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

• No School
• AA 1
• 3

This preview shows page 2 - 3 out of 3 pages.

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

#### You've reached the end of your free preview.

Want to read all 3 pages?

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern

Stuck? We have tutors online 24/7 who can help you get unstuck.
Ask Expert Tutors You can ask You can ask You can ask (will expire )
Answers in as fast as 15 minutes
Ask Expert Tutors