# Time Series - Autoregressive process and Probability Limit

1. Oct 12, 2015

### frenchkiki

1. The problem statement, all variables and given/known data

Calculate: PLIM (probability limit) $\frac{1}{T} \sum^T_{t=2} u^2_t Y^2_{t-1}$
2. Relevant equations

$Y_t = \rho Y_{t-1} + u_t, t=1,...T, |\rho| <1$ which the autoregressive process of order 1

$E(u_t) = 0, Var(u_t) = \sigma^2$ for t

$cov(u_j, u_s) = 0$ for j $\neq s$

3. The attempt at a solution

I know that PLIM $\frac{1}{T} \sum^T_{t=2} u^2_t Y^2_{t-1} = E[u^2_t Y^2_{t-1}]$

I have found $Y_{t-1} = \sum^{T-1}_{j=0} \rho^j u_{t-1-j}$

Plugging in, I get $E[u^2_t Y^2_{t-1}] = E[u^2_t (\sum^{T-1}_{j=0} \rho^j u_{t-1-j})^2]=E[(u_t (\sum^{T-1}_{j=0} \rho^j u_{t-1-j}))^2]=E[(\sum^{T-1}_{j=0} \rho^j u_{t-j} u_{t-1-j})^2]=\sum^{T-1}_{j=0} \rho^j E[(u_{t-j} u_{t-1-j})^2]$

And I am stuck here because I don't know what to do with $E[(u_{t-j} u_{t-1-j})^2]$ ??

Last edited: Oct 12, 2015
2. Oct 12, 2015

### andrewkirk

Perhaps they meant to say that $u_j,u_s$ are independent for $j\neq s$. If That were the case then you could write:

$$E[(u_{t-j} u_{t-1-j})^2]=E[{u_{t-j}}^2 {u_{t-1-j}}^2]=E[{u_{t-j}}^2]\cdot E[{u_{t-1-j}}^2]$$
since ${u_j}^2,{u_s}^2$ will then also be independent.

However they have only given you the weaker condition that $cov(u_j, u_s) = 0$, which I think is not enough to justify that step.

Indeed, I wonder whether it would be possible to construct a counter-example in which the process $u_j$ is conditional heteroscedastic, so that its unconditional variance is $\sigma^2$ and serial correlation is zero but its conditional variance is a mean-reverting random walk, so that successive values are not independent.

You could ask your teacher whether you are allowed to assume serial independence.

3. Oct 12, 2015

### frenchkiki

Thanks andrewkirk. I've seen somewhere in my notes that the errors (u_t's) are i.i.d. I'll use independence then.

4. Oct 12, 2015

### Ray Vickson

I get a different expression from yours, and the difference is substantial. By iterating the recurrence relation I get
$$Y_{t-1} = \rho^{t-1} Y_0 + \sum_{j=0}^{t-2} \rho^j u_{t-1-j}.$$
Thus
$$u_t^2 Y_{t-1}^2 = \rho^{2t-2} u_t^2 Y_0^2 + 2 \rho^{t-1} Y_0 \sum_{j=0}^{t-2} \rho^j u_{t-1-j} u_t^2 \\ + \sum_{j=0}^{t-2} \rho^{2j} u_{t-1-j}^2 u_t^2 + \sum_{k=1}^{t-2} \sum_{j=0}^{k-1} \rho^{j+k} u_{t-1-j} u_{t-1-k} u_t^2$$
In order to be able to compute $E(u_t^2 Y_{t-1}^2)$ you need to assume independence of $u_1, u_2, \ldots u_T$, and you need to make some assumptions about the nature of $Y_0$ and its relation to the $u_t$ sequence.

5. Oct 13, 2015

### andrewkirk

As long as the $u_k$ are i.i.d, when we take the expected value, all the terms that have a factor $u_\alpha$ will become zero and any factors of the form ${u_\alpha}^2$ will become $\sigma^2$. I think that will get rid of the double sum and the sum on the first line. There will still be a $E[{Y_0}^2]$ factor in the first term, but that's OK because it's not entangled with anything else distributionwise.

6. Oct 13, 2015

### Ray Vickson

I wanted the OP to deal with these issues, so I deliberately refrained from saying more about them in my message.

7. Oct 13, 2015

### andrewkirk

Oh, fair enough then. I didn't realise that was what your post was aiming at. Sorry for mucking it up.

8. Oct 13, 2015

### frenchkiki

Thanks Ray and andrewkirk.

The terms involving $Y_0$ go to 0 as $T$ goes to infinity because $|\rho|$ < 1. Each element ends up being the covariance of the squared errors and the whole thing equals to 0.

Thanks for you help!