I've stumbled upon this question in my econometric textbook and can't work out the right answer. The question:
Consider $$Y_t = B_2Y_{t-1} + u_t\\ \\ u_t = pu_{t-1} + \epsilon_t\\ $$
where $ \epsilon $ ~ N (0,1), absolute values of $B_2$ and p are less than one and
$Y_0 = 0 $ and $u_0 = \epsilon_0$
Show that $$ plim b_2^{OLS} = \frac {B_2 + p}{1+B_2p}$$
I've decomposed it into $$B_2 + \frac{covariance [Y_{t-1}, u_t]}{variance [Y_{t-1}]}$$ but the final answer is different from the textbook's. All help really appreciated.
Edit: my answer so far.
$var(u_t)=var \sum_{i=0}^tp^i\epsilon_{t-i} = \sum_{i=0}^tvar(p^i\epsilon_{t-i}) = \sum_{i=0}^tp^{2i}var(\epsilon_{t-i}) = \sum_{i=0}^tp^{2i}$
As $t \to \infty, $ $$var(u_t) \to \frac{1}{1-p^2}$$
$Cov[Y_{t-1},u_t] = Cov[B_2Y_{t-2} + u_{t-1},pu_{t-1} + \epsilon_t]$
The epsilon has no covariance with the two terms, so the expression changes to $$ Cov[Y_{t-1},u_t] = B_2p Cov[Y_{t-2},u_{t-1}] + pvar(u_{t-1})$$
The two expressions for covariance are the same as t$ \to \infty$ so I end up with
$$ Cov[Y_{t-1},u_t] = \frac{pvar(u_{t-1})}{1-B_2p} = \frac{p}{(1-p^2)(1-B_2p)} $$
For the variance of $Y_{t-1}$ I have $var \sum_{i=0}^tB_2^iu_{t-i} = \sum_{i=0}^tB_2^{2i}var(u_{t-i})$
Yesterday I proceeded by substituting in the value for the variance of $u_t$ which I derived, but I now realise that's wrong since that is the plim of its variance and what I'm looking for is the sum of the individual variances of each $u_t$. Do I have to substitute the epsilons for each $u_t$ like so?
$var \sum_{i=0}^tB_2^iu_{t-i} = \sum_{i=0}^tB_2^{2i}var(\sum_{k=0}^ip^k\epsilon_{i-k})$
Hm, I question whether the answer from your book is correct.
From $|p|<1$ and the AR(1) specification
$$ u_t = p u_{t-1} + \epsilon_t, $$
we get the one-sided MA($\infty$) representation
$$ u_t = \sum_{s=0}^{\infty} p^s \epsilon_{t-s}. $$
If $\epsilon_t$'s have mean $0$ and are uncorrelated, the ACF of $\{ u_t \}$ is
$$ \gamma(m) = \frac{p^m}{1-p^2}. $$
In particular, like you said, the variance of $u_t = \gamma(0) = \frac{1}{1-p^2}$. Similarly,
$$ Y_t = \sum_{s=0}^{\infty} B^s u_{t-s}. $$
$\{ Y_t \}$ is weakly stationary like $\{ u_t \}$, and has variance
$$ \gamma(0) \sum_{s=0}^{\infty} B^{2s}+ \gamma(1) B \sum_{s=0}^{\infty} B^{2s} + \cdots = \frac{1}{1-B^2} \sum_{m=0}^{\infty} \gamma(m) B^m. $$
Plugging in $\gamma(m)$ gives
$$ Var(Y_t) = \frac{1}{1-B^2} \cdot \frac{1}{1-p^2} \cdot \frac{1}{1-Bp}. $$
As you said, the OLS estimator converges in probability to
$$ B + \frac{Cov(u_t, Y_{t-1})}{Var(Y_{t-1})}. $$
(One needs to quote a relevant LLN here but some econometrics books are sloppy about these things.) The term $Cov(u_t, Y_{t-1})$ can be computed again using the MA-representation:
$$ Cov(u_t, Y_{t-1}) = Cov(u_t, \sum_{s=0}^{\infty} B^s u_{t-1-s}) = \sum_{s=0}^{\infty} B^s \gamma(s+1) = \frac{p}{1-p^2} \cdot \frac{1}{1-Bp}. $$
Putting it together, you get
$$ \frac{Cov(u_t, Y_{t-1})}{Var(Y_t)} = p(1-B^2). $$
I don't know where the factor $\frac{1}{1+Bp}$ in your book comes from.