{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW7solution

HW7solution - Econ 3190 Fall 2011 HW#7 Solution 1#4.2 By...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Econ 3190 Fall 2011 HW #7 Solution 1. #4.2 By de&nition, given X is a DRV, we know E ( X ) = P X x & f ( x ) From the pmf, we have f (0) = C 3 & (1 = 4) & (3 = 4) 3 = 27 = 64 ; f (1) = C 1 3 & (1 = 4) 1 & (3 = 4) 2 = 27 = 64 ; f (2) = C 2 3 & (1 = 4) 2 & (3 = 4) 1 = 9 = 64 ; f (3) = C 3 3 & (1 = 4) 3 & (3 = 4) = 1 = 64 ) E ( X ) = P X x & f ( x ) = 0 & f (0)+1 & f (1)+2 & f (2)+3 & f (3) = 48 = 64 = 3 = 4 2. #4.12 By de&nition, given X is a CRV, we know E ( X ) = R 1 &1 x & f ( x ) dx: ) E ( X ) = R 1 x & 2(1 ¡ x ) dx = 2 R 1 x ¡ x 2 dx = 2( x 2 = 2 ¡ x 3 = 3) j 1 = 1 = 3 3. #4.34 (a) To &nd the standard deviation of X; we need to compute the mean of X &rst. E ( X ) = P X x & f ( x ) = ¡ 2 & : 3 + 3 & : 2 + 5 & : 5 = 2 : 5 E ( X 2 ) = P X x 2 & f ( x ) = ( ¡ 2) 2 & : 3 + 3 2 & : 2 + 5 2 & : 5 = 15 : 5 ) & 2 X = E ( X 2 ) ¡ [ E ( X )] 2 = 15 : 5 ¡ 6 : 25 = 9 : 25 ) The standard deivation & X = p & 2 X = p 9 : 25 = 3 : 04 4. #4.43 By de&nition, given X is a CRV, we know E ( X ) = R 1 &1 x & f ( x ) dx = R 1 x & 1 4 e & x= 4 dx = ¡ R 1 xde & x= 4 = ¡ [( xe & x= 4 ) j 1 ¡ R 1 e & x= 4 dx ] = ¡ [0 ¡ ( ¡ 4 e & x= 4 ) j 1 ] = 4...
View Full Document

{[ snackBarMessage ]}

Page1 / 4

HW7solution - Econ 3190 Fall 2011 HW#7 Solution 1#4.2 By...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon bookmark
Ask a homework question - tutors are online