Assignment_4_S07_Solution

# Assignment_4_S07_Sol - Homework 4 Spring 2007 Solutions 1 Question Xt = tet t2 = c0 b1 Xt-12 Show that Xt is leptokurtic Solution E[Xt = 0 E[Xt2 =

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

Spring 2007 Solutions 1) Question: X t = σ t e t σ t 2 = c 0 + b 1 X t-1 2 Show that X t is leptokurtic. Solution: E[X t ] = 0 E[X t 2 ] = c 0 / (1 - b 1 ) E[X t 4 ] =3E(c 0 + b 1 X t-1 2 ) 2= 3(c 0 2 + 2 c 0 b 1 EX t-1 2 + b 1 2 EX t-1 4 ) (*) Case 1: If b 1 < sqrt(1/3), then E[X t 4 ] = 3c 0 2 [(1-b 1 2 )/(1-b 1 ) 2 * 1/(1-3b 1 2 )] and the kurtosis K(X t ) is given by K(X t ) = E(X t 4 ) / E(X t 4 ) 2 = 3c 0 2 [(1-b 1 2 )/(1-b 1 ) 2 * 1/(1-3b 1 2 )] / (c 0 / (1 - b 1 )) 2 = 3[(1-b 1 2 )/(1-3b 1 2 )] In order to show that K(X t ) increases as b 1 increases, we need to take the derivative with respect to b 1 . K’(X t ) = (12b 1 ) / (1-3b 1 2 ) 2 Clearly K’(X t ) > 0, which means that K(X t ) is always increasing. This means that b 1 = 0 is the minimum of this function. Therefore the minimum of K(X t ) over this range is 3. Therefore over this range K(X t ) > 3. Case 2: If sqrt(1/3) =< b 1 < 1, then from (*) we notice that E[X t 4 ] does not exist, i.e. equal to infinity. Thus, K(X

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

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

## This note was uploaded on 01/12/2012 for the course STAT 443 taught by Professor Yuliagel during the Spring '09 term at Waterloo.

### Page1 / 7

Assignment_4_S07_Sol - Homework 4 Spring 2007 Solutions 1 Question Xt = tet t2 = c0 b1 Xt-12 Show that Xt is leptokurtic Solution E[Xt = 0 E[Xt2 =

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

View Full Document
Ask a homework question - tutors are online