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Consider ARMA(1,1)

$y_t=by_{t-1}+u_t$

$u_t=ae_{t-1}+e_t$

$\text{Var}(e_t)=\sigma^2$

Why is $y_{t-1}$ endogenous and why is there an endogenity problem?

How can I provide an instrumental variable (for example $y_{t-2}$) such that this variable satisfies the conditions of validity and relevance?

(This is not homework. I just try to understand this topic. Thank you.)

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Because the error term is serially correlated($a \neq 0$), $$E(Y_{t-1}u_t)=E((bY_{t-1}+ae_{t-2}+e_{t-1})(ae_{t-1}+e_t))=aE(e_{t-1}^2)=a\sigma^2 \neq 0$$

That's why we have the problem of endogeneity, and the usual LS estimators are not consistent.

In fact, in this case, the LS estimator $$\hat b = \frac{\sum y_t y_{t-1}}{\sum y^2_{t_1}}=b+\frac{\sum y_{t-1} u_t}{\sum y^2_{t_1}}$$ where the last term tends in probability to $$\frac{a(1-b^2)}{1+ab}$$

To prove this last limit, there are several ways. One of them, is to use a Law of Large Numbers and notice that $\frac{1}{T}\sum Y_{t-1}u_t\rightarrow^pE[Y_{t-1}u_t]$ and $\frac{1}{T}\sum Y_{t-1}^2\rightarrow^pE[Y_t^2]$.

Then, assuming that processes are stationary and that both have an MA representation, we can write:

$u_t=\sum_{l\geq 0} e_{t-l}a^l$

$y_t=\sum_{i\geq 0}\sum_{j\geq 0}e_{t-i-j}a^jb^i=\sum_{i\geq 0}\sum_{k\geq i+1}e_{t-k}a^{k-i}b^i$.

So, $\begin{split} \displaystyle E[Y_t^2] & = E\left[\sum_{i\geq 0}\sum_{k\geq i+1}e_{t-k}a^{k-i}b^i\sum_{j\geq 0}\sum_{l\geq j+1}e_{t-l}a^{l-j}b^j\right] \\ & =E\left[\sum_{i\geq 0}\sum_{j\geq 0}\sum_{k\geq m_{i,j}+1}e_{t-k}^2a^{2k-(i+j)}b^{i+j}\right]\\ & =\sum_{i\geq 0}\sum_{j\geq 0}\left(\frac{b}{a}\right)^{i+j} \sigma^2 \sum_{k\geq m_{i,j}+1}(a^2)^{k} \end{split}$ where $m_{i,j}:=\max\{i,j\}$,

$\begin{split} & =\frac{\sigma^2 a^2}{1-a^2}\sum_{i\geq 0}\sum_{j\geq 0}b^{i+j} a^{2m_{i,j}-(i+j)} \\ & = \frac{\sigma^2 a^2}{1-a^2}\left(\sum_{i=j\geq 0}b^{2i}+2\sum_{i>j\geq 0}b^{i+j} a^{i-j}\right) \\ & = \frac{\sigma^2 a^2}{1-a^2}\left(\frac{1}{1-b^2}+2\sum_{j \geq 0}\left(\frac{b}{a}\right)^{j}\sum_{i\geq j+1}(ba)^{i}\right) \\ & = \frac{\sigma^2 a^2}{1-a^2}\left(\frac{1}{1-b^2}+\frac{2ba}{1-ba}\frac{1}{1-b^2}\right)=\frac{\sigma^2 a^2}{1-a^2}\frac{1}{1-b^2}\frac{1+ba}{1-ba} \end{split}$

Now, for

$\begin{split} \displaystyle E[Y_{t-1}u_t] & = E\left[\sum_{i\geq 0}\sum_{k\geq i+1}e_{t-1-k}a^{k-i}b^i\sum_{l\geq 0}e_{t-l}a^{l}\right] \\ & =E\left[\sum_{i\geq 0}\sum_{k\geq i+2}e_{t-k}a^{k-i-1}b^i\sum_{l\geq 0}e_{t-l}a^{l}\right]\\ & =E\left[\sum_{k-2\geq i\geq 0}e_{t-k}^2a^{2k-i-1}b^i\right] \\ & = \sigma^2 \sum_{i\geq 0} b^ia^{-(i+1)}\sum_{k\geq i+2}a^{2k} \\ & = \frac{\sigma^2}{1-a^2}\sum_{i\geq 0} b^ia^{i+3} \\ & = \frac{\sigma^2 a^3}{1-a^2}\frac{1}{1-ba} \end{split}$

And this gives the desired result.

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    $\begingroup$ You could mention that there is no usual OLS estimator for an ARMA (1,1) model. The estimator you have here is for an AR(1) model. It is model misspecification that leads to inconsistency of the estimators, not, say, the estimation routine (OLS or other). $\endgroup$ – Richard Hardy Apr 7 '18 at 8:23
  • $\begingroup$ @RichardHardy I've tried to deduce the 'inconsistency term'. Do you think it's ok? ;) $\endgroup$ – An old man in the sea. Apr 8 '18 at 7:36
  • $\begingroup$ Well, you have put in a lot of work which is very nice of you, but I do not think it addresses what I was concerned with. I would have just included the text from my comment (perhaps rephrased) somehwere at the beginning of the answer. $\endgroup$ – Richard Hardy Apr 8 '18 at 8:06

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