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seen Feb 25 at 16:22

Feb
11
comment Picking cases to label for classification
The parameter estimates obtained could then be used to build prior distributions on the model paramters so that you could then estimate within a Bayesian framework the mixture model on the labelled and unlabelled data. In this way once the expert has done their classification, the classification of subjects to diseased/un-diseased would be model-based. The accuracy of this classification could of course be checked against the cases the expert has reviewed.
Feb
11
comment Picking cases to label for classification
You could use mixture models assuming your subjects are drawn from one of 2 subgroups in the population: diseased or not. If your response variable is say normally distributed (maybe some measure of kidney function over time) then you could fit a 2-component normal mixture distribution to the data where each group is modelled with a (for example) linear mixed effects model. You could then incorporate the expert by fitting 2 separate mixed models to only those cases reviewed (hence you know the 2 groups).
Jan
26
asked Relationship between Gumbel and Weibull distribution, accelerated failure time models, and Survreg using R
Jan
12
comment Is it valid to include a baseline measure as control variable when testing the effect of an independent variable on change scores?
measurement as a response, and conditioning on a fixed baseline value, and secondly that the point estimate variance from the ANCOVA model is always greater than or equal to that from the unconditional one. It turns out this variance difference will typically be small due to randomization ensuring baseline mean responses between the groups are small. The authors conclude the unconditional model is appropriate for modelling baseline as a random variable, but ANCOVA as appropriate when viewing it as fixed.
Jan
12
comment Is it valid to include a baseline measure as control variable when testing the effect of an independent variable on change scores?
A clear discussion of this paper can be found in (“Should baseline be a covariate or dependent variable in analyses of change from baseline in clinical trials?” by Liu, Mogg, Mallick and Mehrotra 2009). They refer to this model as an unconditional model (i.e. it does not condition on baseline response). In the Liu (2009) paper they discuss the main results of the Zeger (2000) paper. These are firstly that with no missing data the point estimates of $B_{1}$ from the unconditional model are the same as those from the conditional approach of ANCOVA using the post-baseline
Jan
12
comment Is it valid to include a baseline measure as control variable when testing the effect of an independent variable on change scores?
The model I am talking about does indeed imply $Y_{1}$ is a function of treatment, but only from the viewpoint that despite randomization there will always be slight differences between the treatment and control group with respect to their baseline means. Thus $\beta_{1}$ will capture this difference as well as the effect of treatment. The reference for this is ("Longitudinal Data Analysis of Continuous and Discrete Responses for Pre-Post Designs" by Zeger and Liang, 2000).
Jan
12
comment Is it valid to include a baseline measure as control variable when testing the effect of an independent variable on change scores?
In the equation $Y_2 = \beta_1T + \beta_2X + \beta_3Y_1 + (e + Y_1)$ if, as is standard practice, we assume all the covariates are not random variables, then $Y_1$ is not correlated with $e + Y_1$. Thus I think there is only a problem if you view $Y_1$ as random, in which case (again just my opinion) you should model $(Y_1,Y_2)$ jointly but without $Y_1$ as a covariate. In this respect without missing data I have been informed that this approach is equivalent to $Y_1$ being a fixed covariate (I will try and find some references for this).
Oct
17
awarded  Commentator
Oct
17
comment Conditions on ratio of $\frac{\sigma_{BIB}^2}{\sigma_{RB}^2}$
What do $Var_{RB}(\hat{\alpha}_i)$, $Var_{RB}(\hat{\alpha}_j)$, $Var_{BIB}(\hat{\alpha}_i)$ and $Var_{BIB}(\hat{\alpha}_j)$ equal?
Oct
9
comment Identifiability in linear regression and time series
Thus with X1 as the estimator, we cannot take a higher sample size to be sure we will estimate something close to the truth (given one sample of data), but for the mean of the normally distributed variables we can (this estimator is unbiased and consistent). Thus consistency is more "useful" when we want to be sure we are close to the truth for single dataset (albeit usually with a large sample size needed). Finally it is well known an estimator cannot be consistent if the model is not identifiable. Thus identifiability is required for a model if we want any estimator to be consistent.
Oct
9
comment Identifiability in linear regression and time series
Sorry @vman049 for not replying sooner. It is easier to think first of unbiasedness and consistency which relate to the expected value the estimator is close to the true value and the probability an estimator is close to the true value. You can have one without the other (see stats.stackexchange.com/questions/31036/…). Using the "unbiased not consistent" example in the first answer, we see the prob X1 is close to mu never gets closeer to 1 as n gets larger (it does not depend on n), even though it is unbiased.
Oct
3
comment Identifiability in linear regression and time series
Finally the first model is guaranteed to be identifiable if the minimum number of hyperplanes the covariate points lie on is 2 or greater - see Hennig Theorem 2.2.
Oct
3
comment Identifiability in linear regression and time series
Formally all of the covariate pairs $(x_{1},x_{2})$ lie on a 1-dimensional hyperplane (i.e. a line) in $\mathbb{R}^{3}$, and generally this model would not be identifiable if all your $n$ covariate points $(x_{1},...,x_{p})$ lie on the same $(p-1)$-dimensional hyperplane. Going back to the two column example, if just one of the entries in $\mathbf{x}_{2}$ were not equal to $c$ then this would yield information on the covariate-response relationship, and the model MAY be identifiable. See "Identifiability of Models for Clusterwise Linear Regression", Hennig (2000) for an in-depth look at this.
Oct
3
comment Identifiability in linear regression and time series
The first model would not be identifiable if $\mathbf{X}$ contains say a constant column: if you imagine $\mathbf{X}$ contains a column of 1's ($\mathbf{x_{1}}$ say), and a column ($\mathbf{x_{2}}$) whose entries are all the same (c say). Then if we picture $\mathbf{y},\mathbf{x_{1}}$, and $\mathbf{x_{2}}$ in $\mathbb{R}^{3}$ then all your $n$ triples $(y,x1,x2)=(y,1,c)$ lie on a line parallel to the y-axis regardless of what the $n$ values are for $\mathbf{y}$. Thus no information can be obtained on what happens to $\mathbf{y}$ as $\mathbf{x_{1}}$ and $\mathbf{x_{2}}$ change.
Jul
26
awarded  Scholar
Jul
26
accepted Observed information matrix is a consistent estimator of the expected information matrix?
Jul
25
answered Observed information matrix is a consistent estimator of the expected information matrix?
Jul
25
suggested rejected edit on Observed information matrix is a consistent estimator of the expected information matrix?
Jun
13
comment How to calculate sample size needed for comparing the “change from baseline” scores between two groups?
You are right in that the two groups are probably unrelated and so we can use a T-test for independent samples. If the first group represents baselines measurments and the second post-baselines measurments on the same people then we can use a dependent sample t-test.
Jan
22
comment Contrasts in mixed model
A correction to the above comment: The $L$ vector should sum to $0$, so if you use $L=(3/4,-1/4,-1/4,-1/4)$ then $L\beta=A-(A+B+C+D)/4$ which is group A versus the average of the other four groups.