Instrumental variable estimation of the autoregressive coefficient in ARMA(1,1) model 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.)
 A: 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.
