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I'd like to test if the underlying process of a volatility series is stationary, and thus if the mean and variance of the underlying process hasn't changed over time (some structural change in the underlying process).

To do this, can I run an Augmented Dickey-Fuller test on the volatility series?

The application I had in mind was the volatility (rolling 30 day standard deviation) of the Bitcoin-Euro exchange rate returns.

Thanks.

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  • $\begingroup$ hasn't changed over time... $\endgroup$
    – Amagi
    Jan 7, 2017 at 12:49
  • $\begingroup$ you can edit your question to fix the "hasn't". $\endgroup$
    – mugen
    Jan 7, 2017 at 15:57

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I would argue that you cannot, for a couple of reasons. But before we do, lets notice that you can get rid of the rolling volatility idea. If we had a time series $\{x_1\dots{x_t}\}$ and we chunked them into 30 day rolling intervals, then we would needlessly lose information in the first thirty days. Interval one would contain $\{x_1\dots{x_{30}}\}$, while interval two would contain $\{x_2\dots{x_{31}}\}$. Notice that the difference in the intervals is only one day. The only unique information in interval one is $x_1$, while the only unique information in interval two is $x_{31}$. Otherwise, they share the same information set. You would lose information in your scheme, 29 days worth.

Now lets look at the nature and structure of bitcoin. In the short-run, new coins can be added, but at an exponentially decaying rate. As long as the counter currencies are not in deflation, then whether bitcoin is inflationary or deflationary would depend upon the relative inflation rate, except that the method of creating bitcoins, a form of seigniorage, actually can make the coin more valuable by adding outlets for spending.

Because the long run supply of bitcoins is fixed and it exists in a world of inflationary fiat currencies, the value of a coin, in the absence of mass death such as a 21st century black plague, is increasing across time when compared to fiat currencies. This implies that its time series should have the representation $x_{t+1}=Rx_t+\epsilon_{t+1},R>1$. I am a little concerned about the mechanism of trades, but I will ignore the liquidity issues for now and assume $\epsilon\sim\mathcal{N}(0,\sigma_t)$. Notice that $R$ is unknown, it is the reward for holding a bitcoin.

The distribution of $R$ is not a unit root as $R>1$. The idea of a unit root implies $R=1$.

The likelihood function for $R$ is $$\frac{1}{\pi}\frac{\gamma}{\gamma^2+(x_{t+1}-R)^2}.$$ This likelihood function, or density function if you look at it from a maximum likelihood point of view, has no population mean, so it cannot have a standard deviation. $\gamma$ is a scale parameter and $R$ is a center of location, but the former is not a standard deviation and the later is not a mean.

You can know, ex ante, that your system lacks a variance and so has no volatility in the sense you think about it. You can show, with some headaches, that any sample standard deviation is a random number. The asymptotic measure of volatility is $\gamma$ and your only tool is a Bayesian method. You can test for a structural change in either or both parameters using Bayesian methods by assuming two or more series exist and that they change at points $k_1, k_2\dots{k_n},$ using the Bayesian model selection to determine which model is the best.

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  • $\begingroup$ Volatility isn't directly observable, so either use a rolling window to estimate the sd from the returns or nothing. If I run a ACF on returns, there are low and/or not significant autocorrelations (efficient market hypothesis? how can R be bigger than 1 then?) but there is high and decaying persistence for the absolute value of returns. $\endgroup$
    – Amagi
    Jan 8, 2017 at 2:43
  • $\begingroup$ @Amagi, let us assume you wish to invest \$100 and will receive reward R times your \$100 at a future time, unfortunately there will be appraisal errors in the process and so while R isn't observed, the future price is $100R+\epsilon$. Would you invest \$100 if $R$ was systematically less than 1? You may accept equal to one as a currency, but not less than one. You would trade it out. This does not mean you cannot take losses, just that if you anticipated them then you would avoid the transaction. The ACF in standard software has been shown to be invalid under this likelihood function. $\endgroup$ Jan 8, 2017 at 12:09

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