# Is a linear combination of GARCH processes also a GARCH process?

If two time series follow a GARCH process, and a third is a linear combination of them, is the third also GARCH process?

• The answer probably depends on the type of GARCH process. Combining $\mathrm{GARCH}(1, 1)$ and $\mathrm{eGARCH}(1, 1)$ probably won't work. Do you specifically mean standard GARCH so that your question becomes: is $\mathrm{GARCH}(p, q)$ closed under linear combination? – Bob Jansen Jun 19 '12 at 18:16

I think there are a lot of different ways to specify this problem. For simplicity, consider independent Garch processes $$r_{1,t} \sim N\left(0,\sigma_{1,t}^{2}\right)$$ $$\sigma_{1,t}^{2} = \beta_{1,1}+\beta_{1,2}\varepsilon_{1,t-1}^{2}+\beta_{1,3}\sigma_{1,t-1}^{2}$$ and $$r_{2,t} \sim N\left(0,\sigma_{2,t}^{2}\right)$$ $$\sigma_{2,t}^{2} = \beta_{2,1}+\beta_{2,2}\varepsilon_{2,t-1}^{2}+\beta_{2,3}\sigma_{2,t-1}^{2}$$ where $\left[\begin{array}{cc} \varepsilon_{1,t} & \varepsilon_{2,t}\end{array}\right]\sim N\left(0,\left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right]\right)$.

In this case, the linear combination equals $$r_{3,t} = \alpha_{1}r_{1,t}+\alpha_{2}r_{2,t} \sim N\left(0,\alpha_{1}^{2}\sigma_{1,t}^{2}+\alpha_{2}^{2}\sigma_{2,t}^{2}\right)$$

Assuming the coefficients in the Garch equations are constrained to be positive and sum to less than or equal to one on the lagged values, then $r_{3,t}$ will also follow a Garch process as a result of inheriting the Garch variances of the other variables.

• good answer, but usually we work with logarithmic returns, if 'x1' and'x2' are returns and x3=a1*x1+a2*x2, and r1=ln(x1),r2=ln(x2), r3=ln(x3) then r3 isn't equal to r1+r2 – Qbik Jun 20 '12 at 19:42
• True, but the answer I provided was mainly for illustrative purposes since you weren't particularly clear about under what conditions, what kind of Garch, etc. – John Jun 20 '12 at 21:06
• I once tried deriving the conditional variance equation for a process that is a sum of two GARCH processes, and if I remember correctly, it looked quite different than a GARCH equation (perhaps because I considered dependent processes). So I am not sure your answer is correct. Could you derive the equation to show you are right? – Richard Hardy Sep 3 '16 at 19:39
• @RichardHardy The formula for $r_{3,t}$ follows from the above with basic statistics. The only thing that stands out in reviewing my (more than 4 year old) answer is that I would probably characterize $r_{3,t}$ as following a multivariate GARCH process since the variance depends on the errors of both processes (it's easier to express it this way than trying to only refer to $r_{3,t}$. That being said, I used a simple approach on purpose because the OP left things quite vague. If I were looking at dependent process, I probably would have used a DCC-GARCH model. – John Sep 6 '16 at 17:33
• It's harder than I thought. I can illustrate my point but I cannot prove it (so far). Then I also cannot disprove your claim (which also lacks a proof, to be fair). However, I hope the intuition behind my example will be sufficiently convincing for the time being. – Richard Hardy Sep 7 '16 at 8:39

No, a sum of two GARCH processes is generally not a GARCH process.

(I am not even sure whether there exists a nontrivial special case where the opposite holds.)

By GARCH I mean the classic definition of GARCH due to Bollerslev (1986), not an arbitrary variation like EGARCH, IGARCH, FIGARCH or whatever else.

Let me provide an example. Take two independent zero-conditional-mean processes $e_{1,t}$ and $e_{2,t}$. Let their conditional variances follow GARCH(1,1). Then the conditional variance equations of $e_{1,t}$ and $e_{2,t}$ are

\begin{aligned} \sigma_{1,t}^2 = \omega_1 + a_1 e_{1,t-1}^2 + b_1 \sigma_{1,t-1}^2; \\ \sigma_{2,t}^2 = \omega_2 + a_2 e_{2,t-1}^2 + b_2 \sigma_{2,t-1}^2. \\ \end{aligned}

Take $e_t$ to be the simplest possible linear combination of $e_{1,t}$ and $e_{2,t}$, namely, their sum:

$$e_t := e_{1,t} + e_{2,t}.$$

Will its conditional variance follow a GARCH process? If it would, we could express the conditional variance of $e_t$ as

$$\sigma_t^2 = \omega + \sum_{i=1}^s \alpha_i e_{t-i}^2 + \sum_{i=1}^r \beta_i \sigma_{t-i}^2$$

(a GARCH($s$,$r$) equation). To show the conditional variance of $e_t$ follows GARCH($s$,$r$) we need to find the appropriate $\omega$, $\alpha$s, $\beta$s, $s$ and $r$. Can this be done?

Let us start by writing the conditional variance of $e_t$ explicitly based on the fact that $e_t = e_{1,t} + e_{2,t}$ and the properties of $e_{1,t}$ and $e_{2,t}$. The conditional variance of $e_t$ will be the sum of the conditional variances of $e_{1,t}$ and $e_{2,t}$ (there are no covariances due to the assumed independence):

\begin{aligned} \sigma_t^2 &= \sigma_{1,t}^2 + \sigma_{2,t}^2 \\ &= \omega_1 + a_1 e_{1,t-1}^2 + b_1 \sigma_{1,t-1}^2 \\ &+ \omega_2 + a_2 e_{2,t-1}^2 + b_2 \sigma_{2,t-1}^2 \\ &= (\omega_1+\omega_2) + (a_1 e_{1,t-1}^2+a_2 e_{2,t-1}^2) + (b_1 \sigma_{1,t-1}^2+b_2 \sigma_{2,t-1}^2). \\ \end{aligned}

It does not seem possible to express this in terms of $\sigma_t^2 = \omega + \sum_{i=1}^s \alpha_i e_{t-i}^2 + \sum_{i=1}^r \beta_i \sigma_{t-i}^2$ (but how to prove it formally?). And this is the simple example where $e_{1,t}$ and $e_{2,t}$ are independent (so we spare any covariances that would otherwise appear in the above expressions) and the lag orders of their respective GARCH processes coincide.

Why do I arrive at a different conclusion than @John? His claim

Assuming the coefficients in the Garch equations are constrained to be positive and sum to less than or equal to one on the lagged values, then $r_{3,t}$ will also follow a Garch process as a result of inheriting the Garch variances of the other variables

is unfounded, i.e. there is no proof or derivation supporting it. On the contrary, the above expressions illustrate (admittedly, without a formal proof) that the inheritence from the two component processes does not add up to fit in the form of a GARCH model.

References:

• It was useful. (+1) – user16651 Sep 7 '16 at 10:35
• @BehrouzMaleki, thank you, I appreciate it. I put in quite some effort (but unfortunately failed to come up with a formal proof, at least for now). – Richard Hardy Sep 7 '16 at 11:10
• @RichardHardy I had worked through the same math yesterday when we were discussing it, so I have no disagreements mathematically (although there is a specific case, if a1 equals a2 and b1 equals b2, where it should be exactly GARCH(1,1)). Where we disagree is that I wasn't trying to show that. I consider GARCH to be a large class of models, including both univariate and multivariate types. The fact that it is not GARCH(1,1) does not mean it is not part of the broader class of models. As I noted in the comments to my answer, where I was not specific enough was that it is multivariate GARCH. – John Sep 7 '16 at 12:55
• Thank you for the explanation. To justify your answer, it should include the definition of the broader class of GARCH models you are using. Because without a formal definition one cannot verify that a given model belongs to the given class of models. It could happen (this is my guess) that the class is so broad that it is no longer "interesting". On a somewhat subjective note, I think one would be bending the setup too much w.r.t. the original question, which is formulated in a very simple way. As Bob Jansen put it in a comment, the question is, is GARCH(p,q) closed under linear combination? – Richard Hardy Sep 7 '16 at 13:18
• @John, so far reading your answer my impression is that you define a GARCH model in the beginning and do not redefine it on the way so that your conclusion holds for the same definition (which would be wrong). Sorry for nitpicking, but I am concerned that people would take away the wrong message after reading your answer. On a second thought, I think now I see where you are going. The resulting model might be interpreted as one equation from an unrestricted multivariate vech-GARCH model. (What multivariate process would it correspond to, exactly?) Is that what you mean? – Richard Hardy Sep 7 '16 at 13:20