I am currently working on a project to compare different GARCH(1,1) models on a financial data set. I use the rugarch package in R, and everthing seemed fine at first. However, now that I have started to introduce the actual theory I have run into problems regarding the Fractionally Integrated GARCH (FIGARCH) introduced in R. Baillie et al..

A short recap:

The regular GARCH(1,1) is defined as

$$r_t = \sigma_t\epsilon_t, ~~~ \sigma_t^2 = \omega + \alpha r_{t-1}^2 + \beta\sigma_{t-1}^2.$$

Rewriting this model yields the ARMA representation:

$$r_t =\omega + (\alpha + \beta) r_{t-1}^2 + v_t - \beta v_{t-1}^2,$$

where $v_t = r_t^2 - \sigma_t^2$. Now R. Baillie et al. defines the IGARCH:

$$\phi(L)(1-L)r_t^2 = \omega + [1-\beta(L)]v_t,$$

where $L$ is the backshift operator and $\phi(L)$ defined by $\phi(L) \equiv [1-\alpha(L) - \beta(L)](1-L)^{-1}$ and is of order $m-1$, where $m$ is $\max\{p,q\}$.

They then says that replacing the $(1-L)$ with $(1-L)^d$ for $0<d<1$ yields the FIGARCH.

Now turning our attention to p. 15 in the dokumentation for the rugarch package we see that $\phi(L)$ is defined differently here, namely: $\phi(L) \equiv [1-\alpha(L)].$ Also, they include $0$ and $1$ in $d$ and specify that when $d=0$ is collapses to the regular GARCH and when $d=1$ to the IGARCH.

Now to my confusion/question:

In the FIGARCH(1,1) how should I define $\phi$ since it is of order zero according to R. Baillie et al.? Setting it equal to zero I doesn't do any good as well.

When I use the $\phi$ defined in the rugarch package for a FIGARCH(1,1) and setting $d$ to either 0 or 1, I cannot obtain the original GARCH either way. I simply need a $\beta$ term. Is there a mistake in the rugarch package in terms of the $\phi$? And does setting $d$ to either 0 or 1 actually make sense?

I have tried simply setting $\phi = (1-\alpha L - \beta L)$; however, this does not comply with setting $d=1$, as you then obtain a term containing the second order lagged value.

Has anyone encountered this problem before or are able to cast some light on the subject anyhow? Thank you.


1 Answer 1


The ARMA(m,p) representation of GARCH(p,q) is :

\begin{align*} \left[1-\alpha(L)-\beta(L)\right]r_{t}^{2} = w + [1- \beta(L)] v_{i} \end{align*} where \begin{align} &\alpha (L) =\sum_{i=1}^{q} \alpha_{i} L^{i} \qquad , \alpha (0)=0 \\ &\beta (L) =\sum_{i=1}^{p} \beta_{i} L^{i} \qquad , \beta (0)=0 \\ &m = \text{max}(p,q) \end{align}

Next Engle & Bollerslev (1) developed the IGARCH model using the new polynomial $\Phi (L)$ defined as : \begin{equation} \Phi (L) = 1- \sum_{i=1}^{m-1} \Phi_{i}L^{i} =\left[1-\alpha(L)-\beta(L)\right] (1-L)^{-1} \end{equation} where $\Phi(L) $ is a polynomial of order $m-1$ and $\phi(0)=1$ .

The Igarch is defined as follows :

\begin{align*} \Phi(L) (1-L) r_{t}^{2} = w + \left[1-\beta(L) \right]v_{i} \end{align*}

The figarch model is simply:

\begin{align*} \Phi(L) (1-L)^{d} r_{t}^{2} = w + \left[1-\beta(L)\right] v_{i} \end{align*}

So I think there is a typo in the rugarch documentation: page 15: $\Phi(L)=\sum_{i=1}^{m-1}\Phi_{i}L^{i}$ must be $\Phi(L)=1-\sum_{i=1}^{m-1}\Phi_{i}L^{i}$.

I finally understood the $ \Phi(L)= 1 - \alpha (L) $ (page 16) that is used in equation 60 of the rugarch documentation. I have played a bit with rugarch today and I noticed that:

the alpha coefficient in the output corresponds to the $\Phi_{i}$ coefficient of the formula.

rugarch doesn't print the $\alpha_{i}$ coefficients (despite they are labelled alpha), the definition $ \Phi(L)= 1 - \alpha (L) $ make sense if $\alpha (L)$ corresponds to the polynomial $\alpha (L)=\sum_{i=1}^{m-1} \Phi_{i}L^{i} $ with $\alpha (0)=0$. The problem is that the documentation also uses the symbol $\alpha (L)$ to define the arch polynomial and this is very confusing...

So to sum up the FIGARCH implementation in rugarch corresponds to FIGARCH(p,d,f) where f is the order of $\Phi(L)$ (f=m-1)

So the Figarch(1,d,1) (=p,d,f) corresponds to;

\begin{align*} (1-\Phi_{1} L) (1-L)^{d} \epsilon_{t}^{2} = w + [1-\beta_{1}L] \eta_{i} \end{align*}

Also the documentation does not indicate if the alpha coefficients specify as an input to a FIGARCH corresponds to the $\alpha_{i}$ or $\Phi_{i}$ coefficients. If I'm correct they correspond to the $\Phi_{i}$ coefficients.

Remark: At the time of writing, FIGARCH model is a recent feature of rugarch (the changelogs shows it has been added at 2017-10-30 - one year ago) so it may explain why the documentation is unclear. Also changelog indicates it is restricted to (1,d,1). the rugarch package has a very good reputation. It is a free, open source project and I thank the main author Alexios Ghalanos- and all the contributors !

FIGARCH(p,d,q) is confusing ? Let's use FIGARCH(p,d,f) !

Scholars usually employ FIGARCH(p,d,q) to describe in reality FIGARCH(p,d,f) where f refers to the order of $\Phi(L)$. In my opinion, this is very disturbing because we are used of associating the letter q with the order of the garch polynomial $\alpha(L)$. Unfortunately I think this is due to Baillie himself, because he didn't explicitly say it in his paper (in his paper the letter q corresponds to the order of $\Phi(L)$ and not to the order of $\alpha(L)$). I know it just a letter but it can cause a great misunderstanding...

To be clear, the FIGARCH(p,d,f) corresponds to :

  • Figarch(1,d,1)
    \begin{align*} \Phi(L) (1-L)^{d} \epsilon_{t}^{2} = w + [1-\beta_{1}L] \eta_{i} \\ \end{align*}
  • Figarch(1,d,0)
    \begin{align*} (1-L)^{d} \epsilon_{t}^{2} = w + [1-\beta_{1}L] \eta_{i} \end{align*}
  • Figarch(0,d,1)
    \begin{align*} \Phi(L) (1-L)^{d} \epsilon_{t}^{2} = w + \eta_{i} \end{align*}

So for the Figarch(1,d,1) if $d=0$ then we have a standard garch(1,1) where $ \phi_{1} = \alpha_{1}+ \beta_{1}$ : \begin{align*} \Phi(L) (1-L)^{d} \epsilon_{t}^{2} = w + [1-\beta_{1}L] \eta_{i} \\ (1-\Phi_{1} L) \epsilon_{t}^{2} = w + [1-\beta_{1}L] \eta_{i} \\ \end{align*}

I have written a small code with rugarch that show that Figarch(1,0,1) = Garch(1,1). See below:


# specify GARCH(1,1) model
garch11.spec = ugarchspec(variance.model = list(garchOrder=c(1,1)),
                          mean.model = list(armaOrder=c(0,0)),
                          fixed.pars=list(mu = 0, omega=0.1, alpha1=0.15,beta1 = 0.6))
# simulate GARCH(1,1) process
garch11.sim = ugarchpath(garch11.spec, n.sim=40000)

# specify FIGARCH(1,0,1) 
specFigarch = ugarchspec(mean.model=list(armaOrder=c(0,0)),
                  variance.model = list(model = "fiGARCH",submodel="GARCH", garchOrder = c(1,1)),
                  fixed.pars=list(delta = 0.00001)) # delta must be > 0 in rugarch

# Fit a FIGARCH(1,0,1) to a GARCH(1,1)
FGARCH.fit = ugarchfit(spec=specFigarch, data=garch11.sim@path$seriesSim, solver.control=list(trace = 1))

# estimate FIGARCH(1,0,1) coefficients


# "alpha_{1}"  corresponds to  phi_{1} = alpa_{1} + beta_{1}
# so you should get something close to  phi_{1} = 0.15  + 0.6 = 0.75 for "alpha1" .
# beta1 should be close to 0.6

(1) Engle, R. F., & Bollerslev, T. (1986). Modelling the persistence of Conditional Variances. Econometric Reviews, 5(1), 1–50.

(2) Baillie, R. T., Bolleslev, T., & Ole Mikkelsen, H. (1996). Fractionally integrated generalized autoregressive conditional heteroskedasticity. Journal of Econometrics, 6, 3–30.

PS I choosed the letter f for the order of $\Phi(L)$ because it sounds like the beginning of "figarch" and "phi"... ^^

  • $\begingroup$ Thanks for the answer, I appreciate your help; however, I fail to see how the beta lag operator is defined as you say it is. The original authors states it as $\sum_{i=1}^q \beta_i L^i$. How do you justify your definition? Wouldn't you obtain two $r_t^2$ terms when you move it to the other side of the equal sign? And the second thing: I still fail to see how the order of $m-1$ comes into play, or specifically how you divide by $(1-L)$ in the case of an order of (1,1) and setting $d=0$? Can you clarify? Thank you. $\endgroup$ Nov 27, 2018 at 7:36
  • $\begingroup$ Also your sum for the $\phi$ terms are negative and the author of the rugarch package states it without the 1 and positive. Will you also elaborate this choice? $\endgroup$ Nov 27, 2018 at 7:47
  • $\begingroup$ I have edited my post, I hope I answered your questions. $\endgroup$
    – Malick
    Nov 30, 2018 at 0:46
  • $\begingroup$ This makes much more sense now. Thank you. I don't have much time to go over it at the moment, so if you don't mind, I'll inspect it further during the weekend and accept your answer then if I don't find any questions/problems. $\endgroup$ Nov 30, 2018 at 9:05
  • $\begingroup$ Don't you need a Lag operator on $\beta_1$ in your definition of the FIGARCH$(1,d,1)$, that is: $(1-\Phi_1 L)(1-L)^d \epsilon_t^2 = w + [1-\beta_1 L]\eta_t$ or am I mistaken again? I cannot seem to get the calculations right without adding the lag operator to $\beta_1$. $\endgroup$ Dec 12, 2018 at 11:40

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