Given the Wold representation:

$$X_t = \mu + \sum_{j=0}^{\inf} b_j \epsilon_{t-j}$$

Which is its autocovariance function?

I made some algebra and I ended up with: $$COV(X_s,X_t)=\sigma^2 \cdot \sum_{i=0}^{min(s,t)}b_i$$

Is it correct? Thanks.


1 Answer 1


Assume $s<t$. Then:

$$\text{Cov}(X_s, X_t) = \text{Cov}\left(\sum_{j=0}^\infty b_j \epsilon_{s-j},\sum_{j=0}^\infty b_j \epsilon_{t-j}\right) $$

$$ = \text{Cov}\left(b_0\epsilon_s+b_1\epsilon_{s-1}+\cdots,b_0\epsilon_t+b_1\epsilon_{t-1}+\cdots+b_{t-s}\epsilon_{t-(t-s)}+b_{t-s+1}\epsilon_{t-(t-s+1)}+\cdots\right)$$

$$= \text{Cov}\left(b_0\epsilon_s+b_1\epsilon_{s-1}+\cdots,b_{t-s}\epsilon_{s}+b_{t-s+1}\epsilon_{s-1}+\cdots\right)$$

$$=\sigma^2 \sum_{j=0}^\infty b_jb_{t-s+j}$$

If $s>t$, then we can follow an identical procedure and arrive at

$$\text{Cov}(X_s, X_t)=\sigma^2 \sum_{j=0}^\infty b_jb_{s-t+j}$$

Pulling these together, in general we have

$$\text{Cov}(X_s, X_t)=\sigma^2 \sum_{j=0}^\infty b_jb_{|t-s|+j}$$

Note that this also gives the expected answer if $t=s$, i.e.

$$\text{Cov}(X_t, X_t)= \sigma^2 \sum_{j=0}^\infty b_j^2= \text{Var}(X_t)$$


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