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Consider the MA(1) process:

$$ y_t = \varepsilon_t + \theta_1 \varepsilon_{t-1} $$

where $\varepsilon$ is a white noise process with $\mathbb{E}(\varepsilon_t) = 0$ and $\operatorname{Var}(\varepsilon_t) = \sigma^2$, and $t=1,2,\dotsc,T$.   Assume now that given a realization of the MA(1) process you estimate instead an AR(1) model:

$$ y_t = \varphi_1 y_{t-1} + \varepsilon_t $$

where $0<\varphi_1<1$ is an unknown parameter, $\varepsilon$ is a white noise process, and $t=1,2,\dotsc,T$.

Derive the (large sample) bias in the OLS estimator $\varphi_1$.

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    $\begingroup$ I edited the question considerably introducing $\LaTeX$ and standard notation (standard coefficient names) so as to make it easier for people used to the standard notation. I also stripped the $m$ that was not used. Please see if the idea remains unchanged. $\endgroup$ – Richard Hardy Mar 12 '16 at 10:54
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    $\begingroup$ If this is a homework exercise, please add the self-study tag and read its Wiki. (You will need to remove one of the current tags, I suggest you may remove time-series.) $\endgroup$ – Richard Hardy Mar 12 '16 at 10:58
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Lets write the MA as an AR process, $$ \frac{1}{1-L\theta}y_t=\epsilon_t $$ that is an infinite order AR process, $$ \epsilon_t=\sum_{i=0}^{\infty}\theta^iy_{t-i} $$ that is $y_t= \sum_{i=1}^{\infty}\theta^iy_{t-i} +\epsilon_t$. Then you are are ignoring infinite number of terms as you just consider the first order of AR. Thus the bias is comming from misspecifing model. I show an example as you case is a more general case.

Let the true model be $y=x_1\beta_1+x_2\beta_2+e$ and the analyst considers just one variable, $x_1$ then OLS gives, $$ \hat\beta_1=(x'_1 x_1)^{-1}x'_1y $$ and $$ E(\hat\beta_1)=(x'_1 x_1)^{-1}x'_1(x_1\beta_1+x_2\beta_2+e)=\beta_1+E\{(x'_1 x_1)^{-1}x'_1x_2\}\beta_2 $$ where the second term in the right hand side is bias. Your case is more general to this example by infinite terms. On the other hand it is imposible to use OLS for an $AR(\infty)$ process as the covariance matrix is not invertible except $T\rightarrow \infty$.

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  • $\begingroup$ This was a self-study question, so you were supposed to give hints gradually than provide a complete solution at once. $\endgroup$ – Richard Hardy Mar 13 '16 at 10:57

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