# 2 and σ z cov z cov cx c σx c t 1 1 1 1 σ 11 σ 12

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2 and Σ Z = Cov( Z ) = Cov( CX ) = C Σ x C T = 1 - 1 1 1 σ 11 σ 12 σ 12 σ 22 1 1 - 1 1 = σ 11 - 2 σ 12 + σ 22 σ 11 - σ 22 σ 11 - σ 22 σ 11 + 2 σ 12 + σ 22 Note that if σ 11 = σ 22 , i.e., if X 1 and X 2 have equal variances, the off-diagonal terms in Σ Z vanish. This demonstrates the well-known result that the sum and differ- ence of two random variables with identical variances are uncorrelated . Q3. Suppose p = 3, and that a covariance matrix Σ is given by Σ = 4 1 2 1 9 - 3 2 - 3 25 = σ 11 σ 12 σ 13 σ 21 σ 22 σ 23 σ 31 σ 32 σ 33 Find the matrix V 1 2 and the correlation matrix ρ . STAT 394, 2018 2 Tutorial Solutions: Week 2
A3. Here V 1 2 = σ 11 0 0 0 σ 22 0 0 0 σ 33 = 2 0 0 0 3 0 0 0 5 and ( V 1 2 ) - 1 = 1 2 0 0 0 1 3 0 0 0 1 5 Consequently, the correlation matrix ρ is given by ρ = ( V 1 2 ) - 1 Σ( V 1 2 ) - 1 = 1 2 0 0 0 1 3 0 0 0 1 5 4 1 2 1 9 - 3 2 - 3 25 1 2 0 0 0 1 3 0 0 0 1 5 = 1 1 6 1 5 1 6 1 - 1 5 1 5 - 1 5 1 Q4. Let X T = ( X 1 X 2 X 3 X 4 X 5 ) be distributed as N 5 ( μ, Σ) with means μ T = ( μ 1 μ 2 μ 3 μ 4 μ 5 ) and covariance matrix Σ. Find the distribution of ( X 2 X 4 ) T . A4. Let X * = X 2 X 4 , μ * = μ 2 μ 4 , Σ * = σ 22 σ 24 σ 24 σ 44 and note with this assumed notation, X , μ , Σ can be rearranged and partitioned as X = X 2 X 4 X 1 X 3 X 5 , μ = μ 2 μ 4 μ 1 μ 3 μ 5 , Σ = σ 22 σ 24 σ 12 σ 23 σ 25 σ 24 σ 44 σ 14 σ 34 σ 45 σ 12 σ 14 σ 11 σ 13 σ 15 σ 23 σ 34 σ 13 σ 33 σ 35 σ

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