# OLS slope estimate of AR(1) with autocorrelation

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?

$var \sum_{i=0}^tB_2^iu_{t-i} = \sum_{i=0}^tB_2^{2i}var(\sum_{k=0}^ip^k\epsilon_{i-k})$

• Have you tried finding the variance and covariance you are looking for? Where do you get stuck? Commented Apr 29, 2014 at 14:54
• Yes, of course, but the final result I reach is different from the answer provided. Commented Apr 29, 2014 at 16:24
• I understand that. My point was that it is common practice at Math SE to provide your work so that others can help you find where you go wrong, rather than just providing solutions to your problem. So if you edit your post adding what you tried, you are much more likely to get help. Commented Apr 29, 2014 at 19:41
• I've edited it now. Sorry for the misunderstanding. Commented Apr 30, 2014 at 10:45
• I've just had a very quick look at it, so I haven't checked if this solves it. But when you do the variance of $Y_{t-1}$, note that $V(\sum B_2^iu_{t-i})\neq \sum B_2^{2i}V(u_{t-i})$ since you have correlation between the $u$ terms. Commented Apr 30, 2014 at 11:30

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.

• Thank you for this. The preceding chapter before this question was not very difficult, so I'm presuming that it is a misprint. Commented May 1, 2014 at 7:45