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

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

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

 

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

 

 

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.]

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

Which simplifies to:  (phi²)*(sigma^2_e)*(-1+phi²+phi³....)

Which simplifies to: (phi²)*(sigma^2_e)*(-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 & sigma^2_e they 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.

(phi²)*(sigma^2_e)*(-1+phi²+phi³....)   I don't know if this simiplifies the route you took.

(phi²)*(sigma^2_e)*(-1(-1+1)+phi²+phi³....)    So I added in zero essentially, as -1+1=0

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

(phi²)*(sigma^2_e)*(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

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 ^^

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!

PMActuary, the derivation on p.56 is for the particular case when all summands are positive. In our HW we have alternating signs, so we have to do a different calculation.

Basically, due to alternating signs, the terms in the covariance sum will enter one with +, the other with -, like this:
for the term -phi^2 x e(t-2) the other one will be +phi x e(t-2), and so on.
(However, the term e(t-1) enters both times with +).

For anyone that actually looks @ these forums for the fall course:

I was having the same problem as the quoted text...my issue was that I was not summing the geometric progression correctly...This is the case for both parts B & C.

The trick is that the first term cannot be part of the geometric sum as the sign is different. E.g. in Part B you get to:

phi*Var(et-1)) - phi^3*Var(et-2) - phi^5*Var(et-3)...

The proper sum is:

Var(et)*[phi - (phi^3 / (1-phi^2) ) ]

i.e. the first time is not part of the geometric sum.

Hope this helps someone.

I agree with djfobster except the -.345/9 does not equal what he or she says it equals.  But I could be losing it.

   -    Dave