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?

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

  • $\begingroup$ 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? $\endgroup$
    – yoki
    Jan 14, 2013 at 22:56
  • $\begingroup$ 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. $\endgroup$
    – lord12
    Jan 14, 2013 at 23:00

1 Answer 1


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.

  • $\begingroup$ 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)) $\endgroup$
    – lord12
    Jan 15, 2013 at 5:26
  • $\begingroup$ I have edited the answer. $\endgroup$
    – yoki
    Jan 15, 2013 at 18:22
  • $\begingroup$ 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. $\endgroup$
    – lord12
    Jan 15, 2013 at 20:46
  • $\begingroup$ 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. $\endgroup$
    – yoki
    Jan 15, 2013 at 20:57
  • $\begingroup$ 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. $\endgroup$
    – lord12
    Jan 15, 2013 at 21:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .