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I regard the variance of distribution as the moment of inertia with the axis that at the mean of the distribution and each mass as 1. This intuition would make the abstract concept concrete. The first moment is the mean of the distribution and the second moment is the variance. Reference: A first course of probability 8th edition


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Yes, truncating priors brings additional information into your model. By truncating prior distribution, you assume a priori that everything below (or above, depending if it is left- or right-truncation) has zero probability. After multiplying likelihood by prior, you are zeroing out the posterior in truncation region as well. You cannot make (negative) prior ...


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In R, the procedure var.test implements the standard F-test for equality of variances for data from two normal populations. Here is an example, using samples of size $n = 20$ from two normal populations with respective variances $\sigma_1^2 = 49$ and $\sigma_2^2 = 25.$ The null hypothesis of equal variances is not rejected at the 5% level. set.seed(2019) ...


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Going to post a late answer as it's quite simple but useful trick. The UMVUE is indeed given by $$P(X_i < a|T)$$ The trick is to use Renyi's representation which says that $$\frac{X_1}{\sum_{i=1}^n X_i} =_d U_{(1)}$$ Where $U_{(1)}$ is the 1st order statistic from a uniform(0,1) random sample $$U_1, U_2, \dots U_{n-1}$$ $$P(X_1 < a | \sum_i X_i = ...


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Joint density of $X_1,\ldots,X_n$ is $$f_{\theta}(x_1,\ldots,x_n)=\mathbf1_{\theta-\frac{1}{2}<x_{(1)},x_{(n)}<\theta+\frac{1}{2}}=\mathbf1_{x_{(n)}-\frac{1}{2}<\theta<x_{(1)}+\frac{1}{2}}$$ Clearly $T=(X_{(1)},X_{(n)})$ is a sufficient statistic for $\theta$, so it is expected that a reasonable confidence interval for $\theta$ would involve a ...


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(Answering my own question) In Bayesian analysis the prior represents our belief of the model parameters before we have seen any data and is this sense is assumed independent. To paraphrase Rasmussen, section 2.1: Often Bayes’ rule is stated as $p(a|b) = \dfrac{p(b|a)p(a)}{p(b)}$; here we use it in a form where we additionally condition ...


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If $X \sim \mathcal{Pois}(\lambda)$ then $X^* = \mathcal{1}\{X>0\}$ have a Bernoulli distribution with parameter $p=1-e^{-\lambda}$. If you have observations for $n$ boxes then $\sum_i^n X_i^* \sim \mathcal{Bin}(n,p)$ and you can base inference on that. The conjugate prior for the Binomial is the Beta distribution. An example is developed here.


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My understandings: An estimator is not only a function, which input is some random variable and output another random variable, but also a random variable, which is just the output of the function. Something like $y=y(x)$, when we talk about $y$, we mean both the function $y()$, and the result $y$. Example:an estimator $\overline X=\mu(X_1,X_2,X_3)=\frac{...


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In a Bayesian network, each node represents a random variable. A directed edge from node $A$ to node $B$ means that variable $B$ depends on variable $A$. The Bayesian network can be thought of as visual representation of the joint probability of all the random variables involved. A joint probability of random variables $X_1, X_2, \dots, X_n$ can be ...


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The key I think is to study the likelihood ratio. Pdf of the sample is $$f_a(x_1,\ldots,x_n)=\frac{1}{b^n}\exp\left(-\frac{1}{b}\sum_{i=1}^n (x_i-a)\right)\mathbf1_{x_{(1)}>a}\quad,\,a\in\mathbb R\,,b>0$$ Suppose we rewrite the alternative as $H_1:a=a_1\,(\ne a_0)$. The likelihood ratio is then \begin{align} \Lambda(x_1,\ldots,x_n)&=\frac{f_{...


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You should not base conclusions you make about the estimates of regression coefficients on the predictive performance of your model. You can have a model with a low $R^{2}$ but that still produces an unbiased estimate of the relationship you are studying. Whether a coefficient estimate is unbiased and represents the "true" relationship between $X_{1}$ and $...


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There is a much more general solution to this problem that holds for any distribution whose moments exist. It is a problem known as finding moments of moments. The modus operandi for solving such problems is to work with power sum notation $s_r$, namely: $s_r = \sum_{i=1}^n X_i^r$. We seek $\mathbb{E}[ (\frac{s_1}{n})^3]$ ... which is just the $1^\text{st}$...


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As you give no exact details about your data, I will give an example that you can adjust to however your data looks. I'm basing this on an example from the book 'Statistics' by Freedman et al., chapter 26 'Zero-one boxes'. Example of applying the z-test here Let's investigate the number of people that preferred Facebook in your example and formulate the ...


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Since $X_i-\beta$ are i.i.d $\mathsf{Exp}(1)$, we have $\min_i(X_i-\beta)=X_{(1)}-\beta\sim \mathsf{Exp}$ with mean $1/n$. Your conclusion is correct, but there is no need for guesswork. Just work out $E\left[X_{(1)}\right]$ from the pdf of $X_{(1)}$, i.e. find $\int xf_{X_{(1)}}(x)\,dx$ directly. If you are starting with $E_{\beta}\left[g(X_{(1)})\right]=\...


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Based on the analysis you did, the claim is indeed about individual behavior. In your example, you are interested in the probability of being in B1 vs. B2 for an individual in A1. For that individual, their outcome of B1 or B2 is governed by a parameter that applies to them (it may apply to others as well, but the focus here is on the individual). You are ...


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That is exactly what an statistical test is for, but, you must make sure that your sample of 30 individuals has the power to test your hypothesis. The best is to perform an a-priori power analysis, before you sample 30 individuals. Otherwise, how did you decided 30 was enough? It would also depend on the size of the population you are sampling from, if ...


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As Xi'an correctly points out, this is a maximisation problem over integers, not real numbers. The objective function is quasi-concave, so we can obtain the maximising value by finding the point at which the (forward) likelihood ratio first drops below one. His answer shows you how to do this, and I have nothing to add to that excellent explanation. ...


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Note that in the case of the Beta model/distribution the mean and the variance are related. Hence, you expect to see a particular type of heteroscedasticity. A better way to check the fit of the such a model to your data is to use the simulated residuals from the DHARMa package.


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No, plotting the two variables against each other does not constitute controlling for all other variables. Doing this with two different sets of variables also has no effect on that fact. What you are doing is ignoring all other variables. To better understand the distinction, it may help to read my answer to: Is there a difference between 'controlling ...


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Binomial PMF is a discrete function of $n$, say $f(n)$, given others, i.e. $x,p$. We want to maximize it in terms of $n$. Typically, we would take the derivate and equate it to zero, but in discrete cases we shouldn't do that. This PDF is known to have its peak value(s) around its mean (not exactly but close). Its graph first increases, and then decreases. ...


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In this answer to the question $$\dfrac{{n+1 \choose x}p^x(1-p)^{n+1-x}}{{n \choose x}p^x(1-p)^{n-x}} = \dfrac{n+1}{n+1-x}(1-p)$$ represents the likelihood ratio $$\frac{L(n+1|x,p)}{L(n|x,p)}$$ If this ratio is larger than one (1), $${L(n+1|x,p)}>{L(n|x,p)}$$ $-$ergo the likelihood increases$-$ and if it is smaller than one (1) $${L(n+1|x,p)}<{L(n|x,p)}...


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I did a very rough approximation to $a$ and $b$ via a system of equations. In your data, you sort all of the $n$ values and bin them (in the code below I made bins of same length, it would be probably better to make bins with roughly equal number of $n$'s inside). You count how many $1$'s and $0$'s are in your bins ($\#_1$ and $\#_{1\&0}$), and how many $...


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Question: Do you have any bounds or relation between $a$ and $b$? If you don't, suppose $n = 10$, you have many answers for $a$ and $b$ that satisfies $p = 0.5$: $b = 0$ and $a = 5$ $b = 1$ and $a = -5$ $b = 10$ and $a = -95$ If you do have bounds or relationship between your parameters, since you have clear Bernoulli experiment with a custom parameter, ...


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This seems a problem easily solved by additive risk models. Create the regressor 1/n and estimate the additive risk model relating 1/n to the outcome $y$, you will obtain maximum likelihood estimates of $a$ and $b$. This is one of a general class of "binomial models" which are among a general class of "generalized linear models". The most famous example in ...


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While the phenomenon is distinct from a type I or II error, I think calling it a type III error is misleading, because it suggests some sort of similarity in interpretation. This phenomenon arises only in null hypothesis significance testing. In the absence of a two-tailed null-hypothesis, (e.g., in binary classification) there exist only false positives ...


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You used a one-sided $t$-test to test against the null that $$\text{H}_0: \mu_\text{women} \leq \mu_\text{men},$$ but you observed $\hat{y}_\text{men} \approx 57.8$ and $\hat{y}_\text{women} \approx 60.3$. Therefore, you have very weak evidence against the null (in fact, your evidence points to the contrary), so you cannot reject it ($p \approx 0.993$).


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