Practice Midterm12

# Practice Midterm12 - Practice Midterm 2 Solution J Li Stat...

This preview shows pages 1–3. Sign up to view the full content.

Practice Midterm 2 Solution J. Li Stat 416, Spring 2005 The number of points assigned to each question is indicated before the question. The full score for the exam is 100. The inverse of a 2 × 2 matrix is ± ab cd ² - 1 = ± d ad - bc - b ad - bc - c ad - bc a ad - bc ² 1. The joint density of X and Y is f X,Y ( x, y )= x 2 2 e - xy 0 <x< 2 ,y > 0 . (a) (10) Find the marginal distribution of X , f X ( x ). (b) (10) Find the conditional distribution of Y given X = 1 2 , i.e., f Y | X ( y | 1 2 ). (c) (5) Find E ( Y | X = 1 2 ). (d) (10) Find the probability P ( Y> 1 | X = 1 2 ). Solution: (a) f X ( x Z -∞ f X,Y ( x, y ) dy = Z 0 x 2 2 e - xy dy = x 2 2 · ( - e - xy x ) | 0 = x 2 0 2 (b) f Y | X ( y | 1 2 f X,Y ( 1 2 ) f X ( 1 2 ) = ( 1 2 ) 2 2 e - 1 2 y 1 4 = 1 2 e - 1 2 y y> 0 (c) From part (b), it is shown that Y | X = 1 2 follows the exponential distribution with parameter λ =1 / 2. Hence E ( Y | X = 1 2 )=1 =2 . (d) P ( 1 | X = 1 2 R 1 1 2 e - 1 2 y dy = e - 1 2 . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2. A 4-state Markov chain { X 0 ,X 1 , ..., X n , ... } has a transition probability matrix as follows. The states are 0, 1, 2, and 3.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

Practice Midterm12 - Practice Midterm 2 Solution J Li Stat...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online