# Tag Info

2

You are probably familiar with the rule and just don't recognize it in this setting. It applies to densities... $f(x|y)=\frac{f(x,y)}{f(y)}$ ...but also to sets (which is your case)... $P(A|B)=\frac{P(A \cap B)}{P(B)}$ In words, a conditional thing equals the joint thing over the marginal of the condition. This follows from the "chain rule".

0

Its' correct, technically what you write is equivalent to the formula of the conditional probability. Denoting the events as $XY$ where $X$ is taken from Urn $A$ and $Y$ is taken from Urn $B$: $$P(RB | RB\cup BR)=\frac{P(RB\cap (RB\cup BR))}{P(RB\cup BR)}=\frac{P(RB)}{P(RB)+P(BR)}$$

0

Some nomenclature issues here. A classification is a prediction (I predict case x is class A based on the data/model). But I think what you are getting at is the fact that most implementations of classification in computer software can return a binary classification - but this is simply a categorization based on computed probabilities relative to some ...

34

A DAG is a Directed Acyclic Graph. A “Graph” is a structure with nodes (which are usually variables in statistics) and arcs (lines) connecting nodes to other nodes. “Directed” means that all the arcs have a direction, where one end of the arc has an arrow head, and the other does not, which usually refers to causation. “Acyclic” means that the graph is not ...

6

This is generally a fairly elaborate topic, and may require more reading on your part for better understanding, but I will try to answer a couple of your questions in isolation and leave references for further reading. Confounding Consider the example below: Controlling for the confounding variable "Gender" gives us more information about the relationship ...

1

Adding (or subtracting) a small number to (or from) a not-small number results in a number that is roughly the same as the not-small number, hence in many situations can be skipped, meaning the small number can be ignored. As the small number gets even smaller, the approximation obtained by ignoring it and just using the not-small number gets better and ...

0

Spearman correlation will work with variables that are at least ordinal. That is, if you suspect your scores are really interval or quasi-continuous, there is no issue with using Spearman correlation for this kind of data. Also, with an ordinal scale from 1 to 9, that's probably enough categories where respondents are treating it as interval in nature. ...

1

In a 2019 paper titled Bayes-optimal estimation of overlap between populations of fixed size, Daniel Larremore presents a solution when $N_1$ and $N_2$ are fixed and known. I'm not gonna repeat the whole paper and just present the main result. Without loss of generality, assume that $N_1 \leq N_2$. Further, denote $n_1$ and $n_2$ the number of samples drawn ...

1

My two cents: Use maximum likelihood estimation on K: Likelihood $P(data|N1, N2, K)\propto$ ${K\choose k} {N1-K \choose n-k} {N2-k \choose n-k}$ where nCr is the N-choose-k combinations. Then find: K_optimal = argmax(P w.r.t K) I couldn't find an analytical solution so I wrote a few lines of code to calculate it, with an example like this: from math ...

1

Penalized regression does not do an initial regression and then downweight points "outlying" from this initial regression. $L_1$ and $L_2$-penalized regressions do not weight observations at all (de facto). In fact, $L_1$-penalized regression (LASSO) is more prone to bias d/t outlier because it tends to select regressors based on large coefficient values. ...

1

I agree with @Demetri. Baysian inference is based on cause and effect relationship. $P(effect|cause)$ is usually known. Therefore we try to infer $P(cause|effect)$. Remember effect is observable, hence we try to infer the cause given the effect. $P(data)$ is just a normalizing factor which is total $P(cause)$.

2

It's just a normalizing constant which makes the posterior a valid density. In practice, we don't care so much about it. It should be noted that $$p(x) = \int p(x\vert \theta) p(\theta) \, d\theta$$ So it is as if you are averaging the likelihood over the prior.

1

Unfortunately I think the only concusion you can draw here is that your optimization/selection approach failed. For a technically successful optimization (somewhat regardless of the achieved generalization error), cross validation score and test results should be highly and positiviely correlated and that's not the case here. One reason I can think of is ...

2

I actually just wrote it out. So here goes: Let $$\frac{\partial}{\partial \mu} E(X^{n}) = \int \frac{\partial}{\partial \mu} \frac{1}{\sqrt{2\pi}} x^n e^{-(x-\mu)^2/2} dx$$, then if I expand the square (in the exponent) to get $\mu^2 - 2\mu x - x^2$, I can factor out only the terms the involve $\mu$ and use those in the derivative. This gives, $$\... 0 If I understand your question correctly, your constraint is implied when you take the product to get the complete likelihood. You could write it again explicitly if you like, which would be the best approach to avoid ambiguity. (But it is generally implied given we are talking about a Poisson likelihood.) It will be true that for any data point k_i < 0,... 2 I think I understand your confusion. Typically, Bayes' rule is written as:$$p(\theta |y) = \frac{ p(y|\theta) p(\theta)}{p(y)}where p(\theta |y) is the posterior for the observed data y given unknown parameters \theta, p(\theta) is the prior distribution, and p(y) is the marginal distribution of y. As far as Bayes' rule is concerned, p(y)... 3 The likelihood function L(\theta|\mathbf x) is defined as a function of \theta indexed by the realisation x of a random variable with density f(\mathbf x|\theta): \begin{align}L\,&:\Theta\longmapsto \mathbb R\\ &\ \ \ \ \theta\longmapsto f(\mathbf x|\theta)\end{align} 4 A danger in having powerful tools available in standard software programs is that users don't always understand the underlying hidden assumptions. Even if you will be using Stata for routine work, I recommend getting a copy of An Introduction to Statistical Learning and working through the examples in Chapter 6 of LASSO and ridge regression, with the code ... 0 That n_1\delta_1+n_2\delta_2 has finite second moment has been shown in the other answer. To prove the linear combination is UMVUE, I would use this necessary-sufficient condition which states that an unbiased estimator (with finite second moment) is UMVUE if and only if it is uncorrelated with every unbiased estimator of zero. Let \mathcal U_0 be the ... 2 As a somewhat less technical answer, and not necessarily going into strict definitions and associations of what counts as proper causality what does not etc: The word 'to confound' itself means "to mistake / confuse something for something else". A confounding variable, therefore, putting strict technical contexts aside, is a variable [whose effect] is ... 4 This is six questions--and @StubbornAtom has done a good job outlining answers in one pithy comment. There appear to be two obstacles to going further: one is doing the initial algebra and the other is conceptual. Let's deal with the algebra first, because that's straightforward and doing it doesn't deprive you of the joy of discovery and learning the ... 2 This reminds me of a recent conversation with my dad who was surprised to hear that I still run models taking days which he did decades ago: Are you still doing computations that take longer than a day? Shouldn't everything run fast now? What are you people (physicists) improving on modeling nowadays, when there have already been satisfying models in the ... 11 Here, smoking is the confounder. The exposure is coffee drinking and the outcome is heart attack. To be a confounder the variable has to be a cause, or a proxy for a cause of both the exposure and the outcome. It does not have to be a direct cause. So here, it is sufficient for there to simply be correlation between coffee drinking and smoking, because ... 1 That's exactly what you should do to eliminate the healthy worker bias. Stratification on a confounder (which is what you described) eliminates the bias due to the confounder. The problem is when you don't stratify on the confounder, in which case the bias remains. The reason this is an interesting phenomenon is that one might not immediately think that a ... 0 I am certainly not an expert in cancer research, but have read that genomic markers (also known as genetic markers) have been studied with respect to their relationship to various diseases. To unmask the true strength of such genetic markers, I suspect, one might want to control for exposure to agents likely associated with cancer especially relating to ... 4 One problem with dumping all of your predictors into the model is the invitation to extreme collinearity, which will inflate your standard errors and likely make your results uninterpretable. Judea Pearl has pointed to a second problem, if your inference is aimed at modeling causal relationships. In trying to "control for everything" by including all ... 1 To say that the mean of a variable in one group differs from the mean of the variable in another group is to say that what group you're in has a relationship with the variable. For example, if men and women differed in average income, you would say there is a relationship between income and gender. So a test that compares the mean of the two groups is a ... 1 As @whuber notes in a comment, you do need to deal first with what seem to be incorrect premises in your approach. Most important, if your samples aren't truly random then "you can't use any of these methods to make inferences," as he put it. Fix that first. In terms of mean versus median as a measure of central tendency, the choice is yours based on your ... 4 I would like to add one point to EdM's answer, that has not been mentioned yet. Statistically significant but not important This could be some random feature of the data and because of the multiple testing problem some features are significant in the dataset purely by sampling. However, it could also be that the overall effect of an explanatory variable ... 11 A few general points before answering the individual questions. First, in logistic regression (unlike in linear regression) coefficient estimates will be biased if you omit any predictor associated with outcome whether or not it is correlated with the included predictors. This page gives an analytic demonstration for the related probit regression. Second, ... 6 6) Can you let me know what is the problem here and how can I address this? With all due respect, by reading your post I see only red flags due to misapplication and misunderstanding of the statistical methods. I would suggest employing a statistician (and at the very least, reading a great deal on clinical prediction models/regression modeling from Frank ... 3 You may have different primary interests: peak density (the highest y), delay from injection to peak density, or overall increase in density (difference between areas under the solid and dotted curves). Whatever the measure of interest you choose, it can be compared between the different experiments using analysis of variance for repeated measurements. An ... 0 T1 will be more efficient if it has a small variance compered to T2 8 You appear to be using a Beta(1,1) prior on \theta. Since this is a continuous distribution, the prior (and posterior) probability of the event that the coin is exactly fair, \theta=1/2, is zero. What would perhaps be a more sensible prior (see Lindley 1957 pp. 188-189 for a discussion of similar examples) would be a point mass at \theta=1/2 given ... 4 Indeed, X is not random as @whuber mentioned in the comments. Recall formulation of the Linear Regression problem: \begin{aligned} y_j =& \beta_1 x_{j1} + ... + \beta_n x_{jn} + \epsilon_j\\ \mathbf{y} =& X \boldsymbol{\beta} + \boldsymbol{\epsilon} \end{aligned}  where $\mathbb{E}\epsilon_j=0$. We are given $x_{ij}$ and want to estimate \$\...

1

The following holds generally, but the exact relationships may differ depending on the data. Including a variable that is a predictor of the outcome and the treatment will reduce the bias and variance of the effect estimate. The more related to the outcome and the less related to the treatment, the more variance will be reduced. Including a variable that ...

Top 50 recent answers are included