13A.4 Cointegration and error correction

Motivation

Two random walks can share a random walk. If they do, some combination of them is stationary, and that combination is what a pairs trader wants to hold.

13A.4.1Definition#

DefinitionEngle and Granger 1987

An $n$ -vector $y_t$ whose components are each $I(1)$ is cointegrated if there is a non-zero $\beta\in\mathbb R^n$ with $\beta'y_t$ stationary ( $I(0)$ , after removing a constant). The set of such $\beta$ together with 0 is a linear subspace; its dimension $r$ ( $0<r<n$ ) is the cointegrating rank, and any basis, collected in an $n\times r$ matrix $\beta$ , is a set of cointegrating vectors.

For a pair, $n=2$ and $r\le1$ : cointegration means one stationary combination $p^1_t - b\,p^2_t$ , unique up to scale. The normalisation (coefficient 1 on $p^1$ ) is a choice.

13A.4.2The vector error-correction model#

Write a VAR( $p$ ) in levels, $y_t = \nu + \sum_{i=1}^p A_i y_{t-i} + \varepsilon_t$ , in differences:

\begin{gathered} \Delta y_t = \nu + \Pi\,y_{t-1} + \sum_{i=1}^{p-1}\Gamma_i\,\Delta y_{t-i} + \varepsilon_t, \\ \Pi = -\Big(I - \sum_{i=1}^p A_i\Big),\qquad \Gamma_i = -\sum_{j=i+1}^p A_j. \end{gathered} \tag{13A.9}

(This is Proposition 13A.2 in matrix form; the proof is the same.) If $y_t$ is $I(1)$ and cointegrated with rank $r$ , then $\Pi$ has rank $r$ and factors as

\Pi = \alpha\beta', \qquad \alpha,\beta \in\mathbb R^{n\times r}\ \text{of full column rank}. \tag{13A.10}

$\beta'y_{t-1}$ is the lagged equilibrium error; $\alpha$ holds the adjustment speeds (loadings): how each price responds to last period's error. (13A.9)–(13A.10) is the vector error-correction model (VECM).

13A.4.3The Granger representation, proved for a VAR(1)#

The general theorem (Engle and Granger 1987; Johansen 1991) says that, under a rank condition, a VAR with $\Pi = \alpha\beta'$ is $I(1)$ with exactly $r$ stationary combinations $\beta'y_t$ and $n-r$ common stochastic trends. The VAR(1) case shows every idea.

Proposition 13A.4Granger representation, VAR(1)

Let $y_t = Ay_{t-1}+\varepsilon_t$ with $A - I = \alpha\beta'$ , $\alpha,\beta$ of rank $r$ , and suppose all eigenvalues of $I_r + \beta'\alpha$ lie strictly inside the unit circle. Let $\alpha_\perp, \beta_\perp$ be $n\times(n-r)$ matrices of full rank with $\alpha_\perp'\alpha = 0$ , $\beta_\perp'\beta = 0$ . Then:

(a) $\beta'y_t$ is a stationary VAR(1):

\beta'y_t = (I_r + \beta'\alpha)\,\beta'y_{t-1} + \beta'\varepsilon_t. \tag{13A.11}

(b) $\alpha_\perp'y_t$ is an $(n-r)$ -dimensional random walk:

\alpha_\perp'y_t = \alpha_\perp'y_{t-1} + \alpha_\perp'\varepsilon_t. \tag{13A.12}

C = \beta_\perp(\alpha_\perp'\beta_\perp)^{-1}\alpha_\perp'. \tag{13A.13}

The first term is a random walk (the common trends), the last term is stationary by (a).

Proof

(a) Premultiply $y_t = y_{t-1} + \alpha\beta'y_{t-1} + \varepsilon_t$ by $\beta'$ . The eigenvalue condition makes the $r$ -dimensional VAR(1) stable.

(b) Premultiply by $\alpha_\perp'$ ; the $\alpha$ term vanishes.

(c) The eigenvalues of $\beta'\alpha$ are those of $I_r + \beta'\alpha$ minus one; they lie inside the disc of radius 1 centred at $-1$ and so are non-zero. Hence $\beta'\alpha$ is invertible. Then $[\alpha, \beta_\perp]$ is non-singular: if $\alpha a + \beta_\perp b = 0$ , premultiplying by $\beta'$ gives $\beta'\alpha a = 0$ , so $a=0$ , and then $b=0$ . Similarly $\alpha_\perp'\beta_\perp$ is invertible (if $\alpha_\perp'\beta_\perp b = 0$ then $\beta_\perp b \in \operatorname{span}(\alpha)$ , say $=\alpha a$ , and the previous argument gives $a = b = 0$ ). Now check the identity

I_n = \beta_\perp(\alpha_\perp'\beta_\perp)^{-1}\alpha_\perp' + \alpha(\beta'\alpha)^{-1}\beta'

on the basis $[\alpha,\beta_\perp]$ : applied to $\alpha$ the right side gives $0 + \alpha$ ; applied to $\beta_\perp$ it gives $\beta_\perp + 0$ . Apply the identity to $y_t$ and use (b), $\alpha_\perp'y_t = \alpha_\perp'y_0 + \alpha_\perp'\sum_{s\le t}\varepsilon_s$ .∎

Reading the result for a pair Take $y = (p^1, p^2)'$ , $\beta = (1, -b)'$ , $\alpha = (\alpha_1,\alpha_2)'$ . The spread $s_t = p^1_t - b\,p^2_t$ satisfies, from (13A.11),

s_t = (1 + \alpha_1 - b\,\alpha_2)\,s_{t-1} + (\varepsilon_{1t} - b\,\varepsilon_{2t}).

It mean-reverts iff $-2 < \alpha_1 - b\alpha_2 < 0$ . Typically $\alpha_1 < 0$ (asset 1 falls when the spread is high) and $\alpha_2 > 0$ (asset 2 rises). The common trend is $\alpha_\perp'y_t$ with $\alpha_\perp = (\alpha_2, -\alpha_1)'$ : the price to which both are tied is a weighted average, with weights set by the loadings, not by market capitalisation.

Who adjusts matters for execution If $\alpha_2 = 0$ , asset 2 is weakly exogenous: it does not respond to the spread, all correction happens in asset 1, and asset 2 "leads". For MCX gold against COMEX-implied parity (13A.12.3) one expects the COMEX loading to be close to zero, since India is a price taker in gold; that is a hypothesis to test, not a fact. Gonzalo and Granger (1995) use $\alpha_\perp$ to define permanent–transitory decompositions; Hasbrouck (1995) defines "information shares" for the same purpose in market microstructure.

Long-run impact $C$ in (13A.13) is the long-run response of the price levels to a shock. Problem 13A.4 computes it for a numerical VAR(1) and checks it against $\lim_h A^h$ .