# Independence vs Autocorrelation

In Section 3.2 of R. S. Tsay, Analysis of Financial Time Series, I read:

The basic idea behind volatility study is that the series {r_t} is either serially uncorrelated or with minor lower order serial correlations, but it is a dependent series.

and a little further:

...it seems that the returns are indeed serially uncorrelated, but dependent.

I'm very confused by these statements, because I thought that serial correlation (autocorrelation) and dependence were basically the same thing. Here, for instance, the coin tossing game is given as an example of independence, as a series where each throw has no memory of the previous throws. Therefore, it seems to me that instead a not independent variable should be serially correlated.

Can you give me an example of a time series that is not serially correlated but dependent, and another one that is serially correlated but independent? Is that possible? Or perhaps am I wrongly assuming that causation implies correlation?

• Independence implies zero correlation. But zero correlation on its own DOES NOT imply independence! There are special cases, such as $\mathbf{x}$ following the multivariate normal distribution, where zero correlation between $x_i$ and $x_j$ implies they are independent, but these are special cases. – Matthew Gunn Jan 17 '18 at 16:12

Just elaborating on Matthew Gunn answer, we have that, due to the independence of the various components,

$$E[x_tx_{t-1}] = E[z_t\sigma_tz_{t-1}\sigma_{t-1}] = E(z_t)E(z_{t-1})E(\sigma_t\sigma_{t-1})$$

$$=0\cdot0\cdot E(\sigma_t\sigma_{t-1}) = 0$$

Since the $X$-process has zero mean, the above implies that there is no autocorrelation.

But

$$E[x_t^2x_{t-1}^2] = E[z_t^2\sigma_t^2z_{t-1}^2\sigma_{t-1}^2] = E(z_t^2)E(z_{t-1}^2)E(\sigma_t^2\sigma_{t-1}^2)$$

$$=1\cdot 1 \cdot E(\sigma_t^2\sigma_{t-1}^2) = E(\sigma_t^2\sigma_{t-1}^2)$$

while

$$E[x_t^2]\cdot E[x_{t-1}^2] = ...=E(\sigma_t^2)\cdot E(\sigma_{t-1}^2)$$

Because the sigmas are autocorrelated, if one carries out the multiplications, one will find that

$$E[x_t^2x_{t-1}^2] \neq E[x_t^2]\cdot E[x_{t-1}^2]$$

which shows that there exists higher-order dependence in the $X$-process.

• Thank you! So, would it be correct to say that "dependence" of a X-process can be mathematically defined as the existence of a continuous function f such that E[f(x_t)f(x_(t-1))] \neq E[f(x_t)]E[f(x_(t-1))] ? – Nonancourt Jan 18 '18 at 9:38
• @Nonancourt There are many ways to define the existence of dependence, this is a reasonable one, but note that it applies only for distributions for which these expected values exist. – Alecos Papadopoulos Jan 18 '18 at 9:48

A simple, autocorrelated volatility structure can generate this pattern.

• Let $z_t$ be IID draws where $z_t \sim \mathcal{N}(0, 1)$
• Let's say $\sigma_t$ follows the AR(1) process $\sigma_t = b\sigma_{t-1} + \epsilon_t$ where $0 < b < 1$ and $\epsilon_t$ are IID $\mathcal{N}(0, s^2)$.
• Then define $x_t = z_t \sigma_t$. (This gives $x_t$ a conditional volatility of $\sigma_t$ in period $t$.)

We have that $x_t$ and $x_{t-1}$ are uncorrelated, but clearly they are dependent.