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"Frontmatter". In: Analysis of Financial Time Series

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84 CONDITIONAL HETEROSCEDASTIC MODELS(a) Percentage changes in exchange ratefx-0.6 -0.2 0.2••• •••• • •• • •• •••• ••• • • • •• ••• •• • • •• •• •• • •• •• • • • •• • • •• • • ••• ••• • • •• • •• • •• •••• •• •• • • • • • •• •• • • •• •• • •• • ••• • •• • •• • • •• • •• • •••• •• • • •• • • • ••• • •• • • • •• • •• • • • • • •• •• •• • • •• ••• • • •• ••• • •• • • •••• • •• • • •• • • •••• •• ••• •• •• •• ••• •• •• • •• • •• • • ••• •• • •• •• ••• • • •• •• • •• •• • • •• • • •• •• •• ••• •• • • •• ••• • •• • • ••• • ••• • • • • • •• •• • ••• •• • •• ••• • •• • •• • •• • • • •• • • •••• ••• • •• • •• •• •••• • •• •• • ••• •• • ••• •• ••• • ••• • ••••• •• •• • •• ••••• • ••• • • • • •• •• •• • • ••0 500 1000 1500 2000 2500time index(b) Squared seriessq-fx0.0 0.1 0.2 0.3 0.4•••••••••• • •• • • ••• • •• • • • • • • • • ••• • •• • • • ••• • • • • • • • •• • • • ••• ••• •• • • • • • • •• • •••0 500 1000 1500 2000 2500time index•Figure 3.2. (a) <strong>Time</strong> plot <strong>of</strong> 10-minute returns <strong>of</strong> the exchange rate between Deutsche Markand Dollar, and (b) the squared returns.Because a t is a stationary process with E(a t ) = 0, Var(a t ) = Var(a t−1 ) = E(a 2 t−1 ).Therefore, we have Var(a t ) = α 0 + α 1 Var(a t ) and Var(a t ) = α 0 /(1 − α 1 ). Becausethe variance <strong>of</strong> a t must be positive, we need 0 ≤ α 1 < 1. Third, in some applications,we need higher order moments <strong>of</strong> a t to exist and, hence, α 1 must also satisfy someadditional constraints. For instance, to study its tail behavior, we require that thefourth moment <strong>of</strong> a t is finite. Under the normality assumption <strong>of</strong> ɛ t in Eq. (3.5), wehaveTherefore,E(a 4 t | F t−1 ) = 3[E(a 2 t | F t−1 )] 2 = 3(α 0 + α 1 a 2 t−1 )2 .E(a 4 t ) = E[E(a4 t | F t−1 )]=3E(α 0 + α 1 a 2 t−1 )2 = 3E[α 2 0 + 2α 0α 1 a 2 t−1 + α2 1 a4 t−1 ].If a t is fourth-order stationary with m 4 = E(at 4 ), then we havem 4 = 3[α0 2 + 2α 0α 1 Var(a t ) + α1 2 m 4](= 3α02 1 + 2 α )1+ 3α1 2 1 − α m 4.1

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