Exam 2. Jeremy Morris. March 23, 2006

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1 Exam Jeremy Morris March 3, Consider a bivariate normal population with µ 0, µ, σ, σ and ρ.5. a Write out the bivariate normal density. The multivariate normal density is defined by the following equation. fx π p/ Σ / e x µ Σ x µ σ σ With Σ and σ σ σ ρ σ σ. For this example, we have the following definitions for Σ, and Σ. Σ Σ / 4 Σ 4 To complete the definition of the density function we will derive the squared generalized distance expression x µ Σ x µ. x µ Σ x µ x x 4 x x x x 4 x x x + x 8 x x 4x x + 4 x + 8x 6 x 4x x x x + 8x 6 ϕx, x 4x + 8 x 4 x x 3x + 8x + 3 8

2 Then the bivariate normal distribution function is defined as follows. fx, x π e ϕx,x b Write out the squared generalized distance expression x µ Σ x µ as a function of x and x. Derived in part a equation. c Determine and sketch the constant-density contour that contains 50% of the probability. To determine the constant-density contour, we need to calculate the eigenvalues and eigenvectors of Σ. This is done by solving for the roots of the equation Σ λi. λ λ / 0 λ λ + λ / 0 λ 3λ + 3/ 0 Using the quadratic equation, we get the eigenvalues Then we find the eigenvector for λ. And the eigenvector for λ. λ λ 3 3 Σe λ e x + x 0 x x 0 3 e 0 3 e 3 Σe λ e x + x 0 x x e e + 3

3 4.6 Let X be distributed as N 3 µ, Σ, where µ [,, ] and 4 0 Σ Which of the following random variables are independent? Explain. a X and X. X and X are independent because covx, X covx, X 0. b X and X 3. X and X 3 are not independent because covx, X 3 covx 3, X. c X and X 3. X and X 3 are independent because covx, X 3 covx 3, X 0. d X, X 3 and X. To determine the answer, we need to rearrange the covariance matrix and partition it. The new covariance matrix follows. Σ We conclude that X, X 3 and X are independent. e X and X + 3X X 3. We have the multivariate normal distribution [ ] X AX X + 3X X 3 Where A X [ ] X X X 3 And AX has the distribution N Aµ, AΣA. Here we show the matrix AΣA to determine independence. [ ] 4 0 AΣA [ ] [ ] And we conclude that X, X 3 and X are not independent. 3

4 4.9 Refer to Exercise 4.8, but modify the construction by replacing the break point by c so that { X if c X X c elsewhere X Show that c can be chosen so that CovX, X 0, but that the two random variables are not independent. First, we note from exercise 4.8 that X N0,. For c 0, we have For c very large we have CovX, X E[X X ] E[X ]E[X ] E[X ] varx + E[X ] CovX, X E[X X ] + E[X ]E[X ] E[X ] varx + E[X ] Since the covariance is a smooth function of c, then by the mean value theorem, CovX, X 0 at some point, but that the two random variables are not independent. 4.0 Show each of the following. a 0 B A B Factor 0 B so that we have the following. 0 B 0 B Then take the determinants to get 0 B AI 00 IB 00 A B b A C 0 A B for A 0. B As in part a, factor the determinant so that we have the following. A C 0 B 0 B I A C Then, by part a, we get 0 B A B, and I A C 4.

5 4.0 Show that if A is square, Let A A A A A A for A 0 A A A A A for A 0 A A A A A and factor so that we have the following. I A A A A A A Finally, we have I A A A A I A A A A 0 A A A A I A A A A 0 0 A A A A A A A I Taking the determinant of both sides, we get the following. I A I A A A A A A A A A A A A A A A 0 0 A For the second relation, we do the same by factoring A so that we have the following. A A I A A A A A A A A I I A A 0 A A A A A A A A I A A I A A Again, taking the determinant of both sides we get the following. I A I A A A A A A A A A A A 4.6 Let X i, i,,..., 4, be independent N p µ, Σ random vectors. 0 A A A A 5

6 a Find the marginal distributions for each of the random vectors V 4 X 4 X + 4 X 3 4 X 4 and V 4 X + 4 X 4 X 3 4 X 4 By result 4.8 in the text, V and V have the following distribution. n n c j µ, Σ N p j j Then we have V N p 0, 4 Σ and V N p 0, 4 Σ. b Find the joint distribution of the random vectors V and V defined in a. Also by result 4.8, V and V are jointly multivariate normal with covariance matrix n c j Σ b cσ j n b cσ Σ With c /4, /4, /4, /4 and b /4, /4, /4, /4. So that we have the covariance matrix [ 4 Σ 0 ] 0 4 Σ 6. A likelihood argument provides aditional support for pooling the two independent sample covariance matrices to estimate a common covariance matrix in the case of two normal populations. Give the likelihood function Lµ, µ, Σ, for the two independent samples of sizes n and n from N p µ, Σ and N p µ, Σ respectively. Show that this likelihood is maximized by the choices ˆµ x, ˆµ x and j c j b j ˆΣ [n S + n S ] n + n n + n n + n S pooled Here we note the density function for the multivariate normal distribution for reference. fx; µ, Σ π p/ Σ / e x µ Σ x µ 6

7 Then we have the likelihood function for the two independent samples as defined below. Lµ, µ, Σ n i n fx i ; µ, Σ fx j ; µ, Σ Lµ, ΣLµ, Σ This statement gives us directly that ˆµ x, ˆµ x. Using equation 4-3 in the text, the likelihood function can defined as Lµ, µ, Σ π Np/ exp n Σ N/ tr n Σ Φ X i + Φ X j Where j Φ i x x µ i x µ i n n By result 4.0 in the text, we let B Φ x i + Φ x j, b N n + n and substitute the MLE s for each mean, then the maximum is reached at i j i j ˆΣ n + n B And we get the result ˆΣ [n S + n S ] n + n n + n n + n S pooled

x. Figure 1: Examples of univariate Gaussian pdfs N (x; µ, σ 2 ).

x. Figure 1: Examples of univariate Gaussian pdfs N (x; µ, σ 2 ). .8.6 µ =, σ = 1 µ = 1, σ = 1 / µ =, σ =.. 3 1 1 3 x Figure 1: Examples of univariate Gaussian pdfs N (x; µ, σ ). The Gaussian distribution Probably the most-important distribution in all of statistics