Timeline for Distribution of a MA(1) process
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 15, 2014 at 20:33 | comment | added | Mark Morrisson | From what I know, the mean and adjusted variance estimators are unbiased and consistent. I'd have to check for high moment estimators. | |
| Aug 15, 2014 at 18:29 | comment | added | Alecos Papadopoulos | What about estimator properties? Shouldn't they be taken into account, perhaps more strongly than computational/algebraic convenience? | |
| Aug 15, 2014 at 18:24 | comment | added | Mark Morrisson | Well, in the case the likelihood is easy to compute, I guess the maximum likelihood is the way to go. But in the case of a MA(1) model, it's not the case so yes, I'm wondering why the method-of-moments is not the standard way to estimate a MA(1). | |
| Aug 15, 2014 at 16:28 | comment | added | Alecos Papadopoulos | I believe you are, in essence, wondering "why use maximum likelihood and not method-of-moments", so perhaps you should contemplate this question more generally, since one can almost always apply method-of-moments estimation: why do we use ML at all, when we have method-of-moments? | |
| Aug 15, 2014 at 15:22 | comment | added | Mark Morrisson | Of course, but I was wondering why the marginal distribution (easily obtained) is not used for estimation purposes. If \sigma is known, the variance of the sample immediately gives the value of $\theta$ (except for the sign). And in case $\sigma$ is unknown, higher moments could be used to determine both $\sigma$ and $\theta$. But when it comes to estimating a MA(1) process, books always present the MLE method with a numerical maximization and this is much more complicated. Am I missing something? | |
| Aug 15, 2014 at 14:54 | comment | added | Alecos Papadopoulos | If it is more than 2-dimensional, you cannot fully plot it (you need the 3d dimension for the value of the joint density). | |
| Aug 15, 2014 at 14:48 | comment | added | Mark Morrisson | Yeah, I meant $y_t$ as a process. The correct notation would be $(y_t)_t$. I also think I am plotting the marginal distribution of $y_t$. So, the question is: how do you plot a joint distribution from a sample? | |
| Aug 15, 2014 at 12:52 | comment | added | Alecos Papadopoulos | The joint distribution is a multivariate density -so the phrase "the joint distribution of $y_t$" makes no sense. Do you mean "the joint distribution of $t$ $y$-random variables"? And what you found, is it perhaps the marginal distribution of the single r.v. $y_t$? Because it appears to be univariate, not multivariate. | |
| Aug 15, 2014 at 10:22 | history | asked | Mark Morrisson | CC BY-SA 3.0 |