Econ 616: Problem Set 1

Problem 1

Let \(\phi (z) \equiv 1 - \phi_1 z - \phi_2 z^2\). What we need to show is the solution of the equation \(\phi ( z) =0\) lies outside of unit circle. Let \(z_1\) and \(z_2\) be the solutions of \(\phi (z) =0\).

• Case 1: Suppose \(\phi _{1}^{2}+4\phi _{2}\leq 0\). Then we have either \(z_{1}=z_{2}\) or that \(z_{1}\) and \(z_{2}\) are complex numbers and conjugate of each other. In any case the norm of the solution is given by \(\sqrt{\left\vert \frac{1}{\phi _{2}}\right\vert }\). Hence the condition is \(|\phi_{2}|<1\).

• Case 2: Suppose \(\phi_1^2 + 4 \phi_2 > 0\). Now both solutions are real number. Suppose \(\phi_2 =0\). This is AR(1) model and the condition is \(|\phi_1 | <1\). Suppose \(\phi_2 \neq 0\). It would be easier to analyze the equation \(\psi(z) =0\) where \(\psi(z) \equiv z^2 + \frac{\phi_1}{\phi_2}z - \frac{1}{\phi_2}\) which has the same solutions as \(\phi(z) =0\). If \(\phi_2 <0\), then the conditions are \(\psi(1 )> 0\) and \(\psi(-1) >0\) which means that \(\phi_2+\phi_1 <1\) and \(\phi_2 - \phi_1 <1\). If $φ_2 > 0 $, then the conditions are \(\psi(1 ) < 0\) and \(\psi(-1) <0\) which means that \(\phi_2+\phi_1 <1\) and \(\phi_2 - \phi_1 <1\).

Combining all, we have

\begin{eqnarray*} \phi _{1}+\phi _{2}<1,\quad \phi _{2}-\phi _{1}<1\mbox{ and }\phi _{2}>-1. \end{eqnarray*}

Problem 2

Multiplying both sides of the equation by \(y_{t-j}\), we have

\begin{eqnarray*} y_{t}y_{t-j}=\sum_{i=1}^{p}\phi _{i}y_{t-i}y_{t-j}+\epsilon _{t}y_{t-j}. \end{eqnarray*}% Taking the expectation, we have \begin{eqnarray*} E(y_{t}y_{t-j})=\sum_{i=1}^{p}\phi _{i}E(y_{t-i}y_{t-j})+E(\epsilon _{t}y_{t-j}). \end{eqnarray*}% Or we can rewrite the above as \begin{eqnarray*} \gamma _{yy,j}=\sum_{i=1}^{p}\phi _{i}\gamma _{yy,|i-j|}+E(\epsilon _{t}y_{t-j}). \end{eqnarray*}

For \(j=0\), \(E(\epsilon _{t}y_{t-j})=E(\epsilon _{t}y_{t})=E(\epsilon _{t}^{2})=\sigma _{\epsilon }^{2}\) which gives the first equation. For \(% j=1,\dots ,p\), \(E(\epsilon _{t}y_{t-j})=0\) which gives the rest.

For the AR(3) process in Problem 2, we have

\begin{eqnarray*} 1 &=& \gamma_{yy,0}-(1.3\gamma_{yy,1}-0.9\gamma_{yy,2}+0.3\gamma_{yy,3}) \\ 0 &=& 1.3\gamma_{yy,0}-\gamma_{yy,1}+(-0.9\gamma_{yy,1}+0.3\gamma_{yy,2}) \\ 0 &=& -0.9\gamma_{yy,0}-\gamma_{yy,2}+(1.3\gamma_{yy,1}+0.3\gamma_{yy,1})\\ 0 &=& 0.3\gamma_{yy,0}-\gamma_{yy,3}+(1.3\gamma_{yy,2}-0.9\gamma_{yy,1}) \end{eqnarray*}

Solutions to this system of equations are

\(\gamma_{yy,0}=3.38\quad \gamma_{yy,1}=2.45\qquad \gamma_{yy,2}=0.88\qquad \gamma_{yy,3}=-0.48\)

Problem 3

See jupyter notebook

Problem 4

• The least squares estimates can be written as:

  \begin{align} \hat\rho_{LS} = \rho + \left(\sum_{t=2}^T y_{t-1}^2\right)^{-1}\sum_{t=2}^T \epsilon_t y_{t-1} \end{align}

  Consider an alternative representation of \(y_t\).

  \begin{align} t \mbox{ even} &:& y_t &= y_t^e = \rho^2 y_t^e + \sigma\epsilon_t + \rho\alpha\sigma\epsilon_{t-1} \nonumber \\ t \mbox{ odd} &:& y_t &= y_t^o = \rho^2 y_t^o + \alpha\sigma\epsilon_t + \rho\sigma\epsilon_{t-1}. \end{align}

  Consider,

  \begin{align} \label{eq:y2} \frac{1}{T}\sum_{t=2}^T y_{t-1}^2 &\approx \frac{1}{2}\frac{1}{T/2} \sum_{t=2}^{T/2} (y_{2(t-1)+1}^o)^2

  • \frac{1}{2}\frac{1}{T/2}\sum_{t=2}^{T/2} (y_{2t}^e)^2 \nonumber \\ &\rightarrow \frac12 \frac{\alpha^2+\rho^2}{1-\rho^4}\sigma^2 + \frac12 \frac{1+\alpha^2\rho^2}{1-\rho^4}\sigma^2. \nonumber \\ &= \frac12\frac{(1+\alpha^2)(1+\rho^2)}{(1-\rho^2)(1+\rho^2)}\sigma^2. \nonumber \\ &= \frac{1 + \alpha^2}{2} \frac{\sigma^2}{1-\rho^2}. \end{align}

  Moreover, The sequence \(\{\epsilon_t y_{t-1}\}\) is a Martingale difference sequence (MDS). If \(|\rho| < 1\), \(\frac1T\sum_{i=1}^T \epsilon_t y_{t-1} \rightarrow 0\) as \(T\rightarrow \infty\). Thus, the least squares estimator is consistent.

• Using arguments along the lines of (\ref{eq:y2}) and the central limit theorem for MDS yields the asymptotic distribution for \(\hat\rho_{LS}\). In particular, The least squares estimator \(\hat\rho_{LS} = \left(\sum_{t=2}^T y_{t-1}^2\right)^{-1}\sum_{t=2}^T y_t y_{t-1}\) in large samples behaves such that

  \begin{align} \sqrt{T}(\hat\rho_{LS} - \rho) \sim N(0, \mathbb V_{\hat\rho}). \end{align}

  This variance take the form:

  \begin{align} \mathbb V_{\hat\rho} = \frac{\mathbb E[\epsilon_t^2 y_{t-1}^2]}{\mathbb E[y_{t-1}^2]^2} \end{align}

  Tedious algebra yields:

  \begin{align} E[y_{t-1}^2] &= \frac{1 + \alpha^2}{2}\frac{\sigma^2}{1-\rho^2} \\ E[\epsilon_t^2 y_{t-1}^2] &= \frac{\alpha^2(1+\alpha^2\rho^2) + (\alpha^2 + \rho^2)}{2}\frac{\sigma^4}{1-\rho^4} \end{align}

  Thus:

  \begin{align} \mathbb V_{\hat\rho} = 2 \left(\frac{\rho^2 + 2\alpha^2 + \alpha^4\rho^2}{1+2\alpha^2+\alpha^4}\right)\left(\frac{1-\rho^2}{1+\rho^2}\right) \end{align}

• It is easy to see that this quantity converges in probability to

  \begin{align} \mathbb V^*_{\hat\rho} = \frac{\mathbb E[\epsilon_t^2]}{\mathbb E[y_{t-1}^2]} = 1-\rho^2 \end{align}

• Tedious algebra shows that: \[ \mathbb V_{\hat\rho} \le \mathbb V_{\hat\rho}^*. \] Thus, the typical estimator for standard errors is inconsistent and in particular it overstates the variance of \(\hat\rho_{LS}\).

Problem 5

Recall from the lecture notes that the HP filter can be written as:

\begin{eqnarray} f^{HP}(\omega) = \left[\frac{16\sin^4(\omega/2)}{1/1600 + 16\sin^4(\omega/2)}\right]^2. \end{eqnarray}

The spectrum for the AR1 can be written as:

\begin{eqnarray} f^{AR}(\omega) = \left[ 1 - 2 \phi \cos \omega + \phi^2( \cos \omega^2 + \sin^2 \omega ) \right]^{-1}. \end{eqnarray}

From the lecture notes, we know that:

\begin{eqnarray} f^{Y}(\omega) = f^{HP}(\omega) f^{AR}(\omega) \end{eqnarray}

The spectrum for \(\phi=0.95\) and \(\phi=0.70\) is plotted in Figure 3. The spectrum peaks at about \(\pi/8\), which is associated with a cycle lasting about 16 quarters \(=(2\pi/(\pi/8))\). As \(\phi\) increases, this peak sharpens. So here the HP filter is introducing as spurious periodicity in our data.

<ipython-input-4-04db9a7f9b5a>:6: RuntimeWarning: divide by zero encountered in divide
  return ( (sigma**2/(2*omega)) /
<ipython-input-4-04db9a7f9b5a>:9: RuntimeWarning: invalid value encountered in multiply
  f = lambda omega, **kwds: f_HP(omega)*f_AR1(omega, **kwds)
<matplotlib.legend.Legend at 0x7f5cf62fac40>

Another way to see this is look at the autocovariance function of the implied process, which we can recover by the inverse fourier transform as discussed in class:

\begin{eqnarray} \gamma_k = \int_{-\pi}^{\pi} f^{Y}(\omega)e^{i\omega k}. \end{eqnarray}

The HP filter induces complex dynamics into the process!

from scipy.integrate import quad
H = 20
really_small = 1e-8
acf = [quad(lambda omega:
                 2*f(omega)*np.cos(omega*k), really_small, np.pi)[0]
            for k in np.arange(H)]
plt.plot(acf)

phi = 0.95
acf = [ phi**j / (1-phi**2) for j in np.arange(H)]
plt.plot(acf, linestyle='dashed')