Cointegration Modelling¶

Error Correction Models¶

\[ \begin{aligned} \Delta m_t &= \lambda_m (u_{t-1}) + \epsilon_{mt} \\ &= \lambda_m (blah blah) + \epsilon_{mt} \\ \Delta p_t &= \\ \Delta y_t &= \end{aligned} \]

$\epsilon$ is white noise error
$\lambda$ are velocity of adjustment parameters

Atleast one of the $\lambda$ should be significant, otherwise there is no error correction $\implies$ no cointegration

Cointegration and error correction are equivalent representation

~ Granger representation theorem

Vector regression/Structural estimation model¶

Used when there is no cointegration

I missed this

If the values of $\lambda$ are zero, then it is a simple VAR model and there is no cointegration.

Example: Quantity Theory of Money¶

\[ MV = \underbrace{PY}_{\text{GDP}} \]

			Integrated of order
$M$	Total quantity of money		$I(1)$
$V$	Velocity of money	Number of times a unit of currency is transferred in a year	N/A (Constant value)
$P$	Price		$I(1)$
$Y$	Real quantity of Output		$I(1)$

As they are $I(1)$, they are not mean-reverting variables. Hence, taking log on both sides of equation, and then transposing

\[ \beta_0 + \beta_1 m_t - \beta_2 p_t - \beta_3 y_t = u_t \]

Velocity is a constant, which is an intercept. Here it is represented by $\beta_0$, but can also represented by $1\cdot V$

If $u_t$ is $I(0) \implies M, V, P, Y$ are cointegrating

Notes¶

There can be multiple cointegrating vectors $\{\beta_0, \beta_1, \beta_2, \beta_3 \} = \{\lambda \beta_0, \lambda \beta_1, \lambda \beta_2, \lambda \beta_3 \} \iff \lambda \ne 0$
If $m$ and $p$ are $I(2)$ whereas $y$ is $I(1)$. The linear combination of these three variables will be $I(2)$, hence the 3 are not cointegrated
However, if a linear combination $\beta_1 m + \beta_2 p$ is $I(1)$, and this is cointegrated with y which is $I(1)$, then we say there is multi-cointegration
if monetary policy folows feedback rule that changes money supply based on inflation, then inflation will be another cointegrated variable

Granger Causality¶

Let’s say we have 2 variables $x, y$. We can check if $x$ granger causes $y$

\[ y_t = \beta_1 y_{t-1} + \beta_2 x_{t-1} + u_t \]

Hypotheses¶

$H_0: \beta_2 = 0$
$y$ is independent of $x$
$x$ does not granger cause $y$
$H_1: \beta_2 \ne 0$
$x \to y$
$x$ granger causes $y$

Procedure¶

We check if the $R_{adj}^2$ has increased by incorporating $x_{t-1}$, when compared to without it $(y_t = \beta_1 y_{t-1} + u_t)$
Do a hypothesis test
If $p \le 0.05,$ reject null hypothesis, and hence conclude that $x \to y$

Spread¶

Credit Spread $$ \begin{aligned} y_{1t} &= x_t + u_{1t} \ y_{2t} &= \gamma x_t + u_{2t} \ z_t &= y_{1t} - y_{2t} \quad \text{(Spread)}\ &= y_{1t} - \gamma y_{2t} \ &= u_{1t} - \gamma u_{2t} \ y_1, y_2 &\text{ are co-integrating} \iff z_t \to \text{Stationary Process} \end{aligned} $$

This mean-reverting tendency of the spread can be used for “pairs trading”/“statistical arbitrage”

Estimating $\gamma$¶

Simple¶

\[ \begin{aligned} z_t \implies y_{1t} - \gamma y_{2t} &= \mu + u_t \\ y_{1t} &= \mu + \gamma y_{2t} + u_t \end{aligned} \]

Perform linear regression for $\gamma$

Kalman¶

\[ \begin{aligned} z_t \implies y_{1t} - \gamma_\textcolor{hotpink}{t} y_{2t} &= \mu_\textcolor{hotpink}{t} + u_t \\ y_{1t} &= \mu_\textcolor{hotpink}{t} + \gamma_\textcolor{hotpink}{t} y_{2t} + u_t \\ \mu_{t+1} &= \mu_t + \eta_{1t} \\ \gamma_{t+1} &= \gamma_t + \eta_{2t} \end{aligned} \]

Last Updated: 2024-05-12 ; Contributors: AhmedThahir