This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: SOLUTION FOR HOMEWORK 2, STAT 4382 Well, welcome to your second homework. Problem 1. Denote the means of U , V and W as μ U , μ V ,and μ W , respectively. Also note that always Cov ( X,Y ) = E ( XY ) E ( X ) E ( Y ) . This formula is useful to simplify calculations. Write, E ( XY ) = E ( U + W )( V W ) = E ( UV + WV UW W 2 ) [using independence we continue] = μ U μ V + μ W μ V μ U μ W [ V ar ( W ) + μ 2 W ] . Similarly, E ( X ) E ( Y ) = E ( U + W ) E ( V W ) = ( μ U + μ W )( μ V μ W ) = μ U μ V + μ W μ V μ U μ W μ 2 W . Plug in and get Cov ( X,Y ) = V ar ( W ) = σ 2 . Remark: Adding a constant to random variables does not change their covariances (vari ances). As a result, we could assume that all means are zero. This would simplify the calculation. But I also think that working with means is a nice exercise. 2. Problem 2. Well, let us understand the Hint. Write A ∪ B = A ∪ ( A c B ) and note that events A and BA c are mutually exclusive. Also, I may write sometimes AB in place of A ∩ B to make expressions shorter. Now, if B = C ∪ D then we can continue A ∪ B = A ∪ ( A c B ) = A ∪ ( A c ( C ∪ D )) = A ∪ ( A c ( C ∪ C c D ) To continue, I should remember DISTRIBUTIVE LAWS ( E ∪ F ) ∩ G = ( E ∩ G ) ∪ ( F ∩ G ) , ( E ∩ F ) ∪ G = ( E ∪ G ) ∩ ( F ∪ G ) . It is a good idea to know these laws because they help a lot in solving different problems....
View
Full Document
 Fall '11
 krwaw
 Conditional Probability, Probability, Probability theory, 75%

Click to edit the document details