Esercizi svolti ar ma arch garch

IV parziale 1-12-2010 19-05-15

Esercizio 13) Identificate tra all'interno della classe dei modelli ARMA

• 1. y = e + theta e |theta| < 1 -> MA(1)
• 2. y = e + theta₁ e + theta₂ e -> MA(2)
• 3. y = alpha y + e -> AR(1)
• 4. y = alpha y + e + theta e -> ARMA(1,1)

3) Calcolare E(y)

• 1. y = e + theta e

E(y) = E(e + theta e)= E(e) + theta E(e)= 0 + 0

E(y) = 0

• 2. y = e + theta₁ e + theta₂ e

E(y) = E(e + theta₁ e + theta₂ e)

E(y) = E(e) + theta₁ E(e) + theta₂ E(e)

E(y) = 0

• 3) y = alpha y + e

E(y) = E(alpha y + e)

E(y) = alpha E(y) + E(e)

E(y) = alpha E(y)

alpha E(y) = alpha E(y)

alpha = 1 ma |alpha| < 1 dunque non si verifica -> stazionario in covarianza

mu = alpha mu

mu = 0

-> E(y) = E(y) = mu

mu = E(y) = 0

II parziale 1-10-2010 19-05-15

Esercizio 13) Identificate tra all'interno delle classi dei modelli ARMA

• 1. y_t = ε_t + θε_t-1 |θ| < 1 -> MA(1)
• 2. y_t = ε_t + θ₁ε_t-1 + θ₂ε_t-2 -> MA(2)
• 3. y_t = α y_t-1 + ε_t -> AR(1)
• 4. y_t = α y_t-1 + ε_t + θε_t-1 -> ARMA(1,1)

3) Calcolare E(y_t)

• 1. y_t = ε_t + θε_t-1

E(y_t) = E(ε_t + θε_t-1)= E(ε_t) + θE(ε_t-1)= 0

E(y_t) = 0

• 2. y_t = ε_t + θ₁ε_t-1 + θ₂ε_t-2

E(y_t) = E(ε_t + θ₁ε_t-1 + θ₂ε_t-2)

E(y_t) = E(ε_t) + θ₁E(ε_t-1) + θ₂E(ε_t-2)

E(y_t) = 0

• 3. y_t = α y_t-1 + ε_t

E(y_t) = E(α y_t-1 + ε_t)

E(y_t) = αE(y_t-1) + E(ε_t)

E(y_t) = αE(y_t-1)

μ = αμ

|α| < 1 -> stazionario in covarianza

μ = 0

E(y_t) = E(y_t-1) = μ

μ = E(y_t) = 0

y_t = alpha y_t-1 + epsilon_t + theta epsilon_t-1

E(y_t) = E(alpha y_t-1 + epsilon_t + theta epsilon_t-1)

E(y_t) = alpha E(y_t-1) + E(epsilon_t) + theta E(epsilon_t-1)

E(y_t) = alpha E(y_t-1)

ricordiamo che:

E(y_t) = E(y_t-1) = mu → ne consegue

mu = alpha mu

alpha = 1 ma |alpha| < 1

mu = 0

mu = E(y_t) = 0

3) Calcolare E(y_t epsilon_t)

• 1) E(y_t epsilon_t) = E[(epsilon_t + theta epsilon_t-1) epsilon_t]

= E[epsilon_t² + theta epsilon_t epsilon_t-1]

= E(epsilon_t²) + theta E(epsilon_t epsilon_t-1)

= sigma² + theta E(epsilon_t epsilon_t-1)

N.B. la prima componente è la varianza sigma²

Var(epsilon_t) = sigma² = E(epsilon_t²) - E(epsilon_t)²

sigma² = E(epsilon_t²)

la seconda componente è la covarianza

Cov(epsilon_t, epsilon_t-1) = E(epsilon_t epsilon_t-1) - E(epsilon_t) E(epsilon_t-1)

E(y_t epsilon_t) = E(epsilon_t²) = sigma²

E(y_t e_t) = 0 + 0 = 0

E(e_t e_t) = sigma²

E(y_t, e_t) = E[(alpha y_t-1 + e_t) e_t]

= alpha E(y_t-1 e_t) + E(e_t²)

= sigma²

E(y_t e_t) = E[(alpha y_t-1 + e_t + theta e_t-1) e_t]

= alpha E(y_t-1 e_t) + E(e_t²) + theta E(e_t-1 e_t)

= sigma²

3) var(y