I understood all of raydhiii's post until:

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

and then I was lost as to where the 2Var(e_t) came from exactly. Any help?

Thanks!

We are given in the assignment that Var(e_t) = 1.  (I assume this means for all t, please correct me if I'm wrong.)  This means that Var(e_t-1) = 1 as well.  Thus:

Var(Y_t) = Var(mu + e_t + e_t-1) = Var(mu) + Var(e_t) + Var(e_t-1).

I simplified (perhaps too quickly) to 2Var(e_t) because Var(e_t) = Var(e_t-1).  Let me know if you have further questions.

RDH

Okay, I get it now.  That's what I thought maybe it was but I wasn't sure if it was safe to assume that e_t-1 was equal to 1 as well.

Thank you!

For process 5, Yt = mu + et - et-1

I use equation 3.2.3, I got result of 2/2500
But when I tried to do some simplification like exercise 3.2 on Page 50,
I got Var(~Y)=1/2500Var[(50.u + e50)], which gave me 1/2500 as result.

Im very confused and didnt see where I went wrong.

Y-bar = (1/n)*sum_{(from t=1 to t=n)}(Y_t).  For our fifth process, this becomes: Y-bar = (1/50)*sum_{(from t=1 to t=50)}(e_t-e_t-1).  We remove mu because we know the variance of a constant is zero. 

The summation yields: (e₁-e₀)+(e₂-e₁)+(e₃-e₂)+...+(e₄₉-e₄₈)+(e₅₀-e₄₉).  Note the cancelations within. 

Plugging this summation back in, we find Y-bar = (1/50)(e₅₀-e₀).  From here we can easily find the variance: 

Var(Y-bar) = (1/50)²Var(e₅₀-e₀).  Recall, the e's are independent and identically distributed, so you should reach your original 2/2500.  Let me know if you have any further questions.

RDH

Where is the square coming from for the gamma over n term?

I just came back to this from not looking at it for a while...seems like there are a lot of different answers for A1 floating around.

I'm getting:

gamma_0 = 2 ;
n = 50 ;
p_0 = 1 ;
p_1 = 1 ;
p_k = 0 for all k > 1


Putting it all together using 3.2.3 I'm getting (2/50)*(1 + 2*(1 - (1/50))) = .1184

Anyone agree or see something I'm doing wrong?

Look to appendix A for the rules on variance:  Var(a+bX) = b²Var(X), where "a" and "b" are constants and "X" a random variable.

Also, rho_k = gamma_k/gamma₀, I would double-check your rho₁ and redo your 3.2.3 to get the correct answer.  Let me know if you have further questions.

RDH

I'm still struggling with how to get rho₁ ; could some one walk me through how they got it for A2.

I think I understand everything else i'm supposed get out of this chapter, just can't figure out the rho.

 

Thanks

Mark, rho_k is the autocorrelation function at lag k, which is given to us to be gamma_k/gamma₀.  Gamma_k is the autocovariance function at lag k (which should be written gamma_0,k, but is simplfied for our convenience), Cov(Y_t,Y_t-k). 

Thus, rho₁ = gamma₁/gamma₀ = Cov(Y_t,Y_t-1)/Cov(Y_t,Y_t-0) = Cov(Y_t,Y_t-1)/Var(Y_t).

For problem 2, we are asked to find Var(Y-bar) of an MA(1) process: Y_t = mu + e_t + .5e_t-1 with 50 observations (so we can't use the estimate formula, as n should be greater than 50), and Var(e_t) = 1.  We will use the formula on page 28: Var(Y-bar) = (gamma₀/n)[1+2(1-1/n)(rho₁)], this is a reduced version of the summation formula above on the same page because an MA(1) process has rho_k=0 for k>1.

First, find the variance of Y_t using the rules of variance and assume the independence of the e's:  Var(Y_t) = Var(mu + e_t+ .5e_t-1) = Var(mu) + Var(e_t) + Var(e_t-1) = 0 + Var(e_t) + Var(e_t) = gamma₀.

Next, in order to find rho₁, we must find gamma₁ = Cov(Y_t,Y_t-1) = Cov(e_t+.5e_t-1,e_t-1+.5e_t-2) = Cov(.5e_t-1,e_t-1) = .5Var(e_t-1).  You now have enough information to complete the given formula for Var(Y-bar).  Let me know if you have further questions.

RDH

Thanks raydhiii, I think I have everything pretty much figured out, except for how you got this:

Cov(e_t+.5e_t-1,e_t-1+.5e_t-2) = Cov(.5e_t-1,e_t-1)

is the Cov(e_t-k,e_t-l) = 0; whenever k does not equal l?