Timeline for Showing that 100 measurements for 5 subjects provide much less information than 5 measurements for 100 subjects
Current License: CC BY-SA 3.0
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| Aug 11, 2017 at 21:05 | history | edited | Jake Westfall | CC BY-SA 3.0 |
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| Aug 8, 2017 at 12:01 | history | bounty ended | amoeba | ||
| Aug 2, 2017 at 19:02 | history | edited | Richard Hardy | CC BY-SA 3.0 |
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| Aug 2, 2017 at 19:00 | history | edited | Jake Westfall | CC BY-SA 3.0 |
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| Aug 2, 2017 at 18:55 | history | edited | amoeba | CC BY-SA 3.0 |
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| Aug 2, 2017 at 13:49 | comment | added | Jake Westfall | @amoeba $m$ as the number of levels for a missing, unobserved factor is a very interesting idea. It's interesting to think about how the fully observed model (including the missing factor) could be parameterized so that removing that factor would result in this model with the negative intra-class correlation | |
| Aug 2, 2017 at 11:42 | comment | added | DeltaIV |
@amoeba that's ok, I don't need it to be exactly equivalent to what lmer does. I just wanted to be able to give a "manual" estimate without having to blindly rely on R. This question comes out surprisingly often in an industrial context.
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| Aug 2, 2017 at 10:52 | comment | added | amoeba | +5 btw, I put up a bounty here. | |
| Aug 2, 2017 at 10:51 | comment | added | amoeba | Jake, yes, it's exactly this hard-coding of $m$ that was bothering me. If this is "sample size" then it cannot be a parameter of the underlying system. My current thinking is that negative $\rho$ should actually indicate that there is another within-subject factor that is ignored/unknown to us. E.g. it could be pre & post of some intervention and the difference between them is so large that the measurements are negatively correlated. But this would mean that $m$ is not really a sample size, but the number of levels of this unknown factor, and that can certainly be hard coded... | |
| Aug 2, 2017 at 10:35 | comment | added | amoeba | @DeltaIV Well, the general principle is en.wikipedia.org/wiki/Inverse-variance_weighting, and the variance of each subject's sample mean is given by $\sigma^2_u + \sigma^2_e/m_i$ (that's why Jake wrote above that the weights have to depend on the estimate of between-subject variance). The estimate of within-subject variance is given by the variance of the pooled within-subject deviations, the estimate of between-subject variance is the variance of subjects' means, and using all that one can compute the weights. (But I am not sure if this is 100% equivalent to what lmer will do.) | |
| Aug 2, 2017 at 9:08 | comment | added | DeltaIV | @JakeWestfall I tried to Google for estimator of sample mean when dataset is unbalanced, but couldn't find anything. All "unbalanced" links I found, refer to Machine Learning two-class classification tasks where one class is much more common than the other in the training sample. I will ask a new question here, I hope you'll take a look at it ;-) | |
| Aug 1, 2017 at 15:44 | vote | accept | DeltaIV | ||
| Aug 1, 2017 at 15:05 | comment | added | Jake Westfall | @amoeba I think you can perhaps still think of $\rho$ as a parameter of the generative model, because I'm guessing (but haven't checked) that the constraint on $\rho$ is equivalent to the constraint that $\textbf{C}$ must be positive semidefinite. So all it really says is that you have to use a valid covariance matrix. But it's like the value $m$ is sort of "hard-coded" as a parameter of the generative model, unlike $n$ which we can generate to be as high as we want for any given set of parameters. This might all be semantics, but it seems to me to still make sense to call $\rho$ a parameter | |
| Aug 1, 2017 at 14:51 | comment | added | Jake Westfall | @DeltaIV If the dataset is unbalanced across subjects, then there is a question of how to weight the individual subject means when combining them to form the estimate. Taking the simple mean would be equivalent to weighting each subject directly in proportion to their number of measurements. Which is not a crazy idea, but it turns out that the MLE of $\beta$ is a weighted average of the subject means where the weights depend on the numbers of measurements and the estimated subject variance, $\hat{\sigma^2_u}$. I can't remember the exact expression, but if you Google around you can find it. | |
| Aug 1, 2017 at 14:27 | history | edited | Jake Westfall | CC BY-SA 3.0 |
incorrect interpretation of m=2
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| Aug 1, 2017 at 9:46 | comment | added | DeltaIV | @amoeba you're right, the covariance matrix would only make sense if I had at least a covariate in the model. No covariates $\Rightarrow$ only one random effect $\Rightarrow$ no $\Sigma$ to speak of. I feel embarassed now :P Probably I just don't have a an intuitive understanding of what this intra-class correlation is. BTW, if you have pointers for the other part of my comment (best estimate of sample mean when dataset is unbalanced), I'm all ears :) | |
| Aug 1, 2017 at 9:37 | comment | added | amoeba | @DeltaIV What is "the covariance matrix of the random effects" in this case? In the mixed model written by Jake above, there is only one random effect and so there is no "covariance matrix" really, but just one number: $\sigma^2_u$. What $\Sigma$ are you referring to? | |
| Aug 1, 2017 at 9:33 | comment | added | DeltaIV | ...diagonal. But suppose we assume a more general mixed effect model, where $\Sigma$ is not diagonal. For example, it's estimated from data as done here, even if the link actually considers GLMMs which are a needless complication here). Then, would the mixed effect model still be unable to reproduce negative intraclass correlation or not? | |
| Aug 1, 2017 at 9:28 | comment | added | DeltaIV | Woah impressive answer! (+1). I will need some time to read through it and understand it. Right now I have two questions: 1. you note that with a balanced dataset, our best estimate of the population mean is simply the sample mean. What's the best estimate if the dataset is unbalanced? I can ask it as a separate question if you prefer, but even just a link to a good answer, or to some material, would be great. 2. you say that the mixed effects model is not able to represent negative intra-class correlation. This is true because we assumed the covariance matrix of the random effects to ... | |
| Aug 1, 2017 at 9:17 | comment | added | amoeba | Ah, no. The above is not correct because as $m$ increases to infinity, $\rho$ cannot stay negative and has to approach zero (corresponding to zero subject variance). Hmm. This negative correlation is a funny thing: it's not really a parameter of the generative model because it's constrained by the sample size (whereas one would normally expect a generative model to be able to generate any number of observations, whatever the parameters are). I am not quite sure what is the proper way to think about it. | |
| Aug 1, 2017 at 8:31 | history | edited | amoeba | CC BY-SA 3.0 |
typo in the formula
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| Aug 1, 2017 at 8:28 | comment | added | amoeba | +1. Great answer. I have to admit that the second part, about $\rho<0$, is quite unintuitive: even with a huge (or infinite) total number $nm$ of observations the best we can do is to allocate all observations to one single subject, meaning that the standard error of the mean will be $\sigma_u$ and it's not possible in principle to reduce it any further. This is just so weird! True $\beta$ remains unknowable, whatever resources one puts into measuring it. Is this interpretation correct? | |
| Aug 1, 2017 at 8:22 | history | edited | amoeba | CC BY-SA 3.0 |
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| Jul 31, 2017 at 23:29 | history | edited | Jake Westfall | CC BY-SA 3.0 |
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| Jul 31, 2017 at 23:23 | history | edited | Jake Westfall | CC BY-SA 3.0 |
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| Jul 31, 2017 at 23:17 | history | edited | Jake Westfall | CC BY-SA 3.0 |
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| Jul 31, 2017 at 23:12 | history | edited | Jake Westfall | CC BY-SA 3.0 |
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| Jul 31, 2017 at 23:06 | history | answered | Jake Westfall | CC BY-SA 3.0 |