2
I implemented the problem myself (building on your code) so that you can compare it with your data. I get good results when I use centroids>50. So I think the implementation is correct. I did not use regularization. I simply use the pseudo-inverse function to do regular linear regression
import numpy as np
from scipy.stats import multivariate_normal
import ...
2
The question stipulates that exactly one of the three prisoners will be pardoned. This means that the sample space $\Omega$ can be partitioned into three disjoint events:
$$\mathscr{P}_A \equiv \{ \text{A pardoned} \},$$
$$\mathscr{P}_B \equiv \{ \text{B pardoned} \},$$
$$\mathscr{P}_C \equiv \{ \text{C pardoned} \}.$$
Letting $\mathscr{W}_B \equiv \{ \...
1
I'll use the boot package to give an example of using bootstrapping to determine a 95% confidence interval for a statistic. There's nothing magic in the package, tho. You could easily write code to obtain the bootstrapped samples, and then get the percentile confidence intervals manually. Note that boot also reports the bias and standard error of the ...
1
A reasonable approach for this kind of problem is to look at the limits of the set difference between the estimator and the true set shrinks to the empty set, in some appropriate probabilistic sense, as $n \rightarrow \infty$. To do this, we define the set $\Delta_n \equiv \tilde{\Gamma}_n - \Gamma_n = \{ \boldsymbol{\gamma} \in \Gamma | 0 \leqslant \mathbf{...
1
Let $X$ denotes the observations and $\theta \in \Theta$ the parameter. In a Bayesian approach, both are considered random quantities.
The first step of modeling is to define a statistical model, i.e. the distribution of $X$ given $\theta$, which can be written as $X \mid \theta \sim p(\cdot \mid \theta)$. This is mainly done by expliciting a likelihood ...
1
This is one of many areas in statistics that uses abysmal terminology. According to most of the rest of the world, an invariant function's values do not change at all when acted on by a group. The first definition you cite decidedly is not invariance in this sense: physicists would call it "covariant." The lesson here is to pay attention (as you are) to the ...
1
Arguments that Bayesians do not need to worry about type I errors are starting from the premise that the type I error rate does not matter/is not a relevant concept* and simply adhere to the likelihood principle**.
I don't think this kind of Bayesian viewpoint is compatible with coercing an inferential threshold, but for taking an action it can work well ...
1
Variance/standard deviation is a concept associated to a random variable. In your case,
If we assume that one data point from AML comes from an independent sampling of a probability distribution, then the sampled mean $\overline{AML}$ is also a random variable, thus it (theoretically) has a standard deviation. Same goes for sampled mean $\overline{ALL}$ of ...
1
The coverage by the normal approximation that you have used is erratic, unreliable and frequently low. That method, often called the Wald method, should not be used despite its prominence in textbooks.
See here for more information: Discrete functions: Confidence interval coverage?
1
Since the marginal density writes as$$m(x)=\int_\Theta f(x|\theta),\pi(\theta)\,\text{d}\theta$$a possible numerical approximation is$$\frac{1}{T}\sum_{t=1}^T f(x|\theta_t)\qquad\theta_1,\ldots,\theta_T\sim\pi(\theta)\tag{1}$$but this Monte Carlo approximation based on simulations does not use kernel estimation. As stressed by Juko Kokkala's comments, the ...
Only top voted, non community-wiki answers of a minimum length are eligible
