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# R16.P1.T2 : Explain how GARCH models perform in volatility forecasting

#### Mat5

##### New Member
Hello,

I don't understand why using this equation E[o²n+t - VL] = ( a+ b ) . E[o²n+t-1 - VL] repeatedly
gives : E[o²n+t - VL] = ( a+ b )^t . ( o²n - VL )

Anyone knows ?

Thanks,
Mat

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
HI @Mat5 Thank you for the challenge, I haven't yet had the chance to try this. Let's do just the second forward day, where t = 2:
• Per E[σ^2(n+t) - V(L)] = (α+β)*E[σ^2(n+t-1) - V(L)], in the case of day t = 2 days forward:
• E[σ^2(n+2) - V(L)] = (α+β)*E[σ^2(n+1) - V(L)]
• We need E[σ^2(n+1)] = ω + α*µ^2(n) + β*σ^2(n), but notice Hull is assuming that the expected value of µ^2(n+t-1) = σ^2(n+t-1), or really E[µ^2(future t)] = E[ σ^2(future t) ], so that:
• E[σ^2(n+1)] = ω + α*µ^2(n) + β*σ^2(n) = ω + α*σ^2(n) + β*σ^2(n) = ω + (α + β)*σ^2(n), so now we have:
• E[σ^2(n+2) - V(L)] = (α+β)*E[σ^2(n+1) - V(L)] = (α+β)*[ω + (α + β)*σ^2(n) - V(L)] = (α+β)*[(α + β)*σ^2(n) + ω - V(L)], and now I would like to replace [ω - V(L)]:
• Since V(L) = ω/(1 - α - β) = ω/[1 - (α + β)] --> ω = V(L)*[1 - (α + β)] = V(L) - V(L)*(α + β) --> ω - V(L) = - V(L)*(α + β) ... which is perfectly convenient because I wasn't sure how the ω was dropping out (!):
• E[σ^2(n+2) - V(L)] = (α+β)*[(α + β)*σ^2(n) + ω - V(L)] = (α+β)*[(α + β)*σ^2(n) - V(L)*(α + β)] = (α+β)^2*[σ^2(n) - V(L)]. I can't promise that's the most elegant way, but it works, I hope that's helpful!

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#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi @Mat5 Right, that step is just the application of the GARCH(1,1). Just as today's updated estimate of variance is given by GARCH(1,1) per σ^2(n) = ω + α*µ^2(n-1) + β*σ^2(n-1), we can say that tomorrow's expected variance will a function of today's variance and return: E[σ^2(n+1)] = ω + α*µ^2(n) + β*σ^2(n), so we can drop the E[.] on the right side.

The linked thread is interesting. I have a simpler reason for accepting Hull's assumption that E[µ^2(n+t-1)] = σ^2(n+t-1). Please note I mistakenly wrote the right-hand side E[.] above, which is a mistake. The assumption that makes total sense to me is E[µ^2(n+t-1)] = σ^2(n+t-1). This is simply to expect that the latest return-squared is equal to the most recent variance estimate. Lacking any other information, our best guess of the squared return should be the most recent variance: that is the one squared return that would not change the variance estimate. We can think of GARCH(1,1) as weighted average of three variancess: γ*variance(long run) + β*(recent variance estimate) + α*(the "innovation" that is basically a one-day variance). The squared return is a one-day variance, so it's slightly re-weighting the average variance (technically, GARCH is more technical etc but IMO this is not "at variance" with the theory pun intended). Thanks!

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#### Mat5

##### New Member
Thanks a lot everything is clear now