(See page 50 of the Cryer and Chan text, Exercise 3.2)

 

 

Shouldn't this homework be moved to a place where the moving average terminology has been introduced? It took me a very long time to do this problem and then I see Chapter 4 gives the terminology and a formula I was searching everywhere for.

[NEAS: See the next comment.]

Well, the HW can be done from a direct application of equation 3.2.3 which is from chapter 3. By this point in the module, deriving pk is not hard. Or if you're algebraically minded, you can avoid the equation and do some simplification.

[NEAS: Cryer and Chan show the intution from first principles in the early chapters before giving formulas. Many homework assignments that seem complex at first are easy after you have completed the course. The homework asks for the relative values in these time series, which relies on the intuition in this module. Many concepts are difficult at first but become clear as you work through the modules.]

1st of all p1 for an MA(1) process is not 1.  It can never be one. 

p1 for an MA(1) process is -θ/(1+θ²) which is always between -.5 and.5.

For A1 θ = -1.  θ is always to opposite sign of what is in front of the e_i terms.  I know this not from this module three, but from future modules.  I have learned a few parts of this course puts the chicken before the egg unfortunately.  So for A1, Y_t = mu +e_t - θe_t-1,  θ = -1, to make itY_t = mu +e_t + 1e_t-1

So p1 = -(-1)/(1+(-1)²) = .5  I do not know why you say p1 = 1.  p₀ = 1 but not p1. 

To adhere to module 3 methods you'd have to find the Cov(Y_t,Y_t-1) and the Var(Y_t) = Var(Y_t-1). This is the long way though to get p1 = Cov(Y_t,Y_t-1)/(sqrt(Var(Y_t)xVar(Y_t-1))

Equation (3.2.3) will yield the correct answer (I think, remember none of us has the solution) of .0792 for the Var(Ybar).

 

Tom McNamara III

For part B, what does "pattern" mean? Are we supposed to graph a few of these?

Without doing any graphing, it seems like it depends heavily on the distribution of the e's what the random walks would look like. If e is always positive there will be a clear difference, but if e is 1 half the time and -1 the other half I think the first Y and last Y would behave the same.

[NEAS: For Part B: If Y(j) is higher than average, do you expect Y(j+1) to be higher or lower than average?]

Mr. McNamara III,

The answer for A1, using the 3.2.3, should be .0396.
Var(Y-bar) = (1/50)*(1+2(1-1/50)*.5) = 0.0396.
Perhaps you misplaced a factor of 2 somewhere?

Tell me if I am wrong anyone.

thanks for your time and patience ^^

I don't have my hw in front of me but I got .0792 too. Where is the 2 coming from?

The equation is from the book (3.2.3).

Am I using the wrong equation?



thanks for your time and patience ^^

Dude! I was just using the upper version and ignoring that k can be -1 and 1 (and have pk be non-zero).

I agree with you. I think I'm missing a factor of 2. Thanks!

If I am not mistaken, you are using the correct equation, just not using it correctly.  You wrote

(gamma₀ / n) = 1/50

 as the first part of equation 3.2.3.  My understanding (from 2.2.5 and 2.3.2) is that gamma₀ is the autocovariance function of zero lag, which is

Cov(Y_t,Y_t-0) = Var(Y_t) = Var(mu + e_t + e_t-1) = Var(mu) + Var(e_t) + Var(e_t-1)

 = 0 + 2Var(e_t) due to indepence of the e's.

Hopefully this solves your missing 2.  Please comment on any of my misunderstandings or misconceptions.

RDH