1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Time Series - Autoregressive process and Probability Limit

  1. Oct 12, 2015 #1
    1. The problem statement, all variables and given/known data

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

    3. 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!
     
    Last edited: Oct 12, 2015
  2. jcsd
  3. Oct 12, 2015 #2

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  4. Oct 12, 2015 #3
    Thanks andrewkirk. I've seen somewhere in my notes that the errors (u_t's) are i.i.d. I'll use independence then.
     
  5. Oct 12, 2015 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
  6. Oct 13, 2015 #5

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  7. Oct 13, 2015 #6

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I wanted the OP to deal with these issues, so I deliberately refrained from saying more about them in my message.
     
  8. Oct 13, 2015 #7

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Oh, fair enough then. I didn't realise that was what your post was aiming at. Sorry for mucking it up.
     
  9. Oct 13, 2015 #8
    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Time Series - Autoregressive process and Probability Limit
  1. The limit of a series (Replies: 5)

  2. Limit of a Series (Replies: 4)

Loading...