Ernest A

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	234 reputation	bio	website		visits	member for	1 year, 3 months
27 badges		location	Barcelona, Spain		seen	Jan 11 at 20:55

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Dec 16	comment	Detect trend in time series Using the time variable centered around the mean solves the multicollinearity problem as ct and ct^2 are orthogonal, ct being the centered variable.
Dec 15	comment	Detect trend in time series Thanks. I'm not familiar with orthogonal polynomials. Looks like I have reading to do. With this method, it seems that the values of the parameter estimates are meaningless. Will I still be able the determine the maximum point of the series, i.e. the point where the slope of the fitted function is zero?
Dec 14	comment	Detect trend in time series @StephanKolassa The model that you suggest has colinearity problems, no?
Dec 13	comment	Detect trend in time series @IrishStat Done.
Nov 26	comment	Can we conclude structural break from the following results? Nevermind, the analysis is wrong.
Nov 10	comment	Accurate estimates of the variance of maximum likelihood estimators No, I mean to improve the accuracy for any sample size. Bootstrap can be an alternative in certain circumstances, but it's not feasible when the ML estimation is computationally expensive. E.g., if it takes 30 secs to estimate the model, 10000 repetitions would take over 80 hours!
Nov 1	comment	Normalization of likelihood This paper is a little dense for me given that I'm not into Bayesian statistics, and the frequentist methods it discusses are of little help. I think I can simply evaluate the likelihood function with fixed parameters individually for every observation and work something out from there.
Nov 1	comment	Normalization of likelihood Considering that likelihoods are probabilities, wouldn't they always have a maximum of one?
Oct 31	comment	Normalization of likelihood I am using it in the sense of probability of observing $x$ given a parameter value $\theta$. I think this is the common meaning.
Apr 19	comment	testing that mean(x) = k with a small sample and unknown distribution of x thanks for the answer, though I think I'll use the Wilcoxon test because it makes less assumptions
Apr 18	comment	Testing for a significant difference between ML estimates: Likelihood ratio or Wald test? @LPP: I think you should find when and why your llhood function produces a non-finite result and then fix it. You can use `print` / `cat` statements to see the value of the variables. The usual suspects are logarithms of non-positive numbers and wrong initial values passed to `optim()`.
Apr 17	comment	Testing for a significant difference between ML estimates: Likelihood ratio or Wald test? It means that L2 returned a non-finite value (Inf, Nan...) most likely because P11, P22 or P12 were non-positive. On trick is to replace all values that are equal or below zero with a very small positive number (e.g 0.000001) before calculating the logerithms.
Apr 17	comment	Testing for a significant difference between ML estimates: Likelihood ratio or Wald test? If I understand @guest correctly, you need two log-likelihood functions: L1(s1, s2, d; data) and L2(s, d; data). L2 is just like the `llhood2` function that you already have. And L1, if your log-likelihood is an aggregation of terms, I think you can do this: L1(s1, s2, d; data)=llhood2(s1, d; subset1)+llhood2(s2, d; subset2), so you only need a wrapper function that splits data into subset1 and subset2 and then call `llhood2`.
Apr 16	comment	further hypothesis testing for the difference in ML estimates after optim in R I think you could try to post another question explaining the hypothesis that you want to test and asking what test would be appropriate. As @DWin suggested you could use the hessian matrix of the likelihood function to get a variance estimate, and also there's something called "delta method" but personally I don't have experience with any of this. Other than that we know that MLE are asymptotically unbiased and normally distributed, so the only problem is estimating the variance and finding a suitable test.
Apr 16	comment	further hypothesis testing for the difference in ML estimates after optim in R After reading your update, it looks like I misunderstood your question. I thought you wanted to test whether s=d, but now I see that your hypothesis is s_i=s_j which is completely different. I don't think the LR test applies in this case. I think you need to know the sampling distribution of s for doing such a test.
Apr 16	comment	further hypothesis testing for the difference in ML estimates after optim in R The likelihood ratio test seems like a good idea. You simply have to impose a constraint such that the two parameters that you want to test are equal. In this way you obtain the "reduced model". Once you have estimated the reduced model, doing the test is straightforward.
Apr 10	comment	Removal of statistically significant intercept term boosts $R^2$ in linear model Thanks for such a rigorous answer.
Apr 10	comment	Removal of statistically significant intercept term boosts $R^2$ in linear model @cardinal: the end goal is to express $\alpha$ as a function of $\delta$. To put some perspective, $\alpha_i$ and $\delta_i$ are maximum likelihood estimators from another model (call it model A). What I want is reduce the number of parameters of model A. So if I can say $\alpha = k \delta$, for example, then I can change the specification of model A and cut the number of parameters in half (minus one), which is what I want.
Apr 10	comment	Removal of statistically significant intercept term boosts $R^2$ in linear model @cardinal: Makes sense. So, to summarize, without intercept the $R^2$ is larger but it's not comparable to the $R^2$ of the model with intercept. Instead, the sum of squared residuals can be used to compare the goodness of fit of both models. Correct?
Apr 10	comment	Removal of statistically significant intercept term boosts $R^2$ in linear model @Momo: Good point. I've calculated the residual sums of squares for each model, which seem to suggest that the model with intercept term is a better fit regardless of what $R^2$ says.