# Working out Granger causality of time series where the series are described mathematically

Previously when doing coursework I would be given say two time series $x,y$ which are sampled for $t=1,\ldots,1000$ (or some other length) and I would be asked to calculate Granger causality from them. What I would do is fit two models for each, so for example for $x$ I would fit

\begin{align} x_1(t) &= a_1x(t-1) + a_2x(t-2) + \ldots + a_px(t-p) + \epsilon_{1}\\ x_2(t) &= b_1x(t-1) + b_2x(t-2) + \ldots + b_px(t-p) + \\ &+ c_1y(t-1) + c_2y(t-2) + \ldots + c_py(t-p) + \epsilon_{2} \end{align}

And then I would say that $y$ granger causes $x$ if and only if $\operatorname{var}(\epsilon_1) > \operatorname{var}(\epsilon_2)$ where $\epsilon$ is the estimated noise. (I would do the same to see the other way round).

Now I am looking at an exam question where instead of being given a sample of a time series I am given two time series mathematically and I am asked to calculate the Granger causalities between them so I guess the approach would be different since there is no need to fit the series or estimate the errors. Unfortunately all of the material I find online deals with the case where you have a sample rather than the entire data set. I'll show you what I have done so far.

The time series are

\begin{align} y(t) &= \frac{1}{10}y(t-1) + e_1(t) \\ x(y) &= \frac{8}{10}x(t-1) + \frac{1}{10}y(t-1) + e_2(t) \end{align}

Where $e_1,e_2$ are independent and standard normal random variables and $t$ is time. First I checked if $(x(t),y(t))$ was a stable time series by looking at the eigenvalues of the matrix $$A = \begin{pmatrix} \frac{1}{10} & 0 \\ \frac{1}{10} & \frac{8}{10} \end{pmatrix}$$ which turned out to be $\frac{1}{10},\frac{8}{10}$. Since both lie in the unit circle the time series is stable.

Now I look at the mean and variance of $x(t)$ and $y(t)$. I find $\mathbb{E}(x)=\mathbb{E}(y) = 0$ and then I find

\begin{align} \operatorname{var}(y) &= \frac{1}{100} \operatorname{var}(y) +1 &\implies \operatorname{var}(y) &=\frac{100}{99} \\ \operatorname{var}(x) &= \frac{64}{100} \operatorname{var}(x) + \frac{1}{100} \operatorname{var}(y) &\implies \operatorname{var}(x) &= \frac{2500}{891} \end{align}

Now I have gotten that out the way, I need to calculate the Granger causality and this is where I get confused. I am unsure of the method because it's different to how I usually do this. Here is my attempt, I put error terms in square brackets for clarity.

To consider the Granger causality from $x$ to $y$ I look at

\begin{align} y_1(t) &= \frac{1}{10}y(t-1) + [e_1(t)] \\ y_2(t) &= \frac{1}{10}y(t-1) + cx(t-1) + \left[e_1(t) -cx(t-1)\right] \end{align}

Here $y_1(t)$ is $y$ without lags of $x$ included, $y_2(t)$ is with a lag of $x$ included at amplitude $c$, but then as to not change the time series I remove this bit in the error term (I'm very unsure of this!).

I observe $\operatorname{var}(e_1) = 1$ and $\operatorname{var}(e_1 - cx(t-1)) = 1 - c^2 \frac{2500}{891} < 1$ so there is no Granger causality from $x$ to $y$ (which aligns with what you would expect looking at the series).

Now looking to see if there is Granger causality from $y$ to $x$

\begin{align} x_1(t) &= \frac{8}{10}x(t-1) + \left[ e_2(t) + \frac{1}{10}y(t-1) \right] \\ x_2(t) &= \frac{8}{10}x(t-1) + \frac{1}{10}y(t-1) + [e_2] \end{align}

Then I find $$\operatorname{var}\left( \frac{1}{10}y(t-1) + e_2(t) \right) = \left( \frac{1}{100} \times \frac{100}{99} + 1 \right) > \operatorname{var}(e_2) = 1$$

And using a formula given to me the Granger causality is $$G_{y \rightarrow x} = \log \left( \frac{\frac{1}{100} \times \frac{100}{99} + 1}{1} \right) = \log \left( \frac{100}{99} \right) \approx 0.01$$

I don't really know how to interpret the $0.01$, but at least there is some Granger causality as expected.

My question is, is this the correct way to calculate Granger causality between the two time series given?