Jacob: Formula 3.2.3 applied to the first time series gives y-bar = 0.0792. But mu + e_t + e_t-1 from 1 to 50 is (1/2500) × Var(50 mu + å ₍₅₀₎ + 2 × e₍₁₎ + …+ 2 × e₍₄₉₎ ) =

(1/2500) × (0 + Var(e ₍₅₀₎ + 4 × (e₍₁₎ + …+ e₍₄₉₎ ) ) ) = (1 + 4 × 49) / 2500 = 0.0788

Rachel:The difference is subtle. The formula in the textbook assumes 50 observations are taken from this time series. Your second method assumes the time series starts at observation #1. The difference is one error term. Your second method doesn’t have an error term for period zero. Including this error term raises the variance of y-bar by 0.004.

Intuition:

The formula in the textbook assumes the time series process is infinite. We observe fifty values, such as 1.014 through 1.063, and compute the variance of y-bar. Your computation assumes the series starts at observation #1, so the error term e₀ is zero.

Use the formula in the textbook, since its assumptions are closer to real life. We may observe only certain observations, but the time series may have existed for much longer.

LIAPP,

I used the following 'givens' and I made a table like the one below to help me through the calculations.  Hope it helps.

Givens: N=50; Var(e)=1; Var(Y_t) = Var(mu)+Var(A)+Var(B); rho_1=Cov(Y_t,Y_t-1)/Var(Y_t);  that last 'given' is explained in this forum and in the book (I think)

Table:

#     Var            Cov(Y_t,Y_t-1)            rho_1            Formula from bottom of page 28

1    0+1+1=2;      Var(e)=1;                   1 / 2 = .5;      2/50 * [1 + 2*(50-1)/50*(.5)]=0.0792

2    0+1+1/4=1.25; .5*Var(e)=.5 ;            .5/1.25=.4 ;   1.25/50[1+1.96*(.4)=0.0446

and so on and so forth.  I'd write it al lout, but I don't have the time.    

I have seriously read over every single comment on this thread, and still can't get the answer. I'm stuck on A.2. I got rho as (1/2)(1+.25) = .625. As far as the Gamma coefficient at the front is... I have no idea. I haven't had a stats class in almost 2 years, and I don't see one single example in this or the last chapter that demonstrates how to use coefficients in the covariance term, or whether I'm supposed to use Gamma 0 or Gamma k, or what the difference is. Can someone spell it out for me instead of giving hints that are already given in the book? I need a step by step process that I can extrapolate to other problems.

To help with parts B and C, think of real world applications of The Random Walk.

A random walk (e_t) is defined as a random number of steps in a random direction. Imagine sitting on a park bench watching one pigeon and its aimless wandering in a patch of grass. It moves 2 steps one direction, turns 45 degrees, moves 1 step, turns 240 degrees, moves half a step, turns 1 degree, moves 5 steps, and so on for an hour. There is no discernable pattern or consistency to the pigeon's travels. So after an hour you can expect the pigeon to be exactly where you first observed the pigeon. It is possible for the pigeon to walk a mile away in that time if it would pick that same direction at every time t, but the probability of that happening is close to nil.

Let Y_t be the position of the pigeon at time t and mu be the origin of the pigeon. The variance of the pigeon's position is going to be correlated to the variance of the pigeon's random walks. With every moment, pigeons 1-5 take a random walk defined by the problem. You can draw 5 circles around the origin to depict the area that the different pigeons could possibly end up in, with the larger areas having the larger variances.

Pigeon #3 (Y_t=mu+e_t) is the simplest of the group, he makes one random walk every time t. It is a random direction and a random distance every moment.

Pigeon #5 (Y_t=mu+e_t-e_t-1) has OCD and it really likes its origin. Every moment, it returns along its previous random walk before performing another random walk. So every moment involves #5 returning to its origin before making another random walk. Obviously, the variance of its position is going to be small as it will always be within one random walk of its origin.

Pigeon #4 (Y_t=mu+e_t-.5e_t-1) is similar to #5, but it only returns half the distance of its previous random walk. Its variance will be larger than #5, but smaller than #3.

Pigeon #1 (Y_t=mu+e_t+e_t-1) wanders the most. After every random walk, it continues in that same direction before adding another random walk every moment. Pigeon #1 has the highest area of variance because it has the highest probability of traveling a large distance from the origin.

Pigeon #2 (Y_t=mu+e_t+.5e_t-1) wanders like #1, but not as far. So its area of wandering will be smaller than #1.

It is still possible for Pigeon #1 to be standing at the origin after an hour while Pigeon #5 is 20 steps away, although a very unlikely possibility.

Once you can imagine the different walk observations, you can describe the difference between #1 and #5 easily. Part C is just asking why #4 and #5 have smaller variances than #1 and #2, because every moment partially reduces the total distance travelled from the origin on an oscillating pattern.

Alex, if you still need help, post and I can try to help you, ill continue to check back. I found this homework extremely hard and really had to read everybodys post to understand how to do it. I thought the textbook section was pretty much useless for this section, but Im sure a couple chapters from now it will make more sense, as with most NEAS work. And by second problem, do you mean problem B, or Problem A with the second equation?

How does the homework relate to the textbook. The first few lines has an equation Yt=mean+Xt. However, after that the mean only appears in teh assumption E(sample mean)=mean, after that no more mention of the mean. How in the world is somebody supposed to do the second homework problem. I am completely at a loss. Somebody please help me. Please in less than 3 days , please?  

Nevermind, I'm an idiot and forgot how to take a simple variance.  Var(Yt) is 5/4. 

Number 4 is giving me all kinds of trouble.  I'm getting -.0046 too.  I get Var(Yt) to be 3/4, p1 is then (-1/2)/(3/4), which is -2/3.  Plugging this in to equation 3.2.3 gets me -.0046, what am I doing wrong? 

Also, is there a way to use equation 3.2.2 for all of these problems to by simplifying?  I got it to work for number 3 and 5 but not 1,2,or 4. 

Any help would be appreciated.

Thanks you.

Sure. I wouldn't mind at all. It'll help me refresh.

I also got the same answers.

A1) 0.0792
A2) 0.0446

A3) 0.02
A4) 0.0054
A5) 0.0008