Hey Mark, pages 14-15 on "A Moving Average" might answer that question, and after (2.2.11) on page 13 answers your question as well.  My understanding is that the e's, or "innovation" terms are independent in time.  Which to me means they have absolutely no relationship to eachother unless they are at the same point in time.

So, to answer your question simply, yes.

RDH

raydhiii, maybe this is a typo, but referring to your recent post, doesn't Var(mu + e_t+ .5e_t-1) equal Var(mu) + Var(e_t) + 0.25Var(e_t-1) not Var(mu) + Var(e_t) + Var(e_t-1) ?

What is part B asking for?  What is the first and last time series?

Happy Monday, everyone!

hs, I believe there is already a post about this somewhere, however I shall restate it.  Page 25 (Appendix A), equation (2.A.6):  Var(a + bX) = b²Var(X)  where a and b are constants and X is a random variable.  So, to answer your question, yes, Var(mu) is in there, but we know the variance of a constant is zero.

Your "pattern" question was also in a previous post, NEAS asks to describe how we would predict Y_t+1 if Y_t was above or below the mean.  This is easily determined from the autocorrelation function, recalling that a positive autocorrelation describes a positive increase over time.  Let me know if you have any further questions.

RDH

Sorry for being slow, but I still don't get part B, so are we talking about 1. and 5. as first and last series? Are the mean for all five processes mu right? I am not sure if this is right but if it is, that means Y t+1 should also be mu? I think I might be totally confused.

And can someone explain what C is asking?

Thanks!

Right, so isn't the following true?

Var(Y_t) = Var(mu + e_t+ .5e_t-1) = Var(mu) + Var(e_t) + Var(.5e_t-1) = 0 + Var(e_t) + .25Var(e_t)

Regarding the "pattern" question, if Y(j) is higher than average, I would expect Y(j+1) to be higher than average if they have positive autocorrelation and lower than average if they have negative correlation...

Yes

RDH

just want to check, did anyone get .0495 for the second series?

If Var(Y-bar) = (gamma₀/n)[1+2(1-1/n)(rho₁)], and you got 0.0495 for Var(Y-bar), through some backward calculation this tells me that if you used the same gamma₀ as me then you used 1/2 for rho₁. 

Recall gamma₀ = Var(Y_t) and rho₁ = Cov(Y_t,Y_t-1)/gamma₀ and Cov(a+bY,c+dY) = bdCov(Y_t,Y_t) = bdVar(Y_t).

Double check your rho.

RDH

In problem A, we can consider for all 5 series a general form:
Y(t) = mu + e(t) + q·e(t-1)

Then we need to calculate:
gamma0 = Var(Y(t))
and
gamma1 = Cov(Y(t),Y(t-1))
rho1 = gamma1/gamma0

The calculations are pretty straightforward.
Once you get the general formulas as functions of parameter q, just plug in the relevant values of q for each of the 5 series.