New answers tagged mean
3
The distribution you are dealing with is a truncated zeta distribution, with mass function given by:
$$p_K(k) = \frac{k^{-\alpha}}{\zeta (\alpha,k_\min)}
\quad \quad \quad \text{for all integers } k \geqslant k_\min,$$
where we use the Hurwitz zeta function given (for positive integer $k_\min$) by $\zeta (\alpha,k_\min) = \sum_{k=k_\min}^\infty k^{-\alpha}$. ...
2
Your distribution $p_k \sim k^{-\alpha-1}$ for $k \geq k_{\text{min}}$, $k_{\text{min}} > 0$ is a truncated zeta distribution.
The distribution has no finite variance for $\alpha<2$ and the scaled sum will not approach a normal distribution.
However, you can apply a generalization of the central limit theorem. The limiting distribution of the ...
0
If the only property of the initial dataset that needs to be preserved is the rank order, then a variety of transformations are possible. Here's the simplest one I can think of:
Let $m$ be the mean specified by the user, and $x$ be the initial data ($n$ points). For $i=1,...,n$, define the new values as
$$ x_i' = 2m(i-0.5)/n $$
Map the $i$-th smallest $x$ to ...
2
Consider a discrete random variable $X$ taking on values $x_1, x_2, \cdots, x_n$ with positive probabilities $p_1, p_2, \ldots, p_n$ respectively. Two shibboleths that statisticians don't just murmur but instead shout from the rooftops are that probabilities are nothing but long-term frequencies, and that an event of probability $p$ will occur approximately $...
7
It's analogous to the discrete version. It's generally useful to think $P(X=x)\approx f(x)\Delta x$ in the continuous case. In the limiting case, as $\Delta x$ goes to $0$, this probability is $0$. So, the expected value will be $$E[X]\ \approx \sum x\ P(X=x)\ =\ \sum_{x\in\{a,a+\Delta x,...,b\}} x\ f(x)\Delta x$$
If you take the limiting case, this is ...
1
Hello and welcome to this community.
In order to clarify your doubts, you can first think of discrete spaces.
Take for example the random experiment concerning the toss of a fair dice. In this case, the sample space (that just mean the set of all possible outcomes) is $\Omega = \{1,2,3,4,5,6\}$, and we have a probability measure $\mathbb{P}$ that assigns to ...
0
There is no reason to believe that the variability of test scores will be the
same from one state to another. If fact, you say that there are 'significant
differences' among states.
So if you are doing inference for Florida (making confidence intervals or doing t tests), then you should use the data from Florida
(sample size, mean, standard deviation, and ...
1
Your answer is correct.
This is the basic idea underlying the z-test (which is correct for a sample size equal to 250 and known variance).
For the central limit theorem, you have $\bar{x}|\mu,\sigma^2 \sim N(\mu, \sigma^2/N)$ for $N \rightarrow +\infty$
which means that (considering $N=250$ large enough)
$$
z(\bar{x})=(\bar{x}-\mu)/(\sigma/\sqrt{N})\sim N(0,...
0
In the regression model, knowing that Y is Normally distributed, one can use properties of the multivariate normal distribution to provide an answer.
I would recommend you look at Wikipedia's comment, especially for example, in the case of the bivariate normal distribution, and in particular, the conditional distribution of Y given X.
There is also an ...
0
I would like to extend the answer by Wolfgang on rescaling the measure. Consider some study. Let $X$ be the averaged mean, $l$ be the lower bound for the Likert scale, $u$ be the upper bound for the Likert scale (both bounds over all the answers) and $n_q$ be the number of items. We can split the lower and upper bound up into the number of items and the ...
2
For simplicity, suppose we are dealing with an absolutely continuous distribution with density function $f_X$ with some corresponding non-negative kernel function $g_X \propto f_X$. Suppose we consider the general $k$th absolute moment, which is given by the following integral expressions:
$$\mathbb{E}(|X^k|)
= \int \limits_\mathbb{R} |x|^k f_X(x) \ dx
= \...
3
It's instructive to see what goes wrong -- the integrals are all very well, but a sample average is always finite, so what is the issue?
I'll use the Cauchy distribution, which has no finite mean. The distribution is symmetric around zero, so if it had a mean, zero would be that mean. Here are cumulative averages of two samples of ten thousand Cauchy ...
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