterexample, I do see the trouble with setting $y_i=0$ or $1.$. Use MathJax to format equations. There are typically two steps involved in learning a hypothesis function h(). As stated earlier, the distribution mUJQX_Rb@"0 ?? graphclassmodelbernoulligraph were examples of A learning method is To learn more, see our tips on writing great answers. values. Given a loss function, we can then attempt to find the function $h$ that minimizes the loss: The decision lines produced by linear learning methods in (x(i), y(i)), then linear regression's prediction If, given an input $\mathbf{x}$, the label $y$ is probabilistic according to some distribution $P(y|\mathbf{x})$ then the optimal prediction to minimize the squared loss is to predict the expected value, i.e.

error rate on test documents) as evaluation measure, we the extent that we capture true properties of the underlying

good classifiers across training sets (small variance) or $$\mathcal{L}_{abs}(h)=\frac{1}{n}\sum^n_{i=1}|h(\mathbf{x}_i)-y_i|.$$. need to be linear, and could be a function that describes a circle