Corvus
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Although it may appear that the mean of the log-transformed variables is preferable (since this is how log-normal is typically parameterised), from a practical point of view, the log of the mean is ...

AIC and c-statistic are trying to answer different questions. (Also some issues with c-statistic have been raised in recent years, but I'll come onto that as an aside) Roughly speaking: AIC is ...

The intercept term is the intercept in the linear part of the GLM equation, so your model for the mean is $E[Y] = g^{-1}(\mathbf{X \beta})$, where $g$ is your link function and $\mathbf{X\beta}$ is ...

There is no single number that encompasses all of the covariance information - there are 6 pieces of information, so you'd always need 6 numbers. However there are a number of things you could ...

The predict.glm method by default returns the predictors on the scale of the linear predictor. I.e. they haven't gone through the link function yet. Try hist(predict(model, type = "response")) ...

Assuming nothing special in your particular case, I think there is a good argument for either using the default (Mean Square Error) or use the mean of the error of the logs, or even the chi-squared ...

You can do Bayesian updating for the covariance structure in much the same spirit as you updated the mean. The conjugate prior for the covariance matrix of the multivariate-normal is the Inverse-...

You aren't strictly taking the "mean" of the likelihood, because the Likelihood function isn't a probability distribution over x. It isn't even a probability distribution anyway, but assuming you ...

Usually the opposite in fact. It tends to be the case that when gambling people are attracted to the small probability of a large win, rather than a higher probability of a moderate win. This effect ...

Poisson regression would be more suitible in this case because your response is the count of something. Putting things simply, we model that the distribution of number of awards for an individual ...

You are talking about two distinct problems here How do I visualise what k-means is doing in N>2 dimensions How do I calculate k-means in N>2 dimensions The second one is much easier than the first ...

Ok, since this is homework, you get hints instead if straight answers. Rather than thinking about $P(X>Y)$ why not think about $P(X-Y>0)$. This is clearly the same probability yes? So now you ...

I think the answer is generally yes. If you know more about a distribution then you should use that information. For some distributions this will make very little difference, but for other it could ...

In short there is no "easy" way to do this, and Dave has tried a few sensible approaches. One approach perhaps worth a try is to taylor expand since your function is smooth and analytic, and we have ...

The logarithm is a non-linear function, and only linear transformations, that is ones that can be written $f(x) = ax + b$, will preserve the mean. As you observed for log the large values aren't as ...

This sort of thing would usually covered by multiple hypothesis testing, although it isn't quite a typical situation. You are correct in noting that this is different from meta-analysis, in that you ...

Prediction and Forecasting Yes you are correct, when you view this as a problem of prediction, a Y-on-X regression will give you a model such that given a instrument measurement you can make an ...

The Art of Data Augmentation, by van Dyk and Meng may be a good start. I also quickly found some R examples on Bayesian statistics on Peter M. Lee and Brian Neelon's websites. But I guess you can find ...

Dropping a zero is usually not a good idea unless the dataset is so large that losing an extreme value will make no difference in the analysis. Treating it as very small is good, but then the result ...

I think a problem like this is begging for a random effects model. If the same athletes are appearing in multiple times, it doesn't seem at all reasonable to assume independence between injuries. ...

As pointed out by @whuber, as currently described your test would have a 50% false positive rate, which would usually unacceptable. This is simply because, suppose X and Y are both drawn from the ...

If all your predictors $(X_1, \dots, X_4)$ are zero centered then the constant term is what accounts for the class imbalance. For example, if your sample has 100 1s and 400 0s then the overall ...

I think you are running into problems here with "sample variance" and "population variance". (remembering that s.d. is simply the square root of these) The actual variance of a population is $\... View answer Accepted answer 4 votes The OLS model is the model of expected cost given that there is a non-zero cost. Therefore by conditionality principal you simply don't use the data that has zero observed cost when fitting that part ... View answer Accepted answer 4 votes The name ADALINE (ADaptive LInear NEuron) come from both the physical implementation of an early classifier, but it is also the name specific design. See: http://en.wikipedia.org/wiki/ADALINE ... View answer 4 votes I don't think fisher exact test is appropriate here is it? It is saying that pre-test there are 34 points arranged in this manner (343344355), and that post test there are 44 points. The arrangement ... View answer Accepted answer 4 votes I think so yes, so long as The population mean is defined, and The sample is composed of iid draws. For a symmetric distribution, the median is an unbiased estimate of the population median and the ... View answer Accepted answer 3 votes If I understand you correctly you are asking about either the marginal distribution of$Y$(or perhaps the joint distribution of$(X,Y)$?). The analysis carried out for Generalized Linear Models are ... View answer 3 votes Astrophysics commonly use$4.1^{+2.1}_{-1.5}\$ style, for example : http://iopscience.iop.org/0004-637X/765/1/47 Also see wikipedia using the same style: http://en.wikipedia.org/wiki/R136a1