I give up..I too am stuck with all the same problems as the above posts. How do you get to( b^(2t)-1))/(b^2-1)sigma^2 from formula 5.1.3?

-Damon

I'm confused by the correlation equation.

I derived:

Beta * [ (1-Beta^(2t-2)) / (1-Beta^(2t)) ]^(1/2)

My equation almost matches the book except for the 1/2 power.  It seems like they divided by Var(Y_t) instead of:

Root[ Var(Y_t)*Var(Y_t-1) ]

Can anyone confirm my answer?

I am following your method and have Var(Y_t) = sigma^2( 1 + b^2 + b^4 + b^6 + ... + b^2(t-1) + b^2t x 0) = sigma^2( 1 + b^2 + b^4 + b^6 + ... + b^2(t-1)) = sigma^2( (1/(1-b^2)) - (b^2t + b^2(t+1) + b^2(t+2)+...). This second part of the equation is what we have to subtract out since it is not an infinite series of 1 + b^2 + b^4 + ...

Next I get: = sigma^2( (1/(1-b^2)) - b^2t(1 + b^2 + b^4 + b^6 + ...)) = sigma^2( (1/(1-b^2)) - b^2t(1/(1-b^2))). There is a common denominator of 1- b^2, and in the numerator it's 1 - b^2t. Therefore, we get sigma^2((1-b^2t)/(1-b^2)).

I am so close, and if I multiply by (-1)/(-1) on both sides of the equation, am I good? So many people were saying they had a negative sign problem, but this doesn't cause a problem, does it?

Var(Y_t) * (-1/-1) = sigma^2((1-b^2t)/(1-b^2) * (-1/-1)
Var(Y_t) ) = sigma^2((b^2t-1)/(b^2-1)

Someone please confirm, I have been working on this for way too long.

I'm have the same sign error as previously posted. I get o^2e(1+b^2+b^4+...+b^(2(t-1))+0), then using the geometric formula i get o^2e[(1-b^2t)/(1-b^2)] This times -1 gives you the correct answer...but I can't figure out where the -1 is coming from. I looked at the post that said something like...subtract the rest of the series from the infinite series...1/(1-b^2)...but how do I know what the rest of the series is...isn't that also an infinite series?

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.

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.

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.

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

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.

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!