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So the slight change in slope in the top line segment should not be there....

That is correct - perfect negative correlation would correspond to 2 line segments!

This is more appropriate for Topic I than Topic III no?

I can't understand why anyone would ever pay for an ebook that you cannot own nor print.....just my opinion.

I would also add that I prepared for Part I in 2014....and I tend to read every single resource and much more. Unfortunately, I wasn't able to take Part II yet and since it has been so long, I am now finding that I need to review the Part I material as well! Oh well....I really enjoy studying...

Thanks Nicole. Please send the amazon code to my email.

Fantastic! Both @ami44 and @jairamjana are absolutely fantastic! Thank you both! I can't believe I forgot the covariance term....if I remembered that, I probably would have thought about conditional independence.....

@David Harper CFA FRM @jairamjana and I are trying to decide if the V(aX + bY | Z=z) = a^2Var(X|Z=z) + b^2V(Y|Z=z) approach is sound.

I am relying on V(aX+bY). I do not disagree with you @jairamjana.

I think this works as well.

Interesting....I need to think about this.

Does this make sense? Var(Yt | Omega_t-1) = Var((e_t + theta*e_t-1) | Omega_t-1) = Var(e_t | Omega_t-1) + theta^2Var(e_t-1 | Omega_t-1) = sigma^2 + theta^2(0) = sigma^2

I think it's e_t distributed as WN(0, sigma^2) rather than WN(1, sigma^2) correct?

Along these same lines, is Variance(e_t-1 | e_t-1) = 0 since e_t-1 would be known and constant given itself? Make sense?

Thank you @jairamjana E [ (e_t-1) | e_t-1) ] = the expected value of e_t-1 given e_t-1. This explains it. I was thinking that the expected value of e_t-1 = 0, which is true as an unconditional mean (i think) by covariance stationarity. But if we are given the actual e_t-1, then the expected...

@David Harper CFA FRM Can you explain why the Conditional Mean for an MA(1) is not 0? I see the explanation in the previous chapter regarding: This makes sense. But, if that was the case, then would not be 0 correct? The below seems to indicate to me that =0 and that = . I don't really...

I want to hire @ShaktiRathore!

Expected value of X is the mean of X; they are equivalent. That being said, the Expected Value Function iteself is not the mean, for example, E(X) = the mean of X but E(aX+bX^2) would not be the mean of X, for example. It could be thought of as the mean of aX + bX^2 though!

Thanks for putting it together. Excel is my strongest skill set, so I will take a look and see if I can make it more efficient and/or add to it. If so, I will repost to you. Thanks again.