Statistical Methods in EngineeringStudent Name:University15th January 2018
Task 1: OrthogonalizationAccording to the Gram-Schmidt Process. If we letV to be an inner product spaceand{v1,v2,...,vn}to be a set of linearly independent vectors that are inV. Then there exists anorthonormalsetofvectors{e1,e2,...,en}ofVsuchthatspan(v1,v2,...,vj)=span(e1,e2,...,ej)foreachj=1,2,...,n.The proof is done by induction.Supposej=1; we can then lete1=v1∥v1∥. But ∥v1∥≠0sincev1≠0bearing in mind that a set oflinearly independent vectors has no zero vector. From this it is clear thatspan(v1)=span(e1)based on the fact thatv1ande1differ only by∥v1∥as such they are scalar multiples of eachother. Again,∥e1∥=∥v1∥∥v1∥=1.Next considering the case whenj>1, and assuming a set of orthonormal vectors of j−1,{e1,e2,...,ej−1}such thatspan(v1,v2,...,vj−1)=span(e1,e2,...,ej−1).Now since{v1,v2,...,vn}is a linearly independent set of vectors, we have thatvj∉span(v1,v2,...,vj−1). Thus we definethe vectorejas:ej=vj−¿vj,e1>e1−¿vj,e2>e2−...−¿vj,ej−1>ej−1∥vj−¿vj,e1>e1−¿vj,e2>e2−...−¿vj,ej−1>ej−1∥Clearly∥ej∥=1. Next is to show thatejis orthonormal toe1,e2,...,ej−1. Given any integerksuchthat1≤k<j,weconsidertheinnerproductofejwithek.LetN=∥vj−¿vj,e1>e1−¿vj,e2>e2−...−¿vj,ej−1>ej−1∥. Then we have that:⟨ej,ek⟩=⟨vj−¿vj,e1>e1−¿vj,e2>e2−...−¿vj,ej−1>ej−1N,ek⟩¿1N⟨vj−¿vj,e1>e1−¿vj,e2>e2−...−¿vj,ej−1>ej−1,ek⟩¿1N(⟨vj,ek⟩−⟨vj,ek⟩)=0Now it is clear that the set of vectors{e1,e2,...,ej}is orthonormal. It is therefore evidentthatspan(v1,v2,...,vj)=span(e1,e2,...,ej)
Task 2: Linear models by handa)observation vector ⃗y and evidence matrix XSolutionObservation vector is;⃗y=[0124]Evidence matrix is;X=[1−1101112]b)We sought to find the equation y=β0+β1x of the least-squares line that best fits the given data points.Solution^β=[β0β1]=(X'X)−1X'y=([1111−1012][1−1101112])−1[1111−1012][0124]=[−0.80000.7429]Thus the equation is;y=−0.8000+0.7429x
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