14

Given two samples that have the same mean, standard deviation, and N: are the values in each sample identical? In general, not unless N=2 in both samples. If N is larger than 2, they can differ. You can see this simply by trying it with some simple cases. Perhaps the easiest case is to take an asymmetric sample of size $N=3$ and flip it around its mean ($2,...


14

No. Many data sets can yield the same mean, SD and n. In the graph above, the three data sets on the left of each graph all share the same mean, SD and n. So do the three data sets on the right of each graph. This is Figure 1 from: Weissgerber, T.L., Milic, N.M., Winham, S.J., and Garovic, V.D. (2015). Beyond bar and line graphs: time for a new data ...


4

Yes, a permutation test makes sense here. Strictly speaking, the null hypothesis is that the distributions are the same, not just that they have the same means. (If they had same means but difference variances, the test would have the wrong Type I error rate.) Here's example code > df<-data.frame(group=factor(rep(c(1,2,3),c(11,6,5))), + ...


3

Ultimately, the problem here is that $\sigma \stackrel{P}{=} \sigma' $ does not imply that $\sigma = \sigma'$, so no, you cannot. A couple of notes though: Well normally proving identifiability follows by showing that $p_{\theta}(x)=p_{\theta'}(x)$ implies $\theta=\theta'$. Since the definition of identifiability is that the function $\theta \mapsto p_\...


3

Since $$\frac{1}{1+x} \, \frac{\beta^\alpha \, x^{\alpha-1} \, e^{-\beta x}}{\Gamma(\alpha)}\le \frac{\beta^\alpha \, x^{\alpha-1} \, e^{-\beta x}}{x\Gamma(\alpha)}=\frac{\beta}{\alpha}\frac{\beta^{\alpha-1} \, x^{\alpha-2} \, e^{-\beta x}}{\Gamma(\alpha-1)}$$ another possibility is to accept/reject with a Gamma $\mathcal G(\alpha-1,\beta)$ proposal, ...


2

Yes, this is the only difference - though sometimes you can simplify the calculation if you can phrase the dependency as $f_{X,Y}=f_X(x)\cdot f_Y(y|x)$, using the conditional probability density function, yielding: $$ f_Z(z)=\int_{-\infty}^{\infty}f_X(x)f_Y(z/x | x)\frac{1}{|x|}dx, $$ (In the case of independence, $f_Y(z/x | x) = f_Y(z/x)$, returning to your ...


2

I wouldn't use the p-value to choose; I'd use the test statistic itself. More specifically, I'd be quite unlikely to do something like this in the first place, and if I did I'd be quite unlikely to choose the Kolmogorov-Smirnov to do it wth, but if for some reason I did do it, I'd be looking at the actual measure of discrepancy rather than its p-value, since ...


1

This is an open problem in statistics and machine learning. Several methods to approximate the KL divergence have been proposed. For instance, take a look at the FNN R package: https://cran.r-project.org/web/packages/FNN/FNN.pdf It miserably fails sometimes, but it works in simple cases and with large samples (for samples smaller than 100 it can behave ...


1

In $$\DeclareMathOperator{\E}{\mathbb{E}} D_{KL}(P || Q) = \int_{-\infty}^{\infty}p(x)\log\left(\frac{p(x)}{q(x)}\right)\;dx = \E_{P}\log\left(\frac{p(X)}{q(X)}\right) $$ we see this is te expectation of the loglikelihood ratio when $P$ is the truth, see Intuition on the Kullback-Leibler (KL) Divergence. If, in hypothesis test language, $P$ is the null ...


1

You could look at the beta-binomial distribution, $X\sim\text{BB}(n,\alpha,\beta)$. If $\alpha=\beta=1$, then it is a discrete uniform distribution, if $\alpha\ge 1$ and $\beta<1$ then it is a discrete left-skewed distribution. In R you could use the extraDistr package. For example: > library(extraDistr) > x <- 0:100 > y <- dbbinom(x, 100, ...


1

Many simulation problems seek a large random sample of independent and identically distributed (IID) random observations, all from the same population. The built-in simulation procedures in R are mainly designed with such IID samples in mind. So it is necessary to do some extra work to change the distribution during the sampling process. 100 Coin Tosses: ...


1

TL;DR: Don't confuse the probability density with the probability. In the given example, the probability is zero: $\mathrm{Pr}(m=120\,\mathrm{g})=0$, but the probability density is non-zero: $p_M(m=120\,\mathrm{g}) \approx 0.0299\,\mathrm{g^{-1}}$. There have already been quite a few answers, but I think that visualizing things might help understanding, ...


1

If $X \sim \mathsf{Pois}(\lambda)$ and you have random observations $X_1, \dots, X_n,$ then $T = X_1 + \cdots + X_n \sim \mathsf{Pois}(n\lambda).$ If your goal is to make confidence intervals, then the following may be relevant for making CIs for Poisson means. The logic is somewhat similar that of the Agresti-Coull binomial 95% CI based on $\hat p = (x+2)/(...


1

If a unit-length rod is broken into $n=k+u$ segments, where the breaks occur with uniform probability along the rod, then the total end-to-end length $x$ of $k$ randomly-selected segments (discarding the other $u$ segments) is described by the Beta($k,u$) distribution, with probability density function: $$\rho(k,u,x) = \frac {\Gamma(k+u)} {\Gamma(k)\Gamma(u)...


1

Question 2. You want the stationary distribution of the Gaussian AR process $X_t$ $$ (1 - \phi_1 B - \dots - \phi_p B^p) \, X_t = (1 + \theta_1 B + \dots + \theta_q B^q) \,\varepsilon_t $$ for the special case $p=q=2$. This distribution a.k.a. the invariant distribution is a Gaussian distribution: its mean $\mu_X$ and sd $\sigma_X$ can be found. ...


1

HDRs for standard probability distributions can also be computed using the stat.extendpackage. This gives a user friendly output for the region in set form. Here is an example of computing the region for a particular choice of parameters. #Compute and print HDR for the chi-squared distribution DF <- 10 NCP <- 1 HDR <- stat.extend::HDR.chisq(0.99, ...


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