Timeline for How to calculate the regression variance for a GLS model?
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19 events
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| Apr 19, 2013 at 13:19 | comment | added | IMA | You can't chat yet (need more points) and they don't like these kind of conversations here, because it clutters up the page or something. Besides, I gotta go now. If you have further questions and no one answers here feel free to write me at einsdrei at gmail dot com I will try to answer if I can. | |
| Apr 19, 2013 at 13:14 | comment | added | IMA | This confirms it imo. GLS is not what you need. You are missing the systematic relationship information to use GLS. Use FGLS. Once again, the covariances of $y$ are a function of $X$ and $\epsilon$. As such you still only have $\epsilon$, at best. You are missing information to calculate either sigma OR $C$. To view it differently: You do not have enough information to transform the model as above. Yes, you have unbiased estimators using either OLS or the GLS formula, but they are not EFFICIENT. You are missing the transformed model by knowing either $C$,$P$ or $\sigma^2$. | |
| Apr 19, 2013 at 13:10 | comment | added | IMA | Do you actually have the distribution of $y$, or do you just have empirical values? Because of the latter is the case, I think you need to use FGLS. Your initial question was how to get to $C$. The answer is pretty much that $C$ is the prerequisite to using GLS - which is a way to utilize knowledge of the systemic component in $\epsilon$. You use an additional knowledge to transform the model. Knowledge of values of $y$ does not hold that information. Frankly, I think you can not use GLS. If you are sure about a systematic relationship in $\epsilon$, FGLS can give you good results. | |
| Apr 19, 2013 at 13:05 | comment | added | treemake | Post 2 reply b: What I'd like to know is why you think the use of V instead of C as weight accomplishes this? As I mentioned, $V$ is available from empirical measurements which is why I use it. It makes sense to weight the regression by the variances of the response variable $y$ and to account for the correlation between them. As discussed, the GLS regression coefficients can be readily calculated using $V$ as $V$ and $C$ are related by a constant scaling factor that does not affect the equation for $\beta$. Where further information ($C$) is needed, is in the estimation of $\sigma^{2}$. | |
| Apr 19, 2013 at 13:05 | comment | added | treemake | Post 2 reply a: But remember that the objective of GLS is to negate a systematic component in the ϵ by transforming the model into a model with homoscedastic and uncorrelated error terms. Agreed. | |
| Apr 19, 2013 at 13:04 | comment | added | treemake | Post 3 reply: I think that here we agree and that I was confused only about the notation that was used: $\epsilon = \epsilon^{OLS}$ and are the heteroskedastic (and correlated) residuals in the untransformed coordinates. $\epsilon^{GLS} = P\epsilon$ and are the homoscedastic (and uncorrelated) residuals in the transformed coordinates. Correct? | |
| Apr 19, 2013 at 13:03 | comment | added | treemake | Post 1 reply: It is true that the unbiasedness is not in question when using any weighting factor. Agreed. It is not relevant to your question because you are looking for the covariances. I am not sure if I understand this second sentence. I am looking for the regression error $\sigma^2$, but I know the covariances (of the response variable $y$). | |
| Apr 19, 2013 at 12:56 | comment | added | IMA | Followup to first Post: You also assume that V is proportional to C? Is this necessarily the case? In your previous answer you indicated that you believe you were able to use $V$ to get estimators because proportional/constants cancel out. For this to work, once again, the multiplication with $V$ needs to result in the transformed model. This implies proportion between $V$ and $C$. The issue is that GLS is not a way to find $C$, it is a way to use it. If I understand your issue correctly, I don't think your knowledge of $V$ helps in this case. | |
| Apr 19, 2013 at 12:45 | comment | added | IMA | Post 2: What do you know about the distribution of the $X$? I mean you could try to find out about your error terms by using the fact that $Y = f(X,\epsilon)$. But remember that the objective of GLS is to negate a systematic component in the $\epsilon$ by transforming the model into a model with homoscedastic and uncorrelated error terms. So if you can not separate these effects, you can not transform the model and regain BLUE. What I'd like to know is why you think the use of $V$ instead of $C$ as weight accomplishes this? I dunno maybe it does but I don't see it right now. | |
| Apr 19, 2013 at 12:38 | comment | added | IMA | Post 3: It is in fact the OLS estimator. This is also the case in your formula(s) for $\sigma^2$. This is a bit confusing because the transformed model is not the OLS model, it is the GLS model.$P\epsilon$ is the GLS error term which is homoscedastic, whereas $\epsilon$ is not. You can use your GLS residuals but then you can do away with the weighting matrix. In that case using $\hat{\epsilon}^{GLS}$ without the $C^{-1}$ gives a good estimate of $\sigma^2$. Using the $C$ in that formula is a direct result of the transformed model and is equivalent to using the GLS residuals in the first place. | |
| Apr 19, 2013 at 12:29 | comment | added | IMA | Post 1: It is true that the unbiasedness is not in question when using any weighting factor. It is not relevant to your question because you are looking for the covariances. The GLS formula in question results from the transformed model with the $P$ | |
| Apr 19, 2013 at 12:14 | comment | added | treemake | I would very be interested in your follow-up to my comments ... | |
| Apr 19, 2013 at 12:10 | comment | added | treemake | And then you have a consistent estimator in … I would like to clarify a point that you made the first part of your response (before being able to evaluate it). The equation for $\hat{\sigma }^2$ is expressed in terms of $\hat{\epsilon}^{OLS}$. I do not believe this is correct. In fact $\hat{\epsilon}^{OLS}$ should be replaced by simply $\hat{\epsilon}$, where the latter are in fact the GLS residuals. This follows from examination of the preceding equations where $\epsilon $ represents $\epsilon^{GLS}$. (In fact $\epsilon^{OLS} = P\epsilon$) | |
| Apr 19, 2013 at 12:09 | comment | added | treemake | If you know nothing about the composition of your error terms, you just can't get to that decomposition and the C you need. That is the 'point' of GLS … For my particular problem I know $V$ but not $C$. This is because my empirical measurements of my response variable $y$ include estimates of their variances and covariances. (It is somewhat unusual to know $V$ a priori and not $C$, but nevertheless quite possible, if $y$ represents instrumental measurements or values taken from a complex survey design for example). | |
| Apr 19, 2013 at 12:05 | comment | added | treemake | You could go ahead and plug in $V$ instead of $C$. But of course this is not actually correct because $V$ is not equal to $C$ as you discovered … In fact this is not true. In the equation for $\beta^GLS$ any scaling factor applied to $V$ (in the equation that you showed) will cancel. Thus $V$ or $C$ or any $V$ scaled by any constant will work and yield the same answer for $\beta$. In my case, I used $V$ to compute $\beta$, in the absence of knowing $C$ and $\sigma^2$. | |
| Apr 19, 2013 at 11:12 | history | edited | IMA | CC BY-SA 3.0 |
Cleared up a few things to make it more clear and applicable to the quesrtion
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| Apr 19, 2013 at 11:05 | history | edited | IMA | CC BY-SA 3.0 |
added 238 characters in body
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| Apr 19, 2013 at 10:59 | history | edited | IMA | CC BY-SA 3.0 |
added 238 characters in body
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| Apr 19, 2013 at 10:53 | history | answered | IMA | CC BY-SA 3.0 |