CT6 September 2018 Q9iii Yule-Walker equations

[Edward Smith]

On this question, I can see that for Model X, we can derive an estimate for a1 by calculating:
Autocov(0)=Cov(a0+a1yt-1+et,yt) = a1*autocov(1) + sigma^2
Autocov(1)=Cov(a0+a1yt-l+et,yt-1) = a1*autocov(0), thus a1 = autocov(1)/autocov(0) = r1

However, why can we not solve by:
Autocov(0)=Cov(a0+a1yt-1+et,a0+a1yt-1+et) = a1^2*autocov(0) + sigma^2
Autocov(1)=Cov(a0+a1yt-1+et,a0+a1yt-2+et-1) = a1^2*autocov(1)... here we get a1^2 = autocov(1)/autocov(1) which is not the same as above?

 

[Dave Johnson]

Hi Edward

Your method does give the same answer eventually, but I think it's a bit less efficient.

Using the Yule-Walker equations I can set up the simultaneous equations:

g0 = a1 * g1 + s^2
g1 = a1 * g0

From which I can get:

g0 = s^2 / (1 - a1^2)
g1 = a1 * s^2 / (1 - a1^2)

Using your alternative method for g1 I get (noting that cov(yt, et) = s^2):

g1 = cov(a0 + a1 * yt-1 + et, a0 + a1 * yt-2 + et-1)
= a1^2 * g1 + a1 * s^2

From which:
g1 = a1 * s^2 / (1 - a1^2)

So they do boil down to the same thing eventually, but I think it is harder work to do it the second way.

Hope that helps,

Dave

 

[Edward Smith]

Hi Edward

Your method does give the same answer eventually, but I think it's a bit less efficient.

Using the Yule-Walker equations I can set up the simultaneous equations:

g0 = a1 * g1 + s^2
g1 = a1 * g0

From which I can get:

g0 = s^2 / (1 - a1^2)
g1 = a1 * s^2 / (1 - a1^2)

Using your alternative method for g1 I get (noting that cov(yt, et) = s^2):

g1 = cov(a0 + a1 * yt-1 + et, a0 + a1 * yt-2 + et-1)
= a1^2 * g1 + a1 * s^2

From which:
g1 = a1 * s^2 / (1 - a1^2)

So they do boil down to the same thing eventually, but I think it is harder work to do it the second way.

Hope that helps,

Dave

Great - thanks Dave - I was losing a sigma^2 when taking the covariance of the (yt-1,et-1)