I have two datasets of log returns, one is clearly normal while the other is t-distributed. I want to fit these with a mutlivariate GARCH model. A multivariate GARCH model is defined as $$\mathbf{r}_t=\mathbf{H}_t\boldsymbol{\epsilon}_t$$ Where $\mathbf{H}_t$ is the conditional covariance matrix and $\boldsymbol{\epsilon}_t$ is white noise. In the litterature most authors state that $\boldsymbol{\epsilon}_t$ is IID(0,1). I have not understood the identicality requirement. Is it possible to have different distributions for the elements in the vector $\boldsymbol{\epsilon}_t$? If no, why not and what would be a workaround in this case?


1 Answer 1


To answer your question briefly, yes, it possible to have different distributions for the elements in the vector $\mathbf{\epsilon}_t$. Now an elaboration:

Your model formulation is incomplete.

  1. The mean vector $\mathbf{\mu}_t$ of $\mathbf{r}_t$ may be nonzero and time-varying, turning $\mathbf{r}_t=\mathbf{H}_t\mathbf{\epsilon}_t$ into $\mathbf{r}_t=\mathbf{\mu}_t+\mathbf{H}_t\mathbf{\epsilon}_t$ and adding an equation for $\mathbf{\mu}_t$.
  2. The variance dynamics should be characteristic of GARCH. There should be an equation for $\mathbf{H}_t$ containing an autoregressive component (lag of $\mathbf{H}_t$) and a "moving-average" component (lag of $\mathbf{\epsilon}_t$), or something roughly equivalent to that. Otherwise, multivariate GARCH is not a relevant name for the model.
  3. The distribution of $\mathbf{\epsilon}_t$ must be i.i.d. with zero mean and identity variance matrix. The i.i.d. requirement is stricter than white noise that you refer to in the title of your question.

Referring to point 3., there is no requirement for the components of $\mathbf{\epsilon}_t$ to have identical univariate distributions. E.g. copula-GARCH models build on the multivariate distribution of $\mathbf{\epsilon}_t$ being specified using arbitrary marginal distributions and an arbitrary copula (but still obeying the requirements in point 3.).

  • $\begingroup$ Thanks! Just to clarify, in point 3 you reiterate that the distribution must be i.i.d. do you mean independent? Otherwise, I do not understand how this adds up with the rest of the answer since using multiple distributions would make it not i.i.d. by definition. $\endgroup$
    – Isaac E
    Feb 9 at 13:30
  • $\begingroup$ @IsaacE, the multivariate distribution as a whole must be i.i.d. Regarding "identical", it means $\mathbb{\epsilon}_t$ and $\mathbb{\epsilon}_s$ must have the same distribution for any pair $(t,s)$. That does not mean the univariate distributions of the different components $\mathbb{\epsilon}_{t,i}$, $\mathbb{\epsilon}_{t,j}$ must be the same. One could be N(0,1), the other could be standardized Student-$t$ etc. $\endgroup$ Feb 9 at 13:41
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    $\begingroup$ I see, so I.I.D. means that the vector $\boldsymbol{\epsilon}_t$ has the same distribution over time but the elements of the vector itself do not necessarily have to have the same distribution. That clears it all up. $\endgroup$
    – Isaac E
    Feb 9 at 14:08
  • $\begingroup$ As usual with downvotes, I would appreciate a constructive comment to follow along. That would help me improve my answer and leave it in the best shape for the readers. $\endgroup$ Feb 9 at 14:52

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