8

$$\begin{align}E[X\log X]&=\int_0^\infty \log x \frac{\lambda^\alpha}{\Gamma(\alpha)}x^\alpha e^{-\lambda x}dx\\&=\frac{\Gamma(\alpha+1)}{\Gamma(\alpha)\lambda}\int_0^\infty\log x\frac{\lambda^{\alpha+1}}{\Gamma(\alpha+1)}x^{(\alpha+1)-1}e^{-\lambda x}dx\\&=\frac{\Gamma(\alpha+1)}{\Gamma(\alpha)\lambda}E[\log Y]\end{align}$$ where $Y\sim \text{...


6

Presumably this nomenclature is chosen simply to draw an analogy between the GLM and linear regression. You are correct that this term is not strictly accurate, since the family of distributions chosen for the GLM is not actually the distribution of an "error" quantity in the model. One can construct quantities in the GLM that are essentially measures of ...


4

I made the following figure for your example. The blue curve is the t-distribution under H0 hypothesis: mean = 15.4, from which you have derived the acceptance area $W = \{ 14.541 \leq \bar{x} \leq 16.259\}$. That is, if $\bar{x} \in W$, H0 is not rejected. The yellow curve is the t-distribution under the alternative H1: mean = 15.1. Type 2 error is ...


4

Let $W_{i,j:n} = X_{j:n}-X_{i:n},\; 1\leq i<j\leq n$ be the difference between the $i$th and $j$th order statistics (aka the spacings). The pdf of $W_{i,j:n}$ is then given by: $$ f_{W_{i,j:n}}(w) = \frac{n!}{(i-1)!(j-i-1)!(n-j)!}\times \int_{-\infty}^{\infty}\left\{F(x_{i})\right\}^{i-1}\left\{F(x_{i} + w) - F(x_{i})\right\}^{j-i-1}\times \left\{1-F(x_{...


3

Here is a figure based on a simulation in R that suggests the answer. The simulation uses a million observations of $X \sim \mathsf{Unif}(-2,1).$ Then we show histograms of the samples of $X$ and of $Y = |X|.$ You have made a reasonable start on the formal derivation. Now, you need to express the right-hand side of the CDF of $Y$ as a function of $y$ and ...


3

What you are told is only relevant when you are sampling from a finite population of size $N$, with simple random sampling without replacement. For most applications, there is no definite finite population, so what you are told is irrelevant. It is also irrelevant when you are samling with replacement. When relevant, there is a finite population ...


2

Not that I know of, but there's a related concept: the Nonparametric skew. Let $\mu$ be the mean $\sigma$ the standard deviation and $\nu$ the median. The Nonparametric skew is $$S = \frac{\mu-\nu}{\sigma}$$ Yours is $$S^* = \mu/\nu=(S \sigma + \nu)/\nu=S\sigma/\nu+1$$ And also $$S = (S^*-1)/(\sigma\nu)$$


2

A good example of this is provided by Glen_b -Reinstate Monica at https://stats.stackexchange.com/a/189633/40036 . The question there was slightly different -- asking if normally distributed $X$ and $Y$ must be jointly normal for $X+Y$ to be normal -- but the construction there also answers your question here (because the constructed $X$ and $Y$ happen to be ...


2

I assume $X_i$ are independent. So $Y=\min (X_1,\cdots X_n) $ $$F_Y(y)=1-P(Y>y)=1-P(X_1>y,\cdots, X_n>y)=1-e^{-n(y-\theta)}$$ so $$f_Y(y)=ne^{-n(y-\theta)} \hspace{1cm} \theta < y$$ so $E(Y)=\theta + \frac{1}{n}$. That is, $\min (X_1,\cdots X_n)$ is biased.


2

The "real" way to think of this is in terms of what probability we're assigning to certain sets; formally you could do this with measure theory and use a dominating measure that's a mixture of point masses at $3$ and $6$ plus the Lebesgue measure on $\mathbb R$, but we don't need that formality to reason about this correctly and get the answer. We know that ...


2

tl;dr: There is no known distribution. i.e. it doesn't have a name. You use simulations instead. I found this to give some insight: Basically, Shapiro and Wilk calculated the distribution of their statistic only for $n=3$, to be a truncated $Beta(\frac{1}{2}, \frac{1}{2})$ distribution for $\frac{3}{4} \le w \le 1$ and zero elsewhere. For $3 < n \le ...


2

The mathematical definition of an unbiased estimator is: $E[u(X_1, X_2,\dots,X_n)]=\theta$. In English, this formula means that the expected value of a statistic, generally given as $u(X_1, X_2], \dots, X_n)$, equals the parameter (of the population) value. While a parameter has a single (probably unknown) value, a statistic has a distribution of values (...


2

The simulated method of moments is a way in econometrics to exploit this ability to produce new samples for a given value of the parameter $\theta$. If $\mathbf x^\text{obs}$ is the observed sample and if $\mathbf x(\theta)$ is a simulated sample associated with the value $\theta$ of the parameter, then a consistent estimate of $\theta$ is produced by the ...


2

No, it is (generally) not possible to unequivocally determine the mean from this information, since you could use either the upper or lower tail of the distribution to meet your requirements. If you find that offsetting the distribution to the right of your range by some amount yields the correct probability of falling in that range, offsetting the ...


2

The data is just too noisy to analyze by direct inspection, so I can understand why the question of outliers arose. However, identifying outliers requires that a logical, physical reason for that status to be at least postulated. The only outliers herein are the 99+ answers, which literally lie outside the range of the data. What is occurring with the human ...


2

You've basically got it! Here's some answers to your questions: No; a sample space is not a population. The sample space is the set of all the possible outcomes of an experiment. The population is the set of all events which is of interest. The set of outcomes of a coin flip {heads, tails} is a sample space, but not a population. The population would be ...


2

Having the mean result isn't necessarily the one with the highest probability. Also, the probability of it won't be as high as you'd expect in most cases. For example, think about a fair coin toss 100 times. The probability of having 50 heads is ${100 \choose 50}0.5^{100}\approx 0.08$. It's not that high as you'd expect. It gets even much smaller as $n$ ...


2

There is no general way to generate a set of pseudo-random numbers without specification of a method. Any method you use is going to require you to stipulate what you are doing. So, if you don't want to use real data, and you want to remain ignorant of the generating method, the most obvious solution to your problem is to have someone else generate the ...


2

If the data sample is large enough (I would say at least 10 times the number of categories), then you may apply a chi-square test of homogeneity for an uniform distribution. Regarding the graphical methods, consider a bar chart showing a subset of categories, for instance, the top 5 and the bottom 5 categories.


2

As @linog mentioned in his comment, typing poisson.test into R will print the source code for the function. The relevant bit is: ⋮ m <- r * T PVAL <- switch(alternative, less = ppois(x, m), greater = ppois(x - 1, m, lower.tail = FALSE), two.sided = { if (m == 0) (x == 0) else { relErr <- 1 + 1e-07 ...


1

First, a discrete distribution (or a distribution with atoms) cannot be transformed such to a multinormal, so assume an absolutely continuous distribution. Then, if $p=1$, we can always transform to normal. So assume $p>1$. Then, if the components are independent, again, we can use the univariate solution. So there must be some dependencies between the ...


1

I think the following example could do it : Let $X_1 \sim \mathcal{N}(0, 1)$ and $B$ be a binary variable which is equal to $1$ with probability $0.5$ and $-1$ with probability $0.5$. Define $X_2 = B\times X_1$. Then the marginal distributions of $X_1$ and $X_2$ are obviously $\mathcal{N}(0, 1)$ but $(X_1; X_2)$ is not a multivariate Gaussian. An easy ...


1

Knowing that the pdf is the derivative of the cdf, from $F_Y(y) = F_X(e^y)$, to get the pdf of $Y$ you just need to differentiate with respect to $y$ and you get that $f_Y(y) = f_X(e^y)e^y$. So if $y\in[ln(2),ln(6)]$ it is $e^y/4$ and $0$ otherwise.


1

You can convert any continuous distribution to a uniform, by transforming it by its cdf. i.e. if $X\sim F_X$ then $U=F_X(X)$ has a standard uniform distribution. https://en.wikipedia.org/wiki/Probability_integral_transform#Proof $\:$ You can convert a standard uniform to a standard normal by transforming it by the inverse cdf of the standard normal $Z=\...


1

Your problem is equivalent to randomly filling 20 boxes each with three items either A's and B's. There is no overlap between the boxes (even if there would be overlap it would actually not matter, but it is conceptually much more easy to understand). This should be a binomial distribution. For any particular point/box you draw independently 3 protein ...


1

Welcome to CV! Generally in mathematics, if $f(x)$ denotes a function, $f(y)$ denotes the same function; the input variable is just a dummy variable. You could call it $x$, $y$, or anything else, but the way the function operates on an input is the same. When they say $P(X)$ is usually not the same as $P(Y)$, they're pointing out that this is different ...


1

If $Y=X$, where $X$ is a binomial distribution. Notice that $Y=2X$ is not a binomial distribution. In particular, $Y$ always take even numbers. Its expected value is indeed $2np$ and its variance is $4np(1-p)$ but it does not follow any binomial distribution. Remark: From the first sentence, I am not sure if $2X$ is what you are interested, if a success ...


1

You can find more detail on this issue in a related question here. Absolute continuity of the CDF is a stronger condition than continuity, and essentially just means that the distribution has a valid density function. For example, if the CDF $F$ of a real scalar random variable is absolutely continuous then there exists a real function $f$ (the density) ...


1

One thing that helps is to plot a histogram, which gives a visual clue as to outliers, skewness, and so forth. Some packages, e.g., Mathematica, have find distribution routines that test common distributions and mixtures of distributions to identify better fitting candidate distributions, e.g., see FindDistribution. Next, a priori knowledge about the problem ...


1

The terminology may be the key here: The two $p$ you cite are in fact the same (if we regard the population as infinitely large). In this context, one trial is drawing one element from the population and recording whether it has the property of interest or not. So the probability of success $p$ (the probability of said element having said property) is the ...


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