# What can go wrong using lagged terms as instrumental variables?

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 $X_{t}$?

• 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 $Z_{t-1}$)? – Alexis Jan 28 '16 at 0:55

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.

• 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? – Regis A. Ely Jan 28 '16 at 12:30
• Good point, I made an edit: OLS on the first lag would be inconsistent for $\phi$ as errors and the regressor would be correlated. – Christoph Hanck Jan 28 '16 at 12:37
• Oh, i see. I missed that one could try to estimate it using OLS. – Regis A. Ely Jan 28 '16 at 12:58
• Wow, that's neat! I still have to read your post on weak instruments, and this one is the next one on my list – Jeremias K Jan 28 '16 at 15:14