6
$\begingroup$

Can anybody give one example of when the set of all lagged $X$ can (or can't) be a good choice of IV's for $X_{t}$?

$\endgroup$
1
  • $\begingroup$ When $X_{t-2}$ has direct causal effects on both $X_{t-1}$ and $X_{t}$? When $X_{t-2}$ has direct causal effects on $X_{t-1}$, and also on $Z_{t-1}$, and $Z_{t-1}$ has a direct causal effect on $X_{t}$ that is not mediated by $X_{t-1}$ (and $X_{t-1}$'s effect on $X_{t}$ is not mediated by $Z_{t-1}$)? $\endgroup$ – Alexis Jan 28 '16 at 0:55
2
$\begingroup$

Consider a causal $ARMA(1,2)$ process $$ Y_t=\phi Y_{t-1}+\epsilon_t+\theta_1\epsilon_{t-1}+\theta_2\epsilon_{t-2} $$ Suppose our interest centers on estimating $\phi$, but we are not aware of the $MA$ components (or we just do not know how to fit ARMA models :-)).

One strategy might therefore consist of just running an OLS regression of $Y_t$ on $Y_{t-1}$. That regression would however inconsistently estimate $\phi$, as the regressor $Y_{t-1}$ is correlated with $\epsilon_{t-1}$ and $\epsilon_{t-2}$, which can be seen directly from shifting the ARMA(2,1) model by one period:

$$ Y_{t-1}=\phi Y_{t-2}+\epsilon_{t-1}+\theta_1\epsilon_{t-2}+\theta_2\epsilon_{t-3} $$ (One might work out the exact plim as for IV below.)

Suppose we instead use IV to estimate $\phi$. The IV estimator of $\phi$ using the lag $Y_{t-2}$ as an instrument for $Y_{t-1}$ is $$ \hat{\phi}_{IV}=\frac{\sum_tY_{t-2}Y_{t}}{\sum_tY_{t-2}Y_{t-1}} $$ Its probability limit therefore is $$ \hat{\phi}_{IV}=\frac{\frac{1}{T}\sum_tY_{t-2}Y_{t}}{\frac{1}{T}\sum_tY_{t-2}Y_{t-1}}\to_p\frac{\gamma_2}{\gamma_1}, $$ where $\gamma_j$ denotes the $j$th autocovariance.

We first find the $MA(\infty)$ representation of the process to find the autocovariances necessary for expressing the probability limit.

Matching coefficients in $$ (1-\phi L)(\psi_0+\psi_1L+\psi_2L^2+\psi_3L^3+\ldots)=1+\theta_1L+\theta_2L^2 $$ gives \begin{eqnarray*} \psi_0&=&1\\ -\phi\psi_0+\psi_1&=&\theta_1\quad\Rightarrow\quad\psi_1=\theta_1+\phi\\ -\phi\psi_1+\psi_2&=&\theta_2\quad\Rightarrow\quad\psi_2=\theta_2+\phi(\theta_1+\phi)\\ -\phi\psi_2+\psi_3&=&0\quad\Rightarrow\quad\psi_3=\phi(\theta_2+\phi(\theta_1+\phi))\\ \psi_j&=&\phi^{j-2}(\theta_2+\phi(\theta_1+\phi))\qquad j>1 \end{eqnarray*}

We may now use this to find $\gamma_1$ and $\gamma_2$.

From the general result on the autocovariance of an $MA(\infty)$ process that $\gamma_k=\sigma^2\sum_{j=0}^{\infty}\psi_j\psi_{j+k}$ we conclude that $\gamma_1=\sigma^2\sum_{j=0}^{\infty}\psi_j\psi_{j+1}$. Hence, \begin{eqnarray*} \gamma_1&=&\sigma^2\left[\theta_1+\phi+(\theta_1+\phi)(\theta_2+\phi(\theta_1+\phi))+(\theta_2+\phi(\theta_1+\phi))\sum_{j=2}^\infty\phi^{j-2}\phi^{j-1}\right]\\ &=&\sigma^2\left[\theta_1+\phi+(\theta_1+\phi)(\theta_2+\phi(\theta_1+\phi))+\phi\frac{(\theta_2+\phi(\theta_1+\phi))}{1-\phi^2}\right], \end{eqnarray*} as $\sum_{j=2}^\infty\phi^{j-2}\phi^{j-1}=\sum_{j=2}^\infty\phi^{2j-3}=\phi\sum_{j=0}^\infty\phi^{2j}$. Similarly, \begin{eqnarray*} \gamma_2&=&\sigma^2\sum_{j=0}^{\infty}\psi_j\psi_{j+2}\\ &=&\sigma^2\left[\theta_2+\phi(\theta_1+\phi)+(\theta_1+\phi)\phi(\theta_2+\phi(\theta_1+\phi))+(\theta_2+\phi(\theta_1+\phi))\sum_{j=2}^\infty\phi^{j-2}\phi^{j}\right]\\ &=&\sigma^2\left[\theta_2+\phi(\theta_1+\phi)+(\theta_1+\phi)\phi(\theta_2+\phi(\theta_1+\phi))+\phi^2\frac{(\theta_2+\phi(\theta_1+\phi))}{1-\phi^2}\right] \end{eqnarray*}

The IV estimator therefore converges to $$ \hat{\phi}_{IV}\to_p\frac{\sigma^2\left[\theta_2+\phi(\theta_1+\phi)+(\theta_1+\phi)\phi(\theta_2+\phi(\theta_1+\phi))+\phi^2\frac{(\theta_2+\phi(\theta_1+\phi))}{1-\phi^2}\right]}{\sigma^2\left[\theta_1+\phi+(\theta_1+\phi)(\theta_2+\phi(\theta_1+\phi))+\phi\frac{(\theta_2+\phi(\theta_1+\phi))}{1-\phi^2}\right]} $$ This does not equal $\phi$ in general.

Intuitively, the instrument then is not uncorrelated with the error term, as $E(Y_{t-2}\epsilon_{t-2})\neq0$.

If, however, $\theta_2=0$ (i.e., we have an $ARMA(1,1)$) the IV estimator would be consistent for $\phi$: $$ \hat{\phi}_{IV}\to_p\frac{\phi(\theta_1+\phi)+\phi^2(\theta_1+\phi)^2+\phi^2\frac{(\phi(\theta_1+\phi))}{1-\phi^2}}{\theta_1+\phi+\phi(\theta_1+\phi)^2+\phi\frac{(\phi(\theta_1+\phi))}{1-\phi^2}}=\phi $$

The result shows up similarly in the estimation of dynamic panel data models, namely that there must not be higher-order autocorrelation so that first differencing to remove the fixed effects does not induce correlation between the differenced error terms and the instruments.

$\endgroup$
4
  • $\begingroup$ This is a good example. I'm probably missing something here, but why would $Y_{t-1}$ be endogenous at the first place for someone to use IV. Could you give an example of endogeneity in this ARMA estimation? $\endgroup$ – Regis A. Ely Jan 28 '16 at 12:30
  • $\begingroup$ Good point, I made an edit: OLS on the first lag would be inconsistent for $\phi$ as errors and the regressor would be correlated. $\endgroup$ – Christoph Hanck Jan 28 '16 at 12:37
  • $\begingroup$ Oh, i see. I missed that one could try to estimate it using OLS. $\endgroup$ – Regis A. Ely Jan 28 '16 at 12:58
  • 2
    $\begingroup$ Wow, that's neat! I still have to read your post on weak instruments, and this one is the next one on my list $\endgroup$ – Jeremias K Jan 28 '16 at 15:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.