OLS Estimation of Markov switching VAR models: asymptotics and application to energy use

Abstract

We show that the ordinary least squares (OLS) estimates of population parameters for Markov switching vector autoregressive (MS VAR) models coincide with the maximum likelihood estimates. Then, we propose an algorithm in matrix form for the estimation of model parameters, and derive an explicit expression in closed-form for the asymptotic covariance matrix of the OLS estimator of such models. The obtained characterization of the asymptotic variance is new to our knowledge. It is easier to program than the usual approach based on second derivatives, and more accurate. Our theorems generalize the classical results known for a linear VAR process, and complete those existing in the literature on the estimation of the asymptotic covariance matrix for multivariate stationary time series. Numerical simulations are provided to illustrate the obtained theoretical results. Finally, an application on energy use and economic growth in the Euro area gives some insights on the nonlinear nature of the corresponding time series, and reproduces the major stylized facts.

Appendix

Derivation of (3).

$$\begin{aligned} \begin{aligned} {\text {SSE}}_T&= \sum _{m = 1}^M \, \sum _{t = 1}^T \, {{\varvec{\epsilon }}}_{t, m}^{'} \, {{\varvec{\epsilon }}}_{t, m}\, = \sum _{m = 1}^M \, T \, E_{T}( {{\varvec{\epsilon }}}_{t, s_t}^{'} \, {{\varvec{\epsilon }}}_{t, s_t} | s_t = m) \\&= \sum _{m = 1}^M \, T \, E_T \left[ E({{\varvec{\epsilon }}}_{t, s_t}^{'} \, {{\varvec{\epsilon }}}_{t, s_t} | s_t = m, {\mathbf{Y}}_T)\, {\text {Pr}}(s_t = m| {\mathbf{Y}}_T)\right] \\&= T \, \sum _{m = 1}^M \, E_T\left[ E({{\varvec{\epsilon }}}_{t, m}^{'} \, {{\varvec{\epsilon }}}_{t, m} | {\mathbf{Y}}_T)\, \xi _{m t | T}\right] \\&= \sum _{m = 1}^M \, \sum _{t = 1}^T \, [{\mathbf{y}}_t - ({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, {\varvec{\beta }}_{m}]^{'} \, [{\mathbf{y}}_t - ({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, {\varvec{\beta }}_{m}]\, \xi _{m t | T}. \end{aligned} \end{aligned}$$

Derivation of (7).

The first derivative of $J({{\varvec{\theta }}})$ with respect to $\pi _m$ gives

$$\begin{aligned} \frac{\partial \, J({{\varvec{\theta }}})}{\partial \, \pi _m} \, = \, \sum _{t = 1}^T \, \pi _{m}^{- 1} \, \xi _{m t | T} \, - \, \lambda \, = \, 0 \end{aligned}$$

hence

$$\begin{aligned} \lambda \, = \, \sum _{t = 1}^T \, \pi _{m}^{- 1} \, \xi _{m t | T}. \end{aligned}$$

Summing up over $m = 1, \dots , M$, produces

$$\begin{aligned} \lambda \, = \, \lambda \, \sum _{m = 1}^M \, \pi _m \, = \, \sum _{m = 1}^M \, \sum _{t = 1}^T \, \xi _{m t | T} = \sum _{t = 1}^T \, \left( \sum _{m = 1}^M \, \xi _{m t | T} \right) \, = \, \sum _{t = 1}^T \, 1 \, = \, T. \end{aligned}$$

Now Equation (7) follows. Finally, we have ${\text {plim}}_{T \rightarrow \infty } \, {{\hat{\pi }}}_m \, = \, E(\xi _{m t | T}) \, = \, \pi _m,$ for all $m \in \Xi$. This proves the consistency of ${\hat{\pi }}_m$.

Derivation of (9).

Taking the first derivative of ${\text {SSE}}_T$ with respect to ${\varvec{\beta }}_m$ gives

$$\begin{aligned} \frac{\partial \, {\text {SSE}}_T}{\partial \, {\varvec{\beta }}_m} = \sum _{t = 1}^T \, (- {\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, \left[ {\mathbf{y}}_t \, - \, ({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, {\varvec{\beta }}_m\right] \, {\xi }_{m t | T} = {{\varvec{0}}} \end{aligned}$$

hence

$$\begin{aligned} \left[ \sum _{t = 1}^T \, ({\mathbf{x}}_t \, {\mathbf{x}}_{t}^{'}) \otimes {\mathbf{I}}_K \, {\xi }_{m t | T} \right] \, {\hat{\varvec{\beta }}}_m \, = \, \sum _{t = 1}^T \, ({\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, {\mathbf{y}}_t \, {\xi }_{m t | T} \end{aligned}$$

or equivalently

$$\begin{aligned} ({\mathbf{X}}_{m T} \otimes {\mathbf{I}}_K) \, {\hat{\varvec{\beta }}}_m \, = \, \sum _{t = 1}^T \, ({\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, {\mathbf{y}}_t \, {\xi }_{m t | T}. \end{aligned}$$

Since ${\mathbf{X}}_{m T}$ is invertible, we get

$$\begin{aligned} {\hat{\varvec{\beta }}}_m \, = \, ({\mathbf{X}}_{m T} \otimes {\mathbf{I}}_K)^{- 1} \, \left[ \sum _{t = 1}^T \, ({\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, {\mathbf{y}}_t \, {\xi }_{m t | T}\right] \end{aligned}$$

which gives (9). Since

$$\begin{aligned} \frac{\partial ^2 \, {\text {SSE}}_T}{\partial \, {\varvec{\beta }}_m \, \partial \, {\varvec{\beta }}_{m}^{'}} \, = \, 2 \, \sum _{t = 1}^T \, ({\mathbf{x}}_t \, {\mathbf{x}}_{t}^{'}) \otimes {\mathbf{I}}_K \, {\xi }_{m t | T} \, = \, 2 \, {\mathbf{X}}_{m T} \otimes {\mathbf{I}}_K \end{aligned}$$

is positive definite, the OLS estimates ${\hat{\varvec{\beta }}}_m$ is a minimizer of the function ${\text {SSE}}_T$ for every $m = 1, \dots , M$.

Derivation of (10).

$$\begin{aligned} \begin{aligned} E_T ({\hat{{\varvec{\epsilon }}}}_{t, s_t} \, {\hat{\varvec{\epsilon }}}_{t, s_t}^{'} | s_t = m)\,&= \, E_T\left[ E({\hat{{\varvec{\epsilon }}}}_{t, m} \, {\hat{{\varvec{\epsilon }}}}_{t, m}^{'} | {\mathbf{Y}}_T)\, \xi _{m t | T}\right] \, = \, E_T ( {\hat{\varvec{\Omega }}}_m \, \, \xi _{m t | T})\\&= \, {\hat{\varvec{\Omega }}}_m \, \left( T^{- 1}\, \sum _{t = 1}^T \, \xi _{m t | T}\right) \, = \, E_T\left( {\hat{\varvec{\epsilon }}}_{t, m} \, {\hat{{\varvec{\epsilon }}}}_{t, m}^{'} \, \xi _{m t | T}\right) \\&= \, T^{- 1}\, \sum _{t = 1}^T \, [{\mathbf{y}}_t - ({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, {\hat{\varvec{\beta }}}_{m}] \, [{\mathbf{y}}_t - ({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, {\hat{\varvec{\beta }}}_{m}]^{'}\, \xi _{m t | T}. \end{aligned} \end{aligned}$$

Derivation of (11).

Using (2) for $s_t = m$ with ${{\varvec{\epsilon }}}_{t m} \, = \, {\varvec{\Sigma }}_m \, {\mathbf{u}}_t$, we get

$$\begin{aligned} \begin{aligned} {\hat{\varvec{\beta }}}_m&= \, [{\mathbf{X}}_{m T}^{- 1} \otimes {\mathbf{I}}_K]\, \left[ \sum _{t = 1}^T \, ({\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, {\mathbf{y}}_t \, {\xi }_{m t | T} \right] \\&= \, [{\mathbf{X}}_{m T} \otimes {\mathbf{I}}_K]^{- 1}\, \left\{ \sum _{t = 1}^T \, ({\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, \left[ ({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, {\varvec{\beta }}_m \, + \, {\varvec{\Sigma }}_m \, {\mathbf{u}}_t\right] \, {\xi }_{m t | T} \right\} \\&= \, [{\mathbf{X}}_{m T} \otimes {\mathbf{I}}_K]^{- 1}\, \left[ \sum _{t = 1}^T \, ({\mathbf{x}}_t \, {\mathbf{x}}_{t}^{'}) \otimes {\mathbf{I}}_K) \, {\varvec{\beta }}_m \, {\xi }_{m t | T} \, + \, \sum _{t = 1}^T \, ({\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, {\varvec{\Sigma }}_m \, {\mathbf{u}}_t \, {\xi }_{m t | T} \right] \\&= \, [{\mathbf{X}}_{m T} \otimes {\mathbf{I}}_K]^{- 1}\, \, [{\mathbf{X}}_{m T} \otimes {\mathbf{I}}_K]\, {\varvec{\beta }}_m \, + \, [{\mathbf{X}}_{m T} \otimes {\mathbf{I}}_K]^{- 1}\, \left[ \sum _{t = 1}^T \, ({\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, {\varvec{\Sigma }}_m \, {\mathbf{u}}_t \, {\xi }_{m t | T}\right] \\&= \, {\varvec{\beta }}_m \, + \, [{\mathbf{X}}_{m T}^{- 1} \otimes {\mathbf{I}}_K]\, \left[ \sum _{t = 1}^T \, ({\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, {\varvec{\Sigma }}_m \, {\mathbf{u}}_t \, {\xi }_{m t | T}\right] . \end{aligned} \end{aligned}$$

Proof of Theorem 2

Consistency of ${\hat{\varvec{\beta }}}_m$.

$$\begin{aligned} \begin{aligned}&{\text {plim}}_{T \rightarrow \infty } {\hat{\varvec{\beta }}}_m = \, {\varvec{\beta }}_m \, + \, {\text {plim}}_{T \rightarrow \infty } \left[ {\mathbf{X}}_{m T} \otimes {\mathbf{I}}_K\right] ^{- 1} \, {\text {plim}}_{T \rightarrow \infty } \left[ \sum _{t = 1}^T \, ({\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, {\varvec{\Sigma }}_m \, {\mathbf{u}}_t \, {\xi }_{m t | T}\right] \\&\quad = \, {\varvec{\beta }}_m \, + \, {\text {plim}}_{T \rightarrow \infty } \left[ \frac{1}{T}\, {\mathbf{X}}_{m T} \otimes {\mathbf{I}}_K\right] ^{- 1} \, {\text {plim}}_{T \rightarrow \infty } \left[ \frac{1}{T}\, \sum _{t = 1}^T \, ({\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, {\varvec{\Sigma }}_m \, {\mathbf{u}}_t \, {\xi }_{m t | T}\right] \\&\quad = \, {\varvec{\beta }}_m \, + \, \left[ {\mathbf{Q}}_{m}^{- 1} \otimes {\mathbf{I}}_K\right] \, E\left[ ({\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, {\varvec{\Sigma }}_m \, {\mathbf{u}}_t \, {\xi }_{m t | T}\right] \end{aligned} \end{aligned}$$

where ${\mathbf{Q}}_{m}^{- 1} \otimes {\mathbf{I}}_K$ is finite by (12). Now, the second summand vanishes. In fact, we have

$$\begin{aligned} \begin{aligned} E\left[ ({\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, {\varvec{\Sigma }}_m \, {\mathbf{u}}_t \, {\xi }_{m t | T}\right]&= E\left\{ E\left[ ({\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, {\varvec{\Sigma }}_m \, {\mathbf{u}}_t \, {\xi }_{m t | T} | {\mathbf{Y}}_T \right] \right\} \\&= E\left[ ({\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, {\varvec{\Sigma }}_m \, E({\mathbf{u}}_t \, {\xi }_{m t | T} | {\mathbf{Y}}_T)\right] \\&= E\left[ ({\mathbf{x}}_t \otimes {\mathbf{I}}_K) \, {\varvec{\Sigma }}_m \, E({\mathbf{u}}_t)\, E({\xi }_{m t | T} | {\mathbf{Y}}_T)\right] = {{\varvec{0}}} \end{aligned} \end{aligned}$$

as $E({\mathbf{u}}_t) = {{\varvec{0}}}$. Here, we use the fact that ${\mathbf{u}}_t$ is independent of $s_t$, and hence ${\varvec{\xi }}_{t | T}$. The claim follows.

Consistency of ${\hat{\varvec{\Omega }}}_m$.

$$\begin{aligned} \begin{aligned} {\hat{\varvec{\Omega }}}_m&= \frac{\sum _{t = 1}^{T} \, [{\mathbf{y}}_t \, - \, ({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, {\hat{\varvec{\beta }}}_m] \, [{\mathbf{y}}_t \, - \, ({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, {\hat{\varvec{\beta }}}_m]^{'} \, {\xi }_{m t | T}}{\sum _{t = 1}^{T} \, {\xi }_{m t | T}}\\&= \frac{\sum _{t = 1}^{T} \, [({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, {{\varvec{\beta }}}_m \, - \, ({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, {\hat{\varvec{\beta }}}_m \, + \, {{\varvec{\epsilon }}}_{t, m}] \, [({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, {{\varvec{\beta }}}_m \, - \, ({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, {\hat{\varvec{\beta }}}_m \, + \, {{\varvec{\epsilon }}}_{t, m}]^{'} \, {\xi }_{m t | T}}{\sum _{t = 1}^{T} \, {\xi }_{m t | T}}\\&= \frac{\sum _{t = 1}^{T} \, [({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, ({{\varvec{\beta }}}_m \, - \, {\hat{\varvec{\beta }}}_m) \, + \, {{\varvec{\epsilon }}}_{t, m}] \, [({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, ({{\varvec{\beta }}}_m \, - \, {\hat{\varvec{\beta }}}_m) \, + \, {{\varvec{\epsilon }}}_{t, m}]^{'} \, {\xi }_{m t | T}}{\sum _{t = 1}^{T} \, {\xi }_{m t | T}}\\&= \frac{\sum _{t = 1}^{T} \, ({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, ({{\varvec{\beta }}}_m \, - \, {\hat{\varvec{\beta }}}_m) \, {{\varvec{\epsilon }}}_{t, m}^{'} \, {\xi }_{m t | T} \, + \sum _{t = 1}^{T}\, {{\varvec{\epsilon }}}_{t, m}\, ({{\varvec{\beta }}}_m \, - \, {\hat{\varvec{\beta }}}_m)^{'}\, ({\mathbf{x}}_{t} \otimes {\mathbf{I}}_K) \, {\xi }_{m t | T}}{\sum _{t = 1}^{T} \, {\xi }_{m t | T}}\\&\quad +\frac{ \sum _{t = 1}^{T}\, ({\mathbf{x}}_{t}^{'} \otimes {\mathbf{I}}_K) \, ({{\varvec{\beta }}}_m \, - \, {\hat{\varvec{\beta }}}_m) \, ({{\varvec{\beta }}}_m \, - \, {\hat{\varvec{\beta }}}_m)^{'} \, ({\mathbf{x}}_t \otimes {\mathbf{I}}_K)\, {\xi }_{m t | T} \, + \, \sum _{t = 1}^{T} \, {{\varvec{\epsilon }}}_{t, m} \, {{\varvec{\epsilon }}}_{t, m}^{'}\, {\xi }_{m t | T}}{\sum _{t = 1}^{T} \, {\xi }_{m t | T}}. \end{aligned} \end{aligned}$$

Since ${\text {plim}}_{T \rightarrow \infty } \, {\hat{\varvec{\beta }}}_m \, = \, {{\varvec{\beta }}}_m$, it follows that

$$\begin{aligned} \begin{aligned} {\text {plim}}_{T \rightarrow \infty } \, {\hat{\varvec{\Omega }}}_m&= {\text {plim}}_{T \rightarrow \infty } \, \frac{T^{- 1} \, \sum _{t = 1}^{T} \, {{\varvec{\epsilon }}}_{t, m} \, {\varvec{\epsilon }}_{t, m}^{'}\, {\xi }_{m t | T}}{T^{- 1}\, \sum _{t = 1}^{T} \, {\xi }_{m t | T}}\\&= \frac{E({{\varvec{\epsilon }}}_{t, m} \, {\varvec{\epsilon }}_{t, m}^{'}\, {\xi }_{m t | T})}{E({\xi }_{m t | T})} = \frac{E({\varvec{\Sigma }}_{ m} \, \, {\mathbf{u}}_t \, {\mathbf{u}}_{t}^{'}\, {\varvec{\Sigma }}_{m}^{'}\, {\xi }_{m t | T})}{E[E({\xi }_{m t} | {\mathbf{Y}}_T)]} \\&= \frac{E[E({\varvec{\Sigma }}_{ m} \, \, {\mathbf{u}}_t \, {\mathbf{u}}_{t}^{'}\, {\varvec{\Sigma }}_{m}^{'}\, {\xi }_{m t | T} | {\mathbf{Y}}_T)]}{E({\xi }_{m t})}\\&= \frac{E[{\varvec{\Sigma }}_{ m} \, \, E({\mathbf{u}}_t \, {\mathbf{u}}_{t}^{'})\, {\varvec{\Sigma }}_{m}^{'}\, E( {\xi }_{m t | T} | {\mathbf{Y}}_T)]}{\pi _m}\\&= \frac{E[{\varvec{\Sigma }}_{ m} {\varvec{\Sigma }}_{ m}^{'} \, E( {\xi }_{m t | T} | {\mathbf{Y}}_T)]}{\pi _m} = \frac{{\varvec{\Sigma }}_{ m} {\varvec{\Sigma }}_{ m}^{'} \, E[ E( {\xi }_{m t | T} | {\mathbf{Y}}_T)]}{\pi _m}\\&= \frac{{\varvec{\Omega }}_{ m} \, \pi _m}{\pi _m} \, = \, {\varvec{\Omega }}_{ m}. \end{aligned} \end{aligned}$$

as ${\mathbf{u}}_t$ is independent of $s_t$ (and hence ${\varvec{\xi }}_{t | T})$, $E({\mathbf{u}}_t \, {\mathbf{u}}_{t}^{'}) = {\mathbf{I}}_K$, and $E(\xi _{m t}) = \pi _m$, for every $m = 1, \dots , M$.

Derivation of (13).

We have to prove that the process $\{({\mathbf{x}}_t \otimes {\mathbf{I}}_K)\, {\varvec{\Sigma }}_m \, {\mathbf{u}}_t \, \xi _{m t | T}\}$ has zero mean and asymptotic covariance matrix ${\mathbf{Q}}_m \otimes {\varvec{\Omega }}_m$. For the mean, see the proof of consistency of ${\hat{\varvec{\beta }}}_m$ given above. For the variance, we have

$$\begin{aligned} \begin{aligned}&{\text {var}} \left[ ({\mathbf{x}}_t \otimes {\mathbf{I}}_K)\, {\varvec{\Sigma }}_m \, {\mathbf{u}}_t \, \xi _{m t | T}\right] = E\left[ ({\mathbf{x}}_t \otimes {\mathbf{I}}_K)\, {\varvec{\Sigma }}_m \, {\mathbf{u}}_t \, \xi _{m t | T}^{2}\, {\mathbf{u}}^{'}_{t} \, {\varvec{\Sigma }}^{'}_m \, ({\mathbf{x}}^{'}_t \otimes {\mathbf{I}}_K)\right] \\&\quad = E \left\{ E\left[ ({\mathbf{x}}_t \otimes {\mathbf{I}}_K)\, {\varvec{\Sigma }}_m \, {\mathbf{u}}_t \, \xi _{m t | T}^{2}\, {\mathbf{u}}^{'}_{t} \, {\varvec{\Sigma }}^{'}_m \, ({\mathbf{x}}^{'}_t \otimes {\mathbf{I}}_K) | {\mathbf{Y}}_T \right] \right\} \\&\quad = E \left[ ({\mathbf{x}}_t \otimes {\mathbf{I}}_K)\, {\varvec{\Sigma }}_m \, E({\mathbf{u}}_t \, {\mathbf{u}}^{'}_{t} \, \xi _{m t | T}^{2} | {\mathbf{Y}}_T) \, {\varvec{\Sigma }}^{'}_m \, ({\mathbf{x}}^{'}_t \otimes {\mathbf{I}}_K) \right] \\&\quad = E \left[ ({\mathbf{x}}_t \otimes {\mathbf{I}}_K)\, {\varvec{\Sigma }}_m \, E({\mathbf{u}}_t \, {\mathbf{u}}^{'}_{t}) \, E(\xi _{m t | T}^{2} | {\mathbf{Y}}_T) \, {\varvec{\Sigma }}^{'}_m \, ({\mathbf{x}}^{'}_t \otimes {\mathbf{I}}_K) \right] \\&\quad = E \left[ ({\mathbf{x}}_t \otimes {\mathbf{I}}_K)\, {\varvec{\Sigma }}_m \, {\varvec{\Sigma }}^{'}_m \, E(\xi _{m t | T} | {\mathbf{Y}}_T) \, ({\mathbf{x}}^{'}_t \otimes {\mathbf{I}}_K) \right] \\&\quad = E \left[ ({\mathbf{x}}_t \otimes {\mathbf{I}}_K)\, {\varvec{\Omega }}_m \, E(\xi _{m t | T} | {\mathbf{Y}}_T) \, ({\mathbf{x}}^{'}_t \otimes {\mathbf{I}}_K) \right] \\&\quad = E \left[ E({\mathbf{x}}_t \, {\mathbf{x}}^{'}_t \, \xi _{m t | T} | {\mathbf{Y}}_T)\right] \otimes {\varvec{\Omega }}_m \\&\quad = E ({\mathbf{x}}_t \, {\mathbf{x}}^{'}_t \, \xi _{m t | T}) \otimes {\varvec{\Omega }}_m \, = \, {\mathbf{Q}}_m \otimes {\varvec{\Omega }}_m. \end{aligned} \end{aligned}$$

Here, we use the fact that ${\mathbf{u}}_t$ is independent of $s_t$, and hence $\xi ^{2}_{m t | T}$. Furthermore, $E({\mathbf{u}}_t \, {\mathbf{u}}^{'}_{t}) = {\mathbf{I}}_K$ and $E(\xi ^{2}_{m t | T} | {\mathbf{Y}}_T) = E(\xi _{m t | T}| {\mathbf{Y}}_T)$ as $E(\xi ^{2}_{m t}) = E(\xi _{m t}) = \pi _m$, for every $m = 1, \dots , M$. $\square$