THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS

Transcription

1 THE UNIVERSITY OF HONG KONG DEPARTMENT OF MATHEMATICS MATH853: Linear Algebra, Probability and Statistics May 5, 05 9:30a.m. :30p.m. Only approved calculators as announced by the Examinations Secretary can be used in this examination. It is candidates responsibility to ensure that their calculator operates satisfactorily, and candidates must record the name and type of the calculator used on the front page of the examination script. Answer ALL 7 questions Note: You should always give precise and adequate explanations to support your conclusions. Clarity of presentation of your argument counts. So think carefully before you write. P. of 8

2 . (0 marks) a) Given a = [ ] T, b = [ 5] T, and c = [5 3 4] T, determine the dimension of W = span{a, b, c}. b) Express b = α+β, where α is along the direction of a, and β is orthogonal to a. c) Find a unit vector d such that d is orthogonal to any vector in W. d) Express [ 5 ] T as a linear combination of a, β and d. [ ]. (5 marks) Given A =, determine (A 3 ) (5 marks) Let B = B is It is known that one of the eigenvalues of a) Is B positive definite, negative definite or indefinite? Justify your answer. b) Find x that maximizes the quadratic from x T Ax subject to the constraint x =. c) Prove that for an orthonormal matrix U, we must have U U T =. Use this result to compute B. 4. (0 marks) a) Find the set {z C : z + z + = 0}, and for each element in this set, find its modulus. b) Let z = cos θ + i sin θ, where θ R. Find where n N. z n z n, 5. (0 marks) You know that a certain letter is equally likely to be in any one of three different folders. Let α i be the probability that you will find your letter upon making a quick examination of folder i if the letter is, in fact, in folder i, i =,, 3. Suppose you look in folder and do not find the letter. What is the probability that the letter is in folder? 6. (5 marks) a) Given E(X) = 3 and V ar(x) =, find E[( + X) ] and V ar(4 + X). P. of 8

3 b) For any random variable X following Poisson(λ), λ > 0, find E[X ]. c) If the probability density function of X is equal to { ce x, 0 x < f(x) =, 0, x < 0 find c and for any t > 0, find P (X > + t X > t). 7. (5 marks) The following table gives the values of Φ(z), where Φ(z) = z π e x dx. z P.3 of 8

4 a) Use the linear interpolation method to find k such that P (Z k) = 0.90, where Z follows N(0, ). b) For any X following N(µ, σ ), find P (µ σ X µ + σ). c) To study the proportion of defected electronic devices from an assembly line, a survey has been conducted and a sample of 000 has been obtained, among which 90 are defected. Construct a 90% confidence interval for the proportion of defected devices. Suggested Answers END OF PAPER. a) Put a, b and c as the rows of a matrix. Then we perform Gaussian elimination: Since there are only two non-zero rows, W is of dimension. (use Gaussian elimination 3 points, getting correct answer points). b) We project b onto a, and we have α = bt a a T a a = 3 6 a. component β = b α = = The orthogonal (use of projection 3 points, use subtraction to get the orthogonal component points) c) Let d = [d d d 3 ] T. We need d T a = 0 and d T b = 0. This translates into { d + d + d 3 = 0 { d + d + 5d 3 = 0 d + d + d 3 = 0 d + 3d 3 = 0. P.4 of 8

5 This gives d = 3d 3, and d = d d 3 = 3d 3 d 3 = d 3. Putting together gives d = d 3 3. Normalizing it gives d = 3. (list two equations points, solve the correct direction points, normalization points) d) We need to solve c, c and c 3 such that 5 = c a + c β + c 3 d. Since a, β and d are orthogonal, c, c and c 3 are simply given by c = c = c 3 = [ 5 ]a = a T a [ 5 ]β β T β = 6 6 ( 7 6 ) + ( 6 ) + ( 4 = 66 ) 6 36 [ 5 ]d d T d = = (use projection points, correct coefficients ++). First find the eigenvalues by solving det(a λi) = 0: [ ] [ ] 0 det( λ ) = ( λ)( λ) 6 = 0 λ 3λ 4 = 0 (λ 4)(λ + ) = 0. = Therefore, the eigenvalues are λ = 4, λ =. (obtain correct eigenvalues + points) Now, we find the eigenvectors. The eigenvector corresponding to λ = 4 can [ be obtained ] by solving (A 4I)x = 0. Expanding the equation gives 3 x = 0. The eigenvector is taken as v 3 = [ 3] T. Similarly, we can obtain v = [ ] T. (obtain correct eigenvectors + points) Therefore, we can write (A ) 48 = ((VDV ) ) 48 = VD 48 V P.5 of 8

6 [ ] [ 4 0 where V =, and D = 3 0 expression and correctly distributing the index points) [ ] [ 0 0 Since D 48, and V 0 = 5 3 correct V points) (A ) 48 = [ ] [ ] [ ] 0 0 = (correct multiplications points) ]. (use of the diagonalization ], (correct D 48 points, [ 3/5 /5 3/5 /5 3. a) First we find the eigenvalue of B by solving det(b λi) = 0: 3 λ det( 3 λ ) = 0 3 λ (3 λ)[(3 λ) ] [(3 λ) ] + [ (3 λ)] = 0 λ 3 9λ + 4λ 0 = 0 (λ )(λ )(λ 5) = 0 So, the eigenvalues are,, 5, and they are all positive, and B is positive definite. (4 points for getting the eigenvalues correct, points for positive definite) b) The x is the eigenvector corresponding to the eigenvalue 5. So, x is solved from x = x = 0. So the direction of x is along [ ] T. Normalizing the eigenvector, we get x = 3. (knowing we need to find eigenvector corresponding to maximum eigenvalue point, solving correct eigenvector (including normalization) points) c) For an orthonormal matrix U, we have UU T = I. Taking determinant on both sides, we have U U T =. Furthermore, since B = U Λ U T, we have B = Λ = 0. (successfully prove the first part points, applying it to find B points, getting the final answer point) ]. P.6 of 8

7 4. Remark: Standard teaching materials. Solution: a) It follows from straightforward computations that the desired set contains two elements: z = + 3i each of which has modulus. b) By De Moivre s theorem, we have, z = 3i, z n = cos nθ + i sin nθ, z n = cos nθ i sin nθ, which further lead to z n = (cos nθ + i sin nθ) (cos nθ i sin nθ) = i sin nθ. zn 5. Remark: This is a new problem, but similar problems were discussed in class and in tutorial. Solution: Let F i, i =,, 3, be the event that the letter is in folder i; and let E be the event that a search of folder does not comp up with the letter. We desire P (F E). From Bayes formula, we obtain P (F E) = P (E F )P (F ) 3 P (E F i)p (F i ) = ( α )/3 ( α )/3 + /3 + /3 = α 3 α. 6. Remark: Variants of problems in the lecture notes and tutorials. Solution: a) It follows from the definition of variance that E[X ] = V ar(x) + E [X] = + 9 = 0. Then, straightforward computations give E[(X + ) ] = E[X + X + ] = E[X ] + E[X] + = = 7, and moreover, V ar(4 + X) = V ar(x) = 4V ar(x) = 4. P.7 of 8

8 b) We have E[X] = ( λ j j j=0 j! e λ ) = λ λj (j )! e λ = λ. j= Since E[X(X )] = j= ( ) λ j j(j ) j! e λ = j= λ λ j (j )! e λ = λ, we have E[X ] = E[X(X )] + E[X] = λ + λ. c) Solving the equation f(x)dx =, we obtain c =. And furthermore, applying the memoryless property of exponential distribution, we compute P (X > + t X > t) = P (X > ) = f(x)dx = e x dx = e Remark: Modified problems from lecture notes and other sources. Solution: a) It can be checked that from the table,.8 and.9 are the two closest candidates for k. Then, the linear interpolation method would yield that k is approximately equal to.8 + / b) Let Z = (X µ)/σ. Then Z follows N(0, ). It then follows from the table that P (µ σ X µ + σ) = P ( Z ) = (P (Z ) 0.5) c) We have n = 000, Y = 90 and therefore p = 0.9 and Z α/.645. Thus, the 90% C.I. is [ ] , P.8 of 8