Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $V(t)= \frac{1}{2q+1}\sum_{j=-q}^q X(t-j)$ where $X_t = b_o + b_{1}*t + w_t$ where $w_t$ is ~$N(0,1)$. I know the mean is simply bo + b1t but when I calculate the autocovariance function I get 0. Is this correct?

$V(t)= \frac{1}{2q+1}\sum_{j=-q}^q X(t-j) = X(t+q) + X(t+q-1) + ...X(t) + X(t-1)+...+X(t-q) = b0 + b1(t+q) +w(t+q) + b0 + b1(t+q-1) + w(t+q-1) +...b0 +b1(t-q) + w(t-q) = \frac{1}{2q+1}*2q+1(b0 + b1*t + w(t))$

share|improve this question
    
I don't understand what you mean by $b_{1t}$. Also, since $V(t)$ is the average of $X$'s neighborhood, the adjacent $V$'s should have some correlation so I don't think it's 0. How did you calculate? –  ido Jan 14 '13 at 22:56
    
It should be b1*t. I simply calculated cov(V(t+h),V(t)) and got E[V(t+h)*V(t) - E[V(t+h)]E[V(t)]. All the terms cancel when I calculate this difference. –  lord12 Jan 14 '13 at 23:00

1 Answer 1

up vote 1 down vote accepted

For the cross term:

$E[V(t)V(s)] = E[\frac{1}{2q+1}\sum_j X(t-j)\cdot \frac{1}{2q+1}\sum_j X(s-j)] = $

$ = \frac{1}{(2q+1)^2}E[\sum_j\sum _k X(t-j) X(s-k)] $

$ = \frac{1}{(2q+1)^2}\sum_j\sum _k E[X(t-j) X(s-k)] $

From here, if $|t-s|\leq 2q$ you will have some $j,k$ for which you'll get some $Ew_a^2$ since you can find $j,k$ such that $t-j=s-k$ .

  • edit after your comment:

Let's say I have $v_t$ which contains only the $w_t$ part (the rest can be added later and I think it's removed by the second term of the covariance):

$ R(h) = Ev_tv_{t+h} = E\sum_{j=-1}^q w_{t-j} \sum_{k=-q}^q w_{t+h-k} $

$ = \sum_{j=-1}^q \sum_{k=-q}^q Ew_{t-j}w_{t+h-k} $

Here we will get a nonzero value if $t-j = t+j-k$ since it will yield $Ew_m^2$ for some $m$. Otherwise, it is zero for the iid process $w_t$.

Now, if $|h|>2q$ we won't get $t-j=t+h-k$ so $R(h)=0$. But if $|h|\leq 2q$, then:

For $h=0$ we get the demand $t-j=t-k\Rightarrow j=k$ and there are $2q+1$ options for that, since $j=-q,...,q$ and so does $k$. For each of the options we get a $1$, so $R(h=0)=(2q+1)\cdot 1 = (2q+1)$

For $h=1$ we get $t-j=t+h-k\Rightarrow j=k-1$. Here we lose one option, since for $k=-q$ we would need $j=-q-1$ and we don't have that. So $R(h=1)=2q$. Similarly for all the rest, until we get to $h=2q$ and here the condition is $j=k-2q$. Now there's only one possibility: $j=-q, k=q$.

So finally, the autocorrelation of this modified series (without the deterministic part) is something like $R(h) = (2q+1)-|h|$ when $|h|\leq 2q$ and otherwise 0.

share|improve this answer
    
I still don't quite understand. I'm representing s as t+h. I evaluate the individual products inside the expectation and get E[(b1 +b2t+wt)(b1+b2(t+h)+w(t+h) - (b1+b2t)(b1+b2(t+h)) –  lord12 Jan 15 '13 at 5:26
    
I have edited the answer. –  ido Jan 15 '13 at 18:22
    
Thanks for the edit. But isn't V(t) = b1 + b2*t + w(t) and V(s) = b1 + b2*(s) + w(s). So when you multiply the terms and take expectation you get a different answer. Why can't you use this method. –  lord12 Jan 15 '13 at 20:46
    
because $v$ is not as you wrote, it is the sum of $x$'s specified in the first message, at least as I see it. –  ido Jan 15 '13 at 20:57
    
but when you simplify the sum of the xi's you are left with V(t) = b1 + b2*t + w(t). Maybe I am missing something. –  lord12 Jan 15 '13 at 21:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.