Question

In: Computer Science

Assume that the weak learners are a finite set of decision stumps. We then train a...

Assume that the weak learners are a finite set of decision stumps. We then train a AdaBoost classifier.

Can the boosting algorithm select the same weak classifier more than once? Explain.

Solutions

Expert Solution

First let us try to understand what are weak learners.A weak classifier is simply a classifier that performs poorly, but performs better than random guessing. Now if use use decision stumps as weak learners then the algorithm can select same weak classifier more than once.

During the process of boosting, a weak classifier is first trained with training samples as usual:

As the weak classifier is less powerful, because of this all the training samples cannot be correctly classified. The fundamental idea behind boosting is to assign large weight to the (difficult) training samples that were not correctly classified by the first classifier and train the second classifier with the weights. Due to this weighting scheme, the second classifier trained in this way would correctly classify some of the difficult training samples.

As difficult training samples will have large weights, repeating this weighted learning procedure would lead to a classifier that can correctly classify the most difficult training samples.So yes Adaboost can select same weak classifier more than once.


Related Solutions

Machine learning Adaboost question: Assume that the weak learners are a finite set of decision stumps....
Machine learning Adaboost question: Assume that the weak learners are a finite set of decision stumps. We then train an AdaBoost classifier. Can the boosting algorithm select the same weak classifier more than once? Explain.
Prove or Disprove The set of all finite strings is undecidable. The set of all finite...
Prove or Disprove The set of all finite strings is undecidable. The set of all finite strings is recognizable
Let X be the set of all subsets of R whose complement is a finite set...
Let X be the set of all subsets of R whose complement is a finite set in R: X = {O ⊂ R | R − O is finite} ∪ {∅} a) Show that T is a topological structure no R. b) Prove that (R, X) is connected. c) Prove that (R, X) is compact.
Set up a spreadsheet-based decision model. Assume that they have the following information available:
Set up a spreadsheet-based decision model. Assume that they have the following information available:Jim makes an estimated $41,000 and Shirley makes an estimated $36,000They would like to put away 3% of their total income in a retirement account, up to a maximum of $4,000. Any amount the put in that account can be deducted from their total income for tax purposes.They are entitled to a personal exemption of $3,300 each. There is a standard deduction for married couples of $11,500,...
1) Why can we generally assume that the equilibrium concentration of a weak acid equals its...
1) Why can we generally assume that the equilibrium concentration of a weak acid equals its initial concentration? 2) Why is pure water a poor conductor of electricity? 3) Describe the differences and similarities between weak and strong acids.
Prove that a subset of a countably infinite set is finite or countably infinite
Prove that a subset of a countably infinite set is finite or countably infinite
Now set up the decision criteria. (Assume α = .01.) This question has two parts: a)...
Now set up the decision criteria. (Assume α = .01.) This question has two parts: a) state the number of tails; and b) state the critical value(s) of the test statistic.
8. Definition: A set A is finite if there exists a non-negative integer c such that...
8. Definition: A set A is finite if there exists a non-negative integer c such that there exists a bijection from A to {n ∈ N : n ≤ c}. (The integer c is called the cardinality of A.) (a) Let A be a finite set, and let B be a subset of A. Prove that B is finite. (Hint: induction on |A|. Note that our proof can’t use induction on |B|, or indeed refer to “the number of elements...
Let (G,·) be a finite group, and let S be a set with the same cardinality...
Let (G,·) be a finite group, and let S be a set with the same cardinality as G. Then there is a bijection μ:S→G . We can give a group structure to S by defining a binary operation *on S, as follows. For x,y∈ S, define x*y=z where z∈S such that μ(z) = g_{1}·g_{2}, where μ(x)=g_{1} and μ(y)=g_{2}. First prove that (S,*) is a group. Then, what can you say about the bijection μ?
Recall that the set {0,1}∗ is the set of all finite-length binary strings. Let f:{0,1}∗→{0,1}∗ to...
Recall that the set {0,1}∗ is the set of all finite-length binary strings. Let f:{0,1}∗→{0,1}∗ to be f(x1x2…xk)=x2x3…xkx1. That is, f takes the first bit of a string x and moves it to the end of x, so for example a string 100becomes 001; if |x|≤1, then f(x)=x Also, suppose that g:{0,1}∗→{0,1}∗ is a function such that g(x1…xk)=0x1…xk (that is, gg puts an extra 0 in front of the given string, so for example g(100)=0100. Everywhere in this question we...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT