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 =

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

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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 =

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