2
$\begingroup$

Here is the proof that the limit of GARCH(1,1) $$\sigma_n^2 = \gamma V_L + \alpha u_{n-1}^2 + \beta\sigma_{n-1}^2$$ is equivalent to stochastic process of variance $$d V = \alpha(V_L - V) dt + \xi VdW.$$ We have already assumed return $u_{n-1}$ is $(0,\sigma_{n-1}^2)$-normal, then $u^2_{n-1}$ can't be normal. But author said that $$u^2_{n-1} = \sigma_{n-1}^2 + \sqrt{2}\sigma_{n-1}^2\varepsilon$$ where $\varepsilon$ is a random sample from a standard normal?enter image description here

$\endgroup$
  • $\begingroup$ From which book is your excerpt taken? $\endgroup$ – Gordon Aug 25 '17 at 18:38
  • $\begingroup$ @Gordon this is the solution of John Hull's book Options, Futures and Other Derivatives Solutions Manual Eighth Edition page 171 Problem 22.14 $\endgroup$ – A.Oreo Aug 26 '17 at 8:16
  • $\begingroup$ @Gordon I know that for a $\chi^2$-process $V,$ the limit of $V-E[V]$ trends to a Brownian motion on the distribution, but I don't how to take the limit here? $\endgroup$ – A.Oreo Aug 26 '17 at 8:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.