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canccc1122's Question
Math
Posted 3 months ago
2. Let Xt=σtZtX_{t}=\sigma_{t} Z_{t} be a GARCH(1,1)\operatorname{GARCH}(1,1), where {Zt}\left\{Z_{t}\right\} is a sequence of independent standard normal random variables, and
σt2=ω+αXt12+βσt12,
\sigma_{t}^{2}=\omega+\alpha X_{t-1}^{2}+\beta \sigma_{t-1}^{2},

where ω,α\omega, \alpha and β\beta are positive known constants. Assume that β<1\beta<1.
2.1. Show that the hh- step-ahead forecast σt+ht2\sigma_{t+h \mid t}^{2} of σt2\sigma_{t}^{2} can be written as
σt+ht2=ωj=0h2(α+β)j+(α+β)h1σt+12, for h>1
\sigma_{t+h \mid t}^{2}=\omega \sum_{j=0}^{h-2}(\alpha+\beta)^{j}+(\alpha+\beta)^{h-1} \sigma_{t+1}^{2}, \quad \text { for } \quad h>1
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Answer from Sia
Posted 3 months ago
Solution by Steps
step 1
To show the hh-step-ahead forecast σt+ht2\sigma_{t+h|t}^{2}, we start with the given GARCH(1,1) model: σt2=ω+αXt12+βσt12\sigma_{t}^{2}=\omega+\alpha X_{t-1}^{2}+\beta \sigma_{t-1}^{2}
step 2
We iterate the GARCH(1,1) model for hh steps ahead, starting from time t+1t+1 to t+ht+h
step 3
For h=2h=2, we have σt+2t2=ω+αXt+12+βσt+12\sigma_{t+2|t}^{2}=\omega+\alpha X_{t+1}^{2}+\beta \sigma_{t+1}^{2}. Since Xt+12=σt+12Zt+12X_{t+1}^{2}=\sigma_{t+1}^{2}Z_{t+1}^{2} and Zt+1Z_{t+1} is independent of past information, the expected value of Xt+12X_{t+1}^{2} given information up to time tt is σt+12\sigma_{t+1}^{2}
step 4
Thus, σt+2t2=ω+ασt+12+βσt+12=ω+(α+β)σt+12\sigma_{t+2|t}^{2}=\omega+\alpha \sigma_{t+1}^{2}+\beta \sigma_{t+1}^{2} = \omega+(\alpha+\beta)\sigma_{t+1}^{2}
step 5
For h=3h=3, we have σt+3t2=ω+αXt+22+βσt+22\sigma_{t+3|t}^{2}=\omega+\alpha X_{t+2}^{2}+\beta \sigma_{t+2}^{2}. Applying the expectation and using the result from step 4, we get σt+3t2=ω+(α+β)σt+2t2\sigma_{t+3|t}^{2}=\omega+(\alpha+\beta)\sigma_{t+2|t}^{2}
step 6
Substituting the expression from step 4 into step 5, we get σt+3t2=ω+(α+β)(ω+(α+β)σt+12)\sigma_{t+3|t}^{2}=\omega+(\alpha+\beta)(\omega+(\alpha+\beta)\sigma_{t+1}^{2})
step 7
Simplifying, we obtain σt+3t2=ω+(α+β)ω+(α+β)2σt+12\sigma_{t+3|t}^{2}=\omega+(\alpha+\beta)\omega+(\alpha+\beta)^{2}\sigma_{t+1}^{2}
step 8
Continuing this process, we find that for any h>1h>1, σt+ht2\sigma_{t+h|t}^{2} can be written as σt+ht2=ωj=0h2(α+β)j+(α+β)h1σt+12\sigma_{t+h|t}^{2}=\omega \sum_{j=0}^{h-2}(\alpha+\beta)^{j}+(\alpha+\beta)^{h-1}\sigma_{t+1}^{2}
Answer
The hh-step-ahead forecast σt+ht2\sigma_{t+h|t}^{2} of σt2\sigma_{t}^{2} can be written as σt+ht2=ωj=0h2(α+β)j+(α+β)h1σt+12\sigma_{t+h|t}^{2}=\omega \sum_{j=0}^{h-2}(\alpha+\beta)^{j}+(\alpha+\beta)^{h-1}\sigma_{t+1}^{2}, for h>1h>1.
Key Concept
Iterating the GARCH(1,1) model
Explanation
The hh-step-ahead forecast is derived by iteratively applying the GARCH(1,1) model equation, taking expectations, and simplifying the resulting expression.

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