STT 843 Key to Homework 1 Spring 2018

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1 STT 843 Key to Homework Spring 208 Due date: Feb 4, (a Because σ = 2, σ 22 = and ρ 2 = 05, we have σ 2 = ρ 2 σ σ22 = 2/2 Then, the mean and covariance of the bivariate normal is µ = ( 0 2 and Σ = ( 2 2/2 2/2 Then it follows that Σ = 3/2, Σ /2 = 3/2 and Σ = 2 ( 2/2 3 2/2 2 As a result, the bivariate normal density is f(x = (2π 3/2 exp { ( 2 (x, x 2 2Σ x } x 2 2 (b From part (a, we see that = 6π exp { 3 (x2 2x (x (x } (x µ T Σ (x µ = 2 3 (x2 2x (x (x (c The constant density contours satisfies the equation: (x µ T Σ (x µ = c The area that this contour contains is described by R(c = {(x, x 2 : (x µ T Σ (x µ c} To find the constant density contours that contains 50% probability, we need to determine c such that P {X R(c} = 05 This is equivalent to find c such that P {(X, X 2 : (X µ T Σ (X µ c} = 05 Because (X µ T Σ (X µ follows a chi-square distribution with degrees of freedom 2, the constant c should be the 50% quantile of the chi-square distribution That is c = χ 2 2,05 Thus, the contour that contains 50% probability is (x µ T Σ (x µ = χ 2 2,05 =

2 x x Figure : The density contour plot that contains 50% probability 44 (a Let c = (3, 2, Then, 3X 2X 2 + X 3 = c X is linear combination of normally distributed random vector Thus, c X has a normal distribution with mean c µ and variance c Σc Here c µ = 3 2+( 2 ( 3+ = 3, and c Σc = (3, 2, = 9 (b Let a = (a, a 2 Because X 2 and X 2 a X a 2 X 3 are both normally distributed, X 2 and X 2 a X a 2 X 3 are independent if the covariance between X 2 and X 2 a X a 2 X 3 is 0 Then, Cov(X 2, X 2 a X a 2 X 3 = σ 22 a σ 2 a 2 σ 23 = 3 a 2a 2 = 0 Any vector a = (a, a 2 satisfying condition 3 a 2a 2 = 0 ensures the independence between X 2 and X 2 a X a 2 X 3 46 (a Because σ 2 = 0, X and X 2 are independent (b Because σ 3 =, X and X 3 are not independent (c Because σ 23 = 0, X 2 and X 3 are independent (d Because (σ 2, σ 32 = (0, 0, (X, X 3 and X 2 are independent (e Because Cov(X, X + 3X 2 2X 3 = σ + 3σ 2 2σ 3 = 4 2( = 6, X and X + 3X 2 2X 3 are not independent 2

3 47 (a The joint ( distribution of (X, X 3 is normal with mean (, 2 and covariance By Result 46 in the textbook, we know that X 4 2 given X 3 = x 3 is normally distributed with mean µ + σ 3 σ33 (x 3 µ 3 = + ( 2 (x 3 2 = 05x and covariance σ σ 3 σ33 σ 3 = 4 ( 2 /2 = 35 (b The conditional distribution of X given X 2 = x 2, X 3 = x 3 is normal with mean ( ( 5 0 x2 + + (0, = 05x x 3 2 and covariance ( 5 0 σ (0, 0 2 ( 0 = (a Because V and V 2 are linear combinations of X,, X 4, V and V 2 both have multivariate normal distribution Therefore, we only need to determine the means and covariances of V and V 2 For V, the mean is and covariance is E(V = 4 E(X 4 E(X E(X 3 4 E(X 4 = 0 Cov(V = 6 Cov(X + 6 Cov(X Cov(X Cov(X 4 = 4 Σ Therefore, V is normally distributed with mean 0 and covariance 4 Σ For V 2, the mean is and covariance is E(V 2 = 4 E(X + 4 E(X 2 4 E(X 3 4 E(X 4 = 0 Cov(V 2 = 6 Cov(X + 6 Cov(X Cov(X Cov(X 4 = 4 Σ Therefore, V 2 is normally distributed with mean 0 and covariance 4 Σ (b Because (V, V 2 are linear combinations of X,, X 4, (V, V 2 is multivariate normally distributed From part (a, the mean is 0 The covariance between V and V 2 is Cov(V, V 2 = 6 Cov(X 6 Cov(X 2 6 Cov(X Cov(X 4 = 0 3

4 Therefore, the joint distribution of (V, V covariance ( 4 Σ Σ As a result, the joint density of (V, V 2 is 2 is normal with mean 0 and f(v, V 2 = (2π p 4 Σ exp{ 2 V ( 4 Σ V 2 V 2( 4 Σ V 2 } 48 The maximum likelihood estimator for µ is X n = 4 4 X i = i= The maximum likelihood estimator for Σ is ( 4 6 ˆΣ = 4 X X X n X n = ( /2 /4 /4 3/2 49 (a The distribution of (X µ Σ (X µ is chi-square distribution with degrees of freedom 6 (b The distribution of X is normal with mean µ and covariance Σ/20 The distribution of n( X µ is normal with mean 0 and covariance Σ (c The distribution of (n S is Wishart distribution with degrees of freedom 9 and covariance Σ, namely W 6 (9, Σ 42 (a The distribution of X is normal with mean µ and covariance Σ/60 (b The distribution of (X µ Σ (X µ is chi-square distribution with degrees of freedom 4 (c The distribution of n( X µ Σ ( X µ is chi-square distribution with degrees of freedom 4 (c The approximate distribution of n( X µ S ( X µ is chi-square distribution with degrees of freedom (a The QQ plot is given in Figure 2 Based on the QQ plot, it seems that most data points are along the straight line This might indicate that normal distribution is appropriate for this data set However, due to the small sample size, the sample quantile estimated based on 0 data points might not be very informative Therefore, the conclusion based on the QQ plot might not be reliable for this data set (b Using the formula given in equation (4-3 on page 8, the correlation coefficient r Q is defined by n j= r Q = (x (j x(q (j q n j= (x n (j x 2 j= (q (j q 2 4

5 Normal Q-Q Plot Sample Quantiles Theoretical Quantiles Figure 2: Q-Q plot of the annual rates of return given in question 423 The above coefficient r Q is The critical value for the test of normality can be found in Table 42 For our case, the sample size is 0 and significant level is 0%, the critical value is 0935 This means that r Q is larger than the corresponding critical value Thus, we fail to reject the null hypothesis and conclude that the normality assumption might be appropriate for this data set We still need to be cautious about this conclusion because the data might not be independent Therefore, the critical value used in our test might not be appropriate 426 (a Using R, we obtain the mean x = (5200, 248 and sample covariance as ( Then the squared distances are (b Comparing the squared distances with the 50% quantile of chi-square distribution with degrees of freedom 2 (ie, , we find that the 4-th, 5-th, 6-th, 8-th and 9-th data points fall within the estimated 50% probability contour of a bivariate normal distribution The proportion of the observations falling within the contour is 50% (c A chi-square plot is given in the following Figure 4 5

6 x x Figure 3: The density contour plot that contains 50% probability and scatter plot of data points Chi-square plot for χ 2 2 djs qchisq(ppoints(500, df = 2 Figure 4: The chi-square plot for checking normality 6

7 (d We observe from the chi-square plot given in part (c, most data points except the 0-th data point are along the theoretical line Therefore, these data are approximately bivariate normal 7