TS Module 5: Stationary processes HW


TS Module 5: Stationary processes HW

Author
Message
PMActuary
Forum Newbie
Forum Newbie (2 reputation)Forum Newbie (2 reputation)Forum Newbie (2 reputation)Forum Newbie (2 reputation)Forum Newbie (2 reputation)Forum Newbie (2 reputation)Forum Newbie (2 reputation)Forum Newbie (2 reputation)Forum Newbie (2 reputation)

Group: Forum Members
Posts: 2, Visits: 1

I can't seem to get part B.  I used the forumla in the middle of p.56

(.2)*9/(1+.2^2) = 1.731 but it looks like the true answer is 1.725.  Can someone show me where I am going wrong please?  I used a + in the denominator because the ratio is  (-) theta ^2 now unlike in the book is + phi ^2.

How do the signs alternating in the problem affect our solutions?  I am obviously not doing it correctly.

Thanks for the help!


djfobster
Forum Newbie
Forum Newbie (7 reputation)Forum Newbie (7 reputation)Forum Newbie (7 reputation)Forum Newbie (7 reputation)Forum Newbie (7 reputation)Forum Newbie (7 reputation)Forum Newbie (7 reputation)Forum Newbie (7 reputation)Forum Newbie (7 reputation)

Group: Forum Members
Posts: 4, Visits: 1
Thanks rcoffman

to rcoffman and raydhiii
I agree with A and B also.
for C,
Corr(Yt,Yt-2) = Cov(Yt,Yt-2)/(sqrt(Var(Yt)*Var(Yt-2)) = Cov(Yt,Yt-2)/Var(Yt) (since Var(Yt) = Var(Yt-k) for k = 0, 1, 2, 3, . . . )
I got
Cov(Yt,Yt-2) = -0.345

Derivation:

Cov(Yt,Yt-2) =
Cov(e(t)+phi*e(t-1)-phi^2e(t-2)+ . . ., e(t-2)+phi*e(t-3)-phi^2*e(t-4)+. . . ) =>

skipping some steps =>

-phi^2*Var(e(t-2))+phi^4*Var(e(t-3))+phi^6*Var(e(t-4))+ . . . =>

sigma^2*(-phi^2+phi^4+phi^6+ . . . ) =>

sigma^2*(-phi^2+phi^4*(1+phi^2+phi^4+ . . .)) =>

sigma^2*(-phi^2+phi^4/(1-phi^2))

plug in the sigma^2 = 9 and phi = 0.2

THEN to get Corr we divide by Var(Yt)

to get -.0368

anyone agree?


thanks for your time and patience ^^
RayDHIII
Forum Member
Forum Member (43 reputation)Forum Member (43 reputation)Forum Member (43 reputation)Forum Member (43 reputation)Forum Member (43 reputation)Forum Member (43 reputation)Forum Member (43 reputation)Forum Member (43 reputation)Forum Member (43 reputation)

Group: Forum Members
Posts: 39, Visits: 138

(phi2)*(sigma2e)*(-1+phi2+phi3....)   I don't know if this simiplifies the route you took.

(phi2)*(sigma2e)*(-1(-1+1)+phi2+phi3....)    So I added in zero essentially, as -1+1=0

(phi2)*(sigma2e)*(-2+1+phi2+phi3....)    Now I can simplify this to:

(phi2)*(sigma2e)*(1/(1-phi2)+2)

as 1+x2+x3+x4+... = 1/(1-x2)

rcoffman: I feel that you missed a square on the phi in your denominator of your last equation, correct me if I'm wrong.

RDH


rcoffman
Junior Member
Junior Member (11 reputation)Junior Member (11 reputation)Junior Member (11 reputation)Junior Member (11 reputation)Junior Member (11 reputation)Junior Member (11 reputation)Junior Member (11 reputation)Junior Member (11 reputation)Junior Member (11 reputation)

Group: Forum Members
Posts: 10, Visits: 1

raydhiii: I agree with your A & B, but not C.  For C I got gamma2=-.27.  The formula I got for the Cov was Cov(-phi2*et-2,et-2)+Cov(phi3*et-3,phi*et-3)+Cov(-phi4*et-4,-phi2*et-4).....

Which simplifies to:  (phi2)*(sigma2e)*(-1+phi2+phi3....)

Which simplifies to: (phi2)*(sigma2e)*(-1+phi/(1-phi))

You had indicated on your edit that you revised gamma2 due to a sign error, so I was wondering if maybe I made the same sign error, but I could not find one.  Can you help my figure out why our answers are different from what I've posted?

djfobster:  We are given an infinite series (general linear process), not an n-order moving average process.  I think since they give you values of that phi & sigma2they want an actual figure.  You can find this figure by taking the sum of an infinite geometric series, as shown on Page 56 of the text.


djfobster
Forum Newbie
Forum Newbie (7 reputation)Forum Newbie (7 reputation)Forum Newbie (7 reputation)Forum Newbie (7 reputation)Forum Newbie (7 reputation)Forum Newbie (7 reputation)Forum Newbie (7 reputation)Forum Newbie (7 reputation)Forum Newbie (7 reputation)

Group: Forum Members
Posts: 4, Visits: 1

raydhiii,

how can you get an actual number when the series is a moving average process n-order?
i would think an expression with the n-order would be sufficient.

NEAS, are you looking for an expression of an actual figure?

[NEAS: The variance of a process is the expected variance. This is the same for all stochastic distributions and stochastic processes.]



thanks for your time and patience ^^
RayDHIII
Forum Member
Forum Member (43 reputation)Forum Member (43 reputation)Forum Member (43 reputation)Forum Member (43 reputation)Forum Member (43 reputation)Forum Member (43 reputation)Forum Member (43 reputation)Forum Member (43 reputation)Forum Member (43 reputation)

Group: Forum Members
Posts: 39, Visits: 138

Please remove my post if this is not allowed/encouraged:

I understand this section's homework is testing our knowledge of the derivations of gamma and rho at different lags for a general linear process.  In my opinion NEAS gave us an alternating series to keep us on our toes with silly things such as sign errors.  While I am confident in my derivations, as the formulae are indeed straightfoward, I would like to check my finals answers against either the the back of the book (not an option) or the forum.  My answers will be rounded to three decimal places.

A.) gamma0 = 9.375

B.) gamma1 = 1.725

C.) gamma2 = .345  -->  rho2 = gamma2/gamma0

Lastly, does "one or two lines is sufficient for each part" mean only show some of your work?

RDH

(edited with revised gamma2, guess why: sign error!)

[NEAS: Moving average processes often alternate: a positive residual in one period often causes a lower value in the next period. Sales of durable goods are an example. If sales are high one year, consumers may have lower demand the next year. If sales are low one year, the pent-up demand causes higher sales the next year.

The "one of two lines" is as you show. You need not explain the definitions of the gamma and rho parameters; just show the calculation.]


NEAS
Supreme Being
Supreme Being (6K reputation)Supreme Being (6K reputation)Supreme Being (6K reputation)Supreme Being (6K reputation)Supreme Being (6K reputation)Supreme Being (6K reputation)Supreme Being (6K reputation)Supreme Being (6K reputation)Supreme Being (6K reputation)

Group: Administrators
Posts: 4.5K, Visits: 1.6K

TS Module 5: Stationary processes HW

 

(The attached PDF file has better formatting.)

 

Homework assignment: general linear process

 

A time series has the form Yt = åt + ö × åt-1ö2 × åt-2 + ö3 × åt-3 – …

 

The plus and minus signs alternate.  ö = 0.2 and ó2e = 9.

 


A.     What is ã0, the variance of Yt? Show the derivation.

B.     What is ã1, the covariance of Yt and Yt-1? Show the derivation.

C.    What is ñ2, the correlation of Yt and Yt-2? Show the derivation.

 

(Show the algebra for the derivations. One or two lines is sufficient for each part.)

 

 


Attachments
 
GO
Merge Selected
Merge into selected topic...



Merge into merge target...



Merge into a specific topic ID...





Reading This Topic

2 active, 2 guests, 0 members, 0 anonymous
No members currently viewing this topic!

Login
Existing Account
Email Address:


Password:


Login
Social Logins

  • Login with twitter
  • Login with twitter
Select a Forum....













































































































































































































































Neas-Seminars