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1. Consider the following ARCH(1) model:yt = µ + utut = vtσt, vt is i.i.d. with mean 0 and variance 1σ2t = α0 + α1u2t−1.Please compute the following:(a). E[σ2t+1|Ωt](b). E[σ2t+2|Ωt](c). E[σ2t+3|Ωt](d). Derive a general formula for E[σ2t+s|Ωt] for any s ≥ 2.

Question

  1. Consider the following ARCH(1) model:yt = µ + utut = vtσt, vt is i.i.d. with mean 0 and variance 1σ2t = α0 + α1u2t−1.Please compute the following:(a). Eσ2t+1|Ωt. Eσ2t+2|Ωt. Eσ2t+3|Ωt. Derive a general formula for E[σ2t+s|Ωt] for any s ≥ 2.
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Solution 1

The ARCH(1) model is given by:

yt = µ + ut

ut = vtσt, where vt is i.i.d. with mean 0 and variance 1

σ2t = α0 + α1u2t−1

Let's compute the following:

(a). E[σ2t+1|Ωt]

σ2t+1 = α0 + α1u2t

Since E[ut|Ωt] = 0, we have E[σ2t+1|Ωt] = α0 + α1E[u2t|Ωt] = α0 + α1σ2t

(b). E[σ2t+2|Ωt]

σ2t+2 = α0 + α1u2t+1

Since E[u2t+1|Ωt] = E[v2t+1σ2t+1|Ωt] = σ2t+1 = α0 + α1σ2t, we have E[σ2t+2|Ωt] = α0 + α1E[σ2t+1|Ωt] = α0 + α1(α0 + α1σ2t)

(c). E[σ2t+3|Ωt]

σ2t+3 = α0 + α1u2t+2

Since E[u2t+2|Ωt] = E[v2t+2σ2t+2|Ωt] = σ2t+2 = α0 + α1σ2t+1, we have E[σ2t+3|Ωt] = α0 + α1E[σ2t+2|Ωt] = α0 + α1(α0 + α1(α0 + α1σ2t))

(d). Derive a general formula for E[σ2t+s|Ωt] for any s ≥ 2.

From the above computations, we can see a pattern. The expected value of σ2t+s given Ωt is a sum of terms, each of which is the product of α0 and α1 raised to a power. The power to which α1 is raised increases by 1 with each term, starting from 0 and ending at s-1. Therefore, the general formula is:

E[σ2t+s|Ωt] = α0(1 + α1 + α12 + ... + α1^(s-1)) + α1^s σ2t

This formula is valid for s ≥ 2.

This problem has been solved

Solution 2

The ARCH(1) model is given by:

yt = µ + ut

ut = vtσt

σ2t = α0 + α1u2t−1

where vt is i.i.d. with mean 0 and variance 1.

(a). E[σ2t+1|Ωt]

σ2t+1 = α0 + α1u2t

Since E[ut|Ωt] = 0, we have E[u2t|Ωt] = Var(ut|Ωt) = σ2t.

Therefore, E[σ2t+1|Ωt] = α0 + α1E[u2t|Ωt] = α0 + α1σ2t.

(b). E[σ2t+2|Ωt]

σ2t+2 = α0 + α1u2t+1

Since E[u2t+1|Ωt] = E[σ2t+1|Ωt] = α0 + α1σ2t, we have

E[σ2t+2|Ωt] = α0 + α1E[u2t+1|Ωt] = α0 + α1(α0 + α1σ2t) = α0(1 + α1) + α1^2σ2t.

(c). E[σ2t+3|Ωt]

σ2t+3 = α0 + α1u2t+2

Since E[u2t+2|Ωt] = E[σ2t+2|Ωt] = α0(1 + α1) + α1^2σ2t, we have

E[σ2t+3|Ωt] = α0 + α1E[u2t+2|Ωt] = α0 + α1(α0(1 + α1) + α1^2σ2t) = α0(1 + α1 + α1^2) + α1^3σ2t.

(d). General formula for E[σ2t+s|Ωt] for any s ≥ 2.

By induction, we can see that the pattern continues, and we can write a general formula:

E[σ2t+s|Ωt] = α0(1 + α1 + α1^2 + ... + α1^(s-1)) + α1^sσ2t.

This formula is valid for s ≥ 2.

This problem has been solved

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