Consider a time series $\{r_t\}$ following a standard GARCH(1,1) model, i.e., $$ r_t = \sigma_t \epsilon_t,$$ where $\epsilon_t \sim N(0,1)$ and are i.i.d, and $$\sigma_t^2 = \omega + \alpha_1 r_{t-1}^2 + \beta_1 \sigma_{t-1}^2, $$ where $\omega, \alpha_1, \beta_1$ are constant.

My question is: can we derive analytically the expectation of the random variable $\sigma_t$ for a given $t$, i.e., $\mathbb{E}(\sigma_t)$?

It is easy to derive $\mathbb{E}(\sigma_t^2) = \frac{\omega}{1-\alpha_1 - \beta_1}$, assuming $\mathbb{E}(\sigma_t^2)$ is constant over time, but I did not find any result on $\mathbb{E}(\sigma_t)$?

  • $\begingroup$ possible duplicate of How GARCH/ARCH models are useful to check the volatility? $\endgroup$
    – Quantopik
    Commented Jul 7, 2015 at 14:30
  • $\begingroup$ @Quantopic thanks, but I am looking for the analytical expression (if possible) of the expectation of volatility of a GARCH(1,1) model, not an analysis of the time series of \sigma_t. $\endgroup$
    – vitaly
    Commented Jul 7, 2015 at 15:15
  • 2
    $\begingroup$ My first try would be using the mathematical properties of expectation to try and find an answer. Using: E(X^2)= VAR(X) + (E(X))^2 $\endgroup$
    – FernandoG
    Commented Jul 22, 2015 at 20:02
  • $\begingroup$ If you know $Var(\sigma_t^2)$ then you can backout what you are looking for. $\endgroup$
    – phdstudent
    Commented Jul 25, 2015 at 15:20
  • $\begingroup$ @ volcompt Assume that $Var(\sigma_t^2)$ is available. How do you can calculate $E[\sigma_t]$? $\endgroup$
    – user16891
    Commented Jul 26, 2015 at 8:45

2 Answers 2


You can see fairly quickly that an exact answer to this question is not going to be feasible because your functional transformation is to take the square root of $\sigma_t^2$, and the square root function has a countably infinite number of derivatives. This implies that a Taylor expansion is going to leave us with a countably infinite number of terms, most of which won't vanish.

So is all hope lost? No. Interestingly enough, for an appropriate choice of constant to expand around, the first order term in the Taylor expansion drops out, and the second term is known, so we can get a fairly good approximation.

Specifically, define $c_0 = \mathbb{E} \sigma_t^2$, which, as you point out in the question, is a known quantity. Note also that $\mathbb{V} \sigma_t^2$ is a known quantity (it is a reasonably complicated function of the parameters of the GARCH model - I can't remember the reference off the top of my head but you should be able to google it fairly easily). Using Taylor's theorem, we have:

\begin{equation} \sigma_t = \sqrt{c_0} + \frac{1}{2} c_0^{-\frac{1}{2}}(\sigma_t^2 - c_0) - \frac{1}{8} c_0^{\frac{-3}{2}}(\sigma_t^2 - c_0)^2 + R \end{equation}

where $R$ is the remainder term from the expansion. Taking expectations of both sides, you can see immediately that the first order term is going to vanish, since we will get $(\mathbb{E} \sigma_t^2 - c_0)$ which by definition of $c_0$ will equal zero. Further, note the bracketed portion of the second order term will become $\mathbb{E} (\sigma_t^2 - c_0)^2$ which again, given our initial choice of $c_0$, will simplify to $\mathbb{V} \sigma_t^2$. Thus, assuming that $\mathbb{E} R$ is small, we have:

\begin{equation} \mathbb{E} \sigma_t \approx \sqrt{\mathbb{E} \sigma_t^2} - \frac{1}{8} (\mathbb{E} \sigma_t^2)^{\frac{-3}{2}} (\mathbb{V} \sigma_t^2) \end{equation}

As I said before, $\mathbb{E} \sigma_t^2$ and $\mathbb{V} \sigma_t^2$ are both known expressions w.r.t. to the parameters of the model, and so the right hand side of this equation is known (albeit it would be very messy if I wrote it out with all the parameters from a GARCH model).

So how good is the above approximation? Initially I was tempted to suggest that it might be quite good (I was short on time and it seemed an easy way to tie off the answer). However, the comment made by @ZacharyBlumenfeld is a good one, and now I'm not so sure.

There are many factors at play here, some promising, and some not.

1) The series has alternating signs, i.e. $x_1 + x_2 - x_3 + x_4 - x_5 + ...$. This is good, as the conditions for a series with alternating signs to converge are much weaker than for a series with the same sign.

2) The Taylor coefficients get small quite quickly. They go up in the factorial, i.e. $\frac{1}{2!}, \frac{1}{3!}, \frac{1}{4!}$.

3) If $\omega > 1 - \alpha - \beta$, then $c_0 > 1$ and so $c_0^{k}$ gets small quickly as $k \rightarrow -\infty$. This is good. However, note that if $\omega < 1 - \alpha - \beta$ we get the opposite effect, which would be bad. So the parameters of the model matter.

4) We have moments of $\sigma_t^2$ increasing as the terms increase. This is the one that worries me the most. I'm fairly sure I read a paper a while back on laws of large numbers for GARCH processes that shows that under quite reasonable conditions, higher moments of $\sigma_t^2$ may not exist, i.e. are infinite. This would be very bad for our approximation.

If this question is really important to you, I would recommend leveraging the wonderful amount of computing power we have these days and running some simulations for different input parameter values and seeing how good the above approximation really is. Feel free to report back here with your findings!

Cheers, hope this helps.


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    $\begingroup$ (+1) for a clever use of Taylor expansions. The only thing I would be concerned about is whether or not you can drop the remainder off so easily. If $c_0$ is smaller then 1 and "close" to 0, which is very likely for a stock return, then the the third derivative evaluated at $c_0$, $c_0^{-5/2}$ , could potentially be very large and become increasing large with higher degrees. The rate at which the factorial increases relative to the derivative is figure-out-able, but I still think it would be awfully hard to figure out the convergence properties given only mean and variance. $\endgroup$ Commented Jul 28, 2015 at 6:06
  • $\begingroup$ @ZacharyBlumenfeld Yes, I was a little too hasty to finish my answer. Your concern is a good one. Now that I think about it, I'm also worried about some of the higher moments of $\sigma_t^2$. Existence of these moments is a non-trivial matter. I've adjusted my answer to reflect your concerns. Cheers -colin $\endgroup$ Commented Jul 29, 2015 at 1:38

"conditional volatilities from GARCH models are not stochastic since at time $t$ the volatility is completely pre-determined (deterministic) given previous values"-https://en.wikipedia.org/wiki/Stochastic_volatility

$\sigma_t$ is still a random variable in the sense that it has an unconditional distribution. However, this unconditional distribution is not known in full parametric form (all we know is it's first two moments). As such the GARCH model is not really built to answer the types of questions your interested in like $E[\sigma_t]$, $\mathrm{VAR}[\sigma_t]$, etc. You can approximate these values using calculus which is what @ColinTBrowns does below...but that can get messy.

A class of model better suited to answer your question is the stochastic volatility (SV) model. For example, a simple SV model goes something like

$$ r_t=\varepsilon_t\sigma_t,\;\;\sigma_t=e^{h_t/2}$$ $$ h_t=\alpha_0+\alpha_1h_{t-1}+v_t $$

$$ \varepsilon_t\stackrel{iid}{\sim}N(0,1),\;\;v_t\stackrel{iid}{\sim}N(0,w^2) $$

Assuming $h_t$ is stationary its unconditional distribution is $h_t \sim N \bigg(\frac{\alpha_0}{1-\alpha_1},\frac{w^2}{1-\alpha_1^2} \bigg)$ so $\sigma_t$ is log normally distributed $$ \sigma_t \sim LN \bigg(\frac{\alpha_0}{2(1-\alpha_1)},\frac{w^2}{4(1-\alpha_1^2)} \bigg) $$ Thus $$ E[\sigma_t]=exp\bigg(\frac{\alpha_0}{2(1-\alpha_1)}-\frac{w^2}{2(1-\alpha_1^2)} \bigg) $$ See the Wikipedia pg for the log-normal distribution if the last step is hard to follow (i.e. it comes from the formula of the mean for a log-normal)

Long story short, if you are interested in the statistical properties of the underlying volatility states, it's better to use a model that treats those states as stochastic variables. There are also a lot more SV models to choose from in addition to the one used in this example.

  • $\begingroup$ As a side note, $\sigma_t$ has also a conditional distribution $\sigma_t \sim f(\sigma_t |\alpha_0,\alpha_1,w^2,r_1,..r_t)$ which can be used to form more precise individual expectations of $\sigma_t$ for each $t$. This is calculated with a non-linear filter. $\endgroup$ Commented Jul 27, 2015 at 7:08
  • $\begingroup$ Do you answer this qustion? In the GARCH Model $Var[\sigma_t^2]$ is available. $\endgroup$
    – user16891
    Commented Jul 27, 2015 at 16:12
  • $\begingroup$ @Farahvartish, your right it is available, but so what? I will edit it out of the answer, but I think it's is a minor point. I believe volcompt comment is incorrect, $Var[\sigma^2_t]=E[\sigma^4_t]-E[\sigma^2_t]^2$. In general, it is hard to get the moments for the square root of a random variable when you don't know what distribution that random variable follows in the first place. So yes, it answers the question by saying, that's probably going to be really hard if not impossible to do...here is a better way. $\endgroup$ Commented Jul 27, 2015 at 18:37
  • $\begingroup$ Now I realize your purpose. $\endgroup$
    – user16891
    Commented Jul 27, 2015 at 18:44
  • $\begingroup$ I get your point in this answer, but can I suggest you edit out the second paragraph? As it stands, your second paragraph suggests that $\sigma_t^2$ is not really a random variable in a GARCH framework. This is incorrect. $\sigma_t^2$ has a meaningful unconditional distribution in a GARCH framework, with known unconditional mean and variance. $\endgroup$ Commented Jul 28, 2015 at 4:35

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