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Spring 2007
Solutions
1) Question:
X
t
=
σ
t
e
t
σ
t
2
=
c
0
+ b
1
X
t1
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
t1
2
)
2=
3(c
0
2
+ 2 c
0
b
1
EX
t1
2
+ b
1
2
EX
t1
4
)
(*)
Case 1:
If b
1
< sqrt(1/3), then
E[X
t
4
] = 3c
0
2
[(1b
1
2
)/(1b
1
)
2
* 1/(13b
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
[(1b
1
2
)/(1b
1
)
2
* 1/(13b
1
2
)] / (c
0
/ (1  b
1
))
2
= 3[(1b
1
2
)/(13b
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
) / (13b
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.
 Spring '09
 YuliaGel
 Forecasting

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