# Tag Info

12

The phrase a posteriori is from Latin, the sort of Latin phrase that used to appear in educated British English. Along with a priori, it was already in use in philosophy/logic in a slightly different sense before it got used in statistics. There's not really any significance to the a, just history.

11

The maximum likelihood estimator $$\hat\theta(z_1,\ldots,z_n)$$ is the solution of the maximisation program $$\arg\max_\theta\sum_{i=1}^n \log \{f_Z(z_i;\theta)\}\tag{1}$$ It is therefore a random variable since it depends on one realisation of the sample $(Z_1,\ldots,Z_n)$. The justification in using the maximum likelihood estimator is that, since the true ...

4

The answer is double robustness. Like all doubly-robust methods, with TMLE you get two chances to get the model correct and the treatment effect estimate is consistent if either is correct (or approach at a given rate). With g-computation using a GLM, you only get one chance to get the model right. If that model is wrong (and it almost certainly is), then ...

4

If the dimension of $W$ is large, computing the MLE of the whole GLM and then marginalising need not give the efficient estimator of the ATE. From an asymptotic viewpoint, suppose the dimension of $W$ grows fast enough that the whole parameter vector of the GLM is not estimable at $\sqrt{n}$ rate. It's still possible that the ATE is estimable at $\sqrt{n}$ ...

3

The discreteness does make this potentially interesting. Fortunately, there are only four possible outcomes: 2 white, white first, white second, 0 white. It's pretty obvious that if you get 2 white the MLE is $\hat\theta=10$ (because increasing $\theta$ increases the likelihood) and if you get 0 white the MLE is $\hat\theta=0$. If you get one of each, you ...

3

It appears that at least some of your difficulty arises not because the MLE of the dispersion parameter is relatively inefficient in smallish samples, but because the distribution of the MLE is not well-approximated by the Gaussian distribution when the mean of the data's distribution is small, even with sample sizes that appear quite large. For example, we ...

2

Consider three confidence intervals for a sample of size $n = 25$ from the population $\mathsf{Norm}(\mu=50, \sigma=7),$ where $\sigma=7$ is known to us and $\mu$ is not. (Using R.) set.seed(1114) x = rnorm(25, 50, 7) # 7 known summary(x); length(x); sd(x) Min. 1st Qu. Median Mean 3rd Qu. Max. 37.68 45.58 50.82 51.00 55.78 66.64 [1] ...

1

If your training and test sets differed substantially in terms of average observations, as your question suggests, then perhaps there is a problem with your train/test split. The problem potentially posed by a difference in average observations between the training and test sets is that the two sets might not represent the same underlying population. That ...

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