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Consider a population $Y|X$ that follows some distribution according to a true model, and you have a set of trained models $f(X,\theta)$ that make predictions of $Y$ given $X$ and are parameterized by $\theta$. The goal is to find out what the error of the models is, in making predictions about samples from the population, as function of the parameter $\... 2 The suggestion that Bayesian methods use more information from the data than the sufficient statistic is false. For any Bayesian model with data vector$\mathbf{x}$and sufficient statistic$\mathbf{T}(\mathbf{x})you can use the Fisher-Neyman factorisation for the sufficient statistic to get the posterior form: \begin{align} \pi(\theta | \mathbf{x}) &... 2 The ATE is an estimand involving unseen potential outcomes and is defined at E[Y^1-Y^0], where Y^1 and Y^0 are the potential outcomes under treatment and control. Under the main causal assumptions, the ATE is equal to E[E[Y|A = 1, V]-E[Y|A=0, V]], where V is a valid adjustment set. Let's call E[E[Y|A = 1, V]-E[Y|A=0, V]] the average marginal ... 2 If possible I would show a plot of the distribution, rather than just some summary statistics, since it is more informative. Alternatively, if some summary statistics are required, I would use a five-number summary or similar, rather than trying to find a distribution which fits the data. Fitting a distribution might look like a more elegant solution, since ... 2 The chance of A being >2\sigma from the mean is 2.28\%. pnorm(12, 10, 1, lower.tail = FALSE) Thus the chance of A & B (2 independent events both occurring) is equal to A*B or 0.052\% of the time. Therefore expected frequency is once every 1932 times (agrees with the simulation results reasonably well). The medium and mode is a bit more ... 1 I can propose something, the idea is to use an optimization method based on the KDE function. Here is a short example in R: First I define a function to calculate a KDE (most of this code in taken from a personnal project on github (JeremyGelb/spNetwork) quartic_kernel <- function(d, bw){ u <- d/bw k <- (15/16)*(1-u**2)**2 k <- k / bw k &... 1 A page that you linked explains that, for a smooth continuous-time hazard function \lambda (t) and survival function S(t): \lambda (t) = - \frac{d}{dt} \log S(t).\$ Thus once you have one of those functions you have the other. With that background, all survival model fitting involves both the hazard and the survival functions, although the way that ...