Graphical Interpretation of Autocorrelation - Diebold reading

[chanhoke]

Hi,

On page 14 of Diebold's readings, 4 graphs were given (Figure 2 - Figure 5), I'd like to confirm which of the those graphs represents Covariance stationary?

from Diebold's definition, autocorrelation function should
(1) decay with the increase in displacement
(2) approach 0 with larger displacement
With that said, figure 3 (one-sided gradual damping) should be the only graph representing Covariance stationary, no?

thanks in advance.

 

[Dr. Jayanthi Sankaran]

Hi @chanhoke,

I will hazard a guess here. As you rightly mention from Diebold's definition, autocorrelation function should:

(1) decay with the increase in displacement
(2) approach 0 with larger displacement

• Figure 3 is most definitely not a covariance stationary process because is has a constant autocorrelation function and does not eventually decay.
• Figure 2 shows an autocorrelation function that displays gradual one-sided damping.
• Figure 4 shows an autocorrelation function that displays damped oscillation - the autocorrelations are positive at first, then become negative for a while, then positive again, and so on, while continuously getting smaller in absolute value
• Figure 5 shows an autocorrelation function where the autocorrelations drop abruptly to zero beyond a certain displacement

Other than Figure 3, Figures 2, 4 and 5 are all covariance stationary - their autocorrelation functions approach 0, one way or the other!

Hope that helps
Jayanthi

 

[chanhoke]

Hi Jayanthi,

first off, Thanks so much for your reply.

your answer seems to be what Diebold was implying as he kind of grouped the 3 graphs into one paragraph but never explicitly state they are Covariance stationary.

my followup question though, would be say for figure 4 (an autocorrelation function with sharp cutoff), at displacement >15, the autocorrelation goes to zero. as per Diebold's definition of covariance stationary, the autocorrelation should approach zero but never hit zero even at high displacements?

 

[Dr. Jayanthi Sankaran]

Hi @chanhoke,

Thanks - glad to be of help! Figure 4 is just a special case of an autocorrelation function, approaching zero. As you put it, at displacement > 15, the autocorrelation function falls to zero. I guess approaching zero and abruptly falling to zero are all the same, as far as Diebold states. I think it is just semantics.....

Thanks!
Jayanthi

 

[chanhoke]

Hi Jayanthi,

good point. thanks again for your reply.

 

[Dr. Jayanthi Sankaran]

Thanks - appreciate it! I was re-thinking about your question. I will get back to you, if I think differently as I review Diebold.

Jayanthi