Why Sample Kurtosis Converges Slowly in Persistent GARCH Models

# Kurtosis in GARCH


# Kurtosis in GARCH












In a GARCH(1,1) model

$$ x_t = \sigma_tz_t$$ $$\sigma_{t+1}^2=a_0 + a_1x_t^2 + b_1\sigma_t^2$$

the kurtosis (when it exists) can be shown to be equal to

$$ \kappa_x = \kappa_z \frac{1-(a_1+b_1)^2}{1 - (a_1+b_1)^2 - a_1^2 (\kappa_z - 1) }$$

where $\kappa_z$ is the kurtosis of $z_t$. For standard normal innovations, i.e. when $z_t \sim N(0,1)$, $\kappa_z = 3$, and with $$ a_0 = 0.01, a_1 = 0.09, b_1=0.9$$

this gives

$$ \kappa_x = 3 \frac{1-(0.99)^2}{1 - (0.99)^2 - 0.09^2 (3 - 1) } \approx 16.14$$

However, when I run simulation of this GARCH process, I find that the sample kurtosis is somewhere around 7-8. For example, the plot below shows sample kurtosis calculated on 1,000 simulations of the above GARCH process with 10,000 time steps each. I don't understand where this mismatch between the theoretical value above and estimates is coming from.

Add 1

I have run simulations with different parameter settings to move the process away from IGARCH, but it doesn't seem to have improved much the results. For example, with $$ a_0 = 0.2, a_1 = 0.383, b_1 = 0.417$$, the theoretical kurtosis is about $16.2$ but in the simulations the sample estimates produce a mean kurtosis of about $9.7$ and median of about $6.7$ (plot below).

Add 2

I also run simulations using simple t-distributed random number with 4.45 degrees of freedom (which also gives a kurtosis of around 16.3) and calculated sample kurtosis on these. The plot below shows the results, which are also "bunched up" around 8 with the median of 8.78, but the mean is driven up to 20 by the few of the very large outliers (actually, it is mainly the most extreme one which pushes it up to 20, without it the average kurtosis is 12.3). So it is qualitatively similar to GARCH results and supports the arguments by Matthew Gunn.

## Answer by Matthew Gunn (score 10, accepted)

https://quant.stackexchange.com/a/42009

You've found parameterizations where fantastically long samples are required for sample 4th moments to converge on population 4th moments.

### Quick evidence of imprecise estimation

Let $k_i$ denote your estimated kurtosis in simulation $i$. Looking across your $i = (1,\ldots, 1000)$ simulations, your $k_i$ estimates are all over the place. What's your standard error for $\frac{1}{n} \sum_i k_i$? It's huge.

You're saying your sample kurtosis is around 7-8, but if your standard error is huge, how can you say your result is inconsistent with an actual kurtosis of 16? You can't.

### Quick Theory: $\{x^2_t\}$ is close to non-stationary

A GARCH(1,1) implies an ARMA(1,1) in the squared process. If your GARCH model is: $$ x_t = \sigma_tz_t$$ \begin{align*} \sigma_{t}^2&=\omega + a_1x_{t-1}^2 + b_1\sigma_{t-1}^2 \end{align*}

It implies an ARMA(1,1) representation for $\{x^2_t\}$. Observe that you can write $x_t^2 = \operatorname{E}_{t-1}[\sigma^2_tz_t^2] + u_t = \sigma_t^2 + u_t$. Using the lag operator $L$ we can write $(1 - b_1L) \sigma_t^2 = \omega + a_1 L x_{t}^2$. Combining those equations you get:

$$ x_t^2 = \omega + (a_1 + b_1) x_{t-1}^2 + u_t - b_1 u_{t-1} $$

You can see here that if $a_1 + b_1 = 1$, the model is non-stationary. For you, $a_1+b_1 = .99$ and $\{x_t^2\}$ is incredibly persistent.

Another requirement for the existence of a 4th moment of a GARCH(1,1) is $b_1^2 + 2a_1b_1 + 3a_1^2 < 1$. (See Theorem (2.3) and Example 2.4 from Petra Posedel.) Part of what's happening is your 2nd example is getting close to that constraint. Speaking loosely, you've found parameterizations where there's a small probability of very large volatility shocks.

### Simple example with slow convergence of sample mean on population mean

Let $X$ denote a random variability with a $p=.000001$ (one in a million) probability of 10,000,000 and a $q = 1 - p$ probability of 0.

- Trivially, $\operatorname{E}[X] = 10$.

- In an IID sample of 1000 observations, there's a 99.9 percent probability the sample mean is 0 and approximately .0999 percent probability of a sample mean of 10,000.

The distribution of the sample mean is not symmetric around the expected value: far more observations will be below than above. Also observe that you need extreme sample sizes for the sample mean to converge on the population mean.

More broadly, small measure events that you're unlikely to sample via Monte-Carlo methods can have significant impact on overall population moments if the random variables takes on huge values in those events.