A. What is the variance of Y_t as a function of â and t?

B. What is ñ(y_t,y_t-k) as a function of â, k, and t?

See equations 5.1.4 and 5.1.5 on page 89.

{Note: This homework assignment has been replaced because of an unclear equation in the textbook; see the new homework assignment. If you have submitted this assignment already, you will be given credit.}

I said this in the module 8 section.  The homework for module 9 references material in module 8.  Now I am a few pages into the module 9 reading and it looks like you can answer the module 8 homework with the module 9 reading.  Am I off base with this?

[NEAS: Both homework assignments deal with ARIMA processes. Answer them after working through both modules.]

 

 

Tom McNamara III

Should we assume that Y_t=0 for t<1?  You made it clear in the practice problems that this condition must be given in order for the variance to be defined, but then did not give us this condition in the homework.

[NEAS: Yes; use the same assumptions as in textbook on page 89, equations 5.1.4 and 5.1.5.]

Ouch, zarito.  I really don't like that the textbook highlights formulae for specific problems, and not general cases.  It is fairly straightforward to derive the general case from 5.1.4 and 5.1.5.  The problem does ask to show the derivations though, so I began from 5.1.2 and 5.1.3 on the previous page.  Taking a look at 5.1.4, we know beta = 3, but there is an 8 and a 9 in the formula.

So, ask yourself, how does one derive 8 and 9 from 3?

8 = 3^2 - 1

9 = 3^2

This should very simply solve your dilema.  If you would like to follow the 5.1.2 and 5.1.3 route, it involves geometic series, which also simplify very nicely.  Best of luck!

RDH

Believe me, I got the formulas for the general case, by using the beta=3 scenario The problem was the derivation, since I was not sure which time series to use for the general case. Thank you for the help!

I must be missing a negative sign somewhere in my derivation of the Variance of Equation 5.1.3. I am not sure exactly what to do with the (beta^t)*(Y_0) term at the end when taking the variance, so I'm sure that's where I'm messing up. 

Not taking that term into account, I sum a geometric series to get Var_Yt = (sigma^2)_e * (1 - beta^2t) / (1 - beta^2). Multiplying that by -1 gets me to the variance expressed in Equation 5.1.4, but I don't know how to add the variance of the last term in 5.1.3 to get to the -1 multiplier that I need. Any help/suggestions? Thanks.

rmconrad, the variance of Y₀ is simply the variance of the initial error, e₀.  The variance of beta^tY₀ should easily follow from there, recalling the rules of variance and beta a constant.  Let me know if you have further questions.

RDH

Correlation is

Corr(X,Y) = Cov(X,Y) / (sigma_X * sigma_Y)
= Cov(X,Y) / sqrt( Var(X) * Var(Y) )

Corr(Y_t, Y_t-k) = Cov(Y_t, Y_t-k) / sqrt( Var(Y_t) * Var(Y_t-k) )

In this case, X is Y_t and Y is Y_t-k. For a stationary process, Var(Y_t-k) does not depend on k and the square root drops out. In this case, Var(Y_t-k) depends on k, so the square root does not go away.

I am having trouble getting the exact result for VAR(Y).

In both the text and from first principles, I am arriving easily at Var(Y) = sigma-sq * (1 + B^2 + B^2*2 + B^2*3... + B^2t).  If this were an infinite series, the sum would be [1 / (1 - B^2)].  Don't we want to subtract out the rest of the series, starting at B^2(t+1), thereby making the variance:

 sigma-sq * [B^2(t+1) - 1] / [B^2 - 1]..... rather than the same thing with B^2*t?  I'm sure since at least some of you figured this out that I am just messing something up, but I can't see what I've done wrong.

Ah, I see, Y(0) is assumed to be 0, thereby removing B^2t from the original thing I posted, and making that the subtracted term in the finite series.