QMLE with normal distribution for GARCH - consistent & asymptoticaly normal? (R, fGarch) In the section Details of the documentation of the garchFit function we find the following statemement:

"QMLE stands for Quasi-Maximum Likelihood Estimation, which assumes normal distribution and uses robust standard errors for inference."

But in financial returns modeling we all know that the innovations present fat tails. So, do the GARCH estimation with the garchFit function fail if we assume QMLE as conditional distribution?
In the documentation we find then the following statement:

"Bollerslev and Wooldridge (1992) proved that if the mean and the volatility equations are correctly specified, the QML estimates are consistent and asymptotically normally distributed."

So, if we estimate the GARCH model of a financial time series with the QMLE method, have we a non-efficient BUT consistent and unbiased estimation?
 A: It sounds like your question is about the properties of the QMLE estimates of the $GARCH$ coefficients when the $GARCH$ process is heavy tailed. In general, you will still get consistent and asymptotically normal estimates via the QMLE, but they won't be efficient or unbiased.
If $y_t$ are the innovations in the financial returns, and you assume $y_t$ is a standard $GARCH(p,q)$, so $y_t = h_t \eta_t$, where $\eta_t$ are the innovations to the $GARCH$ process (normalized to $E[\eta_t] = 0$, $E[\eta_t^2] = 1$) and $h_t$ is the volatility process, then even if $\eta_t$ has finite moments (for example a $\eta_t \sim N(0,1)$), then $y_t$ can still have an infinite fourth moment (and hence a fat tail) since $y_t^2$ is regularly varying with index  $\alpha \in (1, \infty)$. The index $\alpha$ depends on the $GARCH$ coefficients and the distribution of $\eta$. For example, in a $GARCH(1,1)$, $\alpha$ is the solution to $E[a_1\eta^2 + b_1] = 1$. 
But the QMLE will still be consistent and asymptotically normal if $\eta$ is not heavy tailed. If $\eta$ is heavy tailed itself, then the QMLE will be converge slower than $\sqrt{n}$, and instead will converge to a stable distribution. 
