Newey-West Covariance Matrix (Part 2)

The multivariate case of the Newey-West covariance matrix.

In the previous part we went over how Newey and West's insight can be used to describe a single random variable across time. More precisely, we produced a unbiased and asymptotically consistent estimate of variance for a single heteroskedastic and autocorrelated variable.

This process was quite involved and involved approximating covariance terms of the variable with itself across time.

In this part, we'll cover how to generalize the intuition for our variance estimation procedure to a group of random variables to produce a covariance matrix that has the properties of being both unbiased and asymptotically consistent.

The Multivariate Case

Instead of having random variable XX suppose we have a vector of random variables x\pmb x with corresponding sample means xˉ\bar{\pmb x} and population means μ\pmb \mu. The CLT implies that:

xˉ−μ→dN(0,Var(xˉ))→dN(0,E[(xˉ−μ)(xˉ−μ)′])→dN(0,Ω)\begin{align} \bar{\pmb x} - \pmb \mu &\overset{d}\to N(0,Var(\bar{\pmb x})) \\ &\overset{d}\to N(0,E[(\bar{\pmb x}-\pmb \mu)(\bar{\pmb x}-\pmb \mu)']) \\ &\overset{d}\to N(0,\Omega) \end{align}
Var(xˉ)=Cov(xˉ,xˉ)=Cov(x1+⋯+xT,x1+⋯+xT)/T2=Cov(x1,x1+⋯+xT)/T2+Cov(x2,x1+⋯+xT)/T2⋮+Cov(xT,x1+⋯+xT)/T2\begin{align} Var(\bar{\pmb x}) &= Cov(\bar{\pmb x}, \bar{\pmb x}) \\ &= Cov(\pmb x_1 + \dots + \pmb x_T, \pmb x_1 + \dots + \pmb x_T)/T^2 \\ &= \quad Cov(\pmb x_1, \pmb x_1 + \dots + \pmb x_T)/T^2 \\ &\quad+ Cov(\pmb x_2, \pmb x_1 + \dots + \pmb x_T)/T^2 \\ &\qquad \quad \vdots \\ &\quad+ Cov(\pmb x_T, \pmb x_1 + \dots + \pmb x_T)/T^2 \\ \end{align}

We assume:

Cov(xt,xs)=0 for ∣t−s∣>q\begin{align} Cov(\pmb x_t,\pmb x_s) = 0 \text{ for } |t - s| > q \end{align}

Which allows us to cross out terms in our covariance expression:

Var(xˉ)=Cov(x1,x1+⋯+x1+q)/T2⏟q+1 terms+Cov(x2,x1+⋯+x2+q)/T2⏟q+2 terms⋮+Cov(xT,xT−q+⋯+xT)/T2⏟q+1 terms\begin{align} Var(\bar{\pmb x})&= \quad \underbrace{Cov(\pmb x_1, \pmb x_1 + \dots + \pmb x_{1+q})/T^2}_{\text{q+1 terms}} \\ &\quad+ \underbrace{Cov(\pmb x_2, \pmb x_1 + \dots + \pmb x_{2+q})/T^2}_{\text{q+2 terms}} \\ &\qquad \quad \vdots \\ &\quad+ \underbrace{Cov(\pmb x_T, \pmb x_{T-q} + \dots + \pmb x_{T})/T^2}_{\text{q+1 terms}} \\ \end{align}

(Note: The given formula for term counts holds when q>0q > 0 but when q=0q = 0 all term counts are 11).

Which we collect like so:

Var(xˉ)=1/T2[∑t=1TVar(xt)+∑t=2TCov(xt+1,xt)+∑t=2TCov(xt,xt−1)⋮+∑t=q+1TCov(xt+q,xt)+∑t=q+1TCov(xt,xt−q)]\begin{align} Var(\bar{\pmb x}) = 1/T^2 \bigg[ &\quad \sum_{t=1}^T Var(\pmb x_t) \\ &+ \sum_{t=2}^T Cov(\pmb x_{t+1},\pmb x_t) + \sum_{t=2}^T Cov(\pmb x_t,\pmb x_{t-1}) \\ &\qquad \vdots \\ &+ \sum_{t=q+1}^T Cov(\pmb x_{t+q}, \pmb x_t) + \sum_{t=q+1}^T Cov(\pmb x_t, \pmb x_{t-q})\quad \bigg] \end{align}

To make this more collection of covariance terms more manageable, we collect them like so:

Var(xˉ)=1/T[Γ0+∑v=1q(Γv+Γ−v)]\begin{align} Var(\bar{\pmb x}) = 1/T \bigg[\Gamma_0 + \sum_{v=1}^q(\Gamma_v+ \Gamma_{-v})\bigg] \end{align}

Where

Γv=1/T∑t=v+1TCov(xt−v,xt)\begin{align} \Gamma_v = 1/T \sum_{t=v+1}^T Cov(\pmb x_{t-v}, \pmb x_t) \end{align}

In order to remove the undesirable properties of the estimator, discussed in the previous part , while preserving the desirable properties of the estimator, we add a weighting term

Var(xˉ)=1/T[Γ0+∑v=1q[1−vq+1](Γv+Γ−v)]\begin{align} Var(\bar{\pmb x}) = 1/T \bigg[\Gamma_0 + \sum_{v=1}^q\left[1 - \frac{v}{q+1}\right](\Gamma_v+ \Gamma_{-v})\bigg] \end{align}

Recalling that σxˉ2=σ2/n  ⟹  σ2=nσxˉ2\sigma^2_{\bar x} = \sigma^2/n \implies \sigma^2 = n \sigma^2_{\bar x} we can write out the variance for our un-aggregated observations:

Var(x)=Γ0+∑v=1q[1−vq+1](Γv+Γ−v)\begin{align} Var(\pmb x) = \Gamma_0 + \sum_{v=1}^q\left[1 - \frac{v}{q+1}\right](\Gamma_v+ \Gamma_{-v}) \end{align}

We observe each variable at each slice of time exactly once, therefore our strategy for computing each (co)variance term is:

Cov^(xt+i,xt)=(xt+i−μ)(xt−μ)′\begin{align} \widehat{ Cov }(\pmb x_{t+i}, \pmb x_{t}) = (\pmb x_{t+i}-\pmb \mu)(\pmb x_{t}-\pmb \mu)' \end{align}
\begin{align} \end{align}