Time Series - Autoregressive process and Probability Limit

[frenchkiki]

Homework Statement

Calculate: PLIM (probability limit) [itex]\frac{1}{T} \sum^T_{t=2} u^2_t Y^2_{t-1} [/itex]

Homework Equations

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

[itex] E(u_t) = 0, Var(u_t) = \sigma^2[/itex] for t

[itex] cov(u_j, u_s) = 0[/itex] for j [itex]\neq s [/itex]

The Attempt at a Solution

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

I have found [itex] Y_{t-1} = \sum^{T-1}_{j=0} \rho^j u_{t-1-j} [/itex]

Plugging in, I get [itex]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][/itex]

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

Thank you in advance!

 

[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.

 

[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.

 

[Ray Vickson]

Homework Statement

Calculate: PLIM (probability limit) [itex]\frac{1}{T} \sum^T_{t=2} u^2_t Y^2_{t-1} [/itex]

Homework Equations

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

[itex] E(u_t) = 0, Var(u_t) = \sigma^2[/itex] for t

[itex] cov(u_j, u_s) = 0[/itex] for j [itex]\neq s [/itex]

The Attempt at a Solution

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

I have found [itex] Y_{t-1} = \sum^{T-1}_{j=0} \rho^j u_{t-1-j} [/itex]

Plugging in, I get [itex]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][/itex]

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

Thank you in advance!

I get a different expression from yours, and the difference is substantial. By iterating the recurrence relation I get
[tex]Y_{t-1} = \rho^{t-1} Y_0 + \sum_{j=0}^{t-2} \rho^j u_{t-1-j}. [/tex]
Thus
[tex] 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 [/tex]
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.

 

[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.

 

[Ray Vickson]

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.

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

 

[andrewkirk]

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

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

 

[frenchkiki]

Thanks Ray and andrewkirk.

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

Thanks for you help!