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?

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 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 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