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Sorting a list of objects based on an accompanying set of IID continuous random variables (such as uniform random variables) is equivalent to shuffling those objects into a random order (i.e., by a random permutation). Since the random values are independent continuous random variables, every possible permutation is equally likely, and that is the ...
4
Any linear regression goes through $\bar x,\bar y$, hence, these two should intersect too.
Consider regressions $y=a+bx+e$ and $x=c+dy+u$, take the expectations of both sides of equations:
$$E[y]=a+bE[x]+E[e]$$
$$E[y]=a+bE[x]$$
similarly
$$E[x]=c+dE[y]$$
4
Let the white balls form a set $\mathcal W$ of size $N_1$ and the black balls form a set $\mathcal B$ of size $N_2,$ both of which are disjoint subsets of the set of all $N$ balls, $\mathcal U.$
For any ball $i\in \mathcal U,$ let $X_i$ be the indicator that ball $i$ is included in the sample $\mathcal S:$ that is, it equals $1$ when $i\in S$ and otherwise ...
4
The idea behind bootstrapping is to get the standard error of an estimator ($\approx$ standard deviation of the sampling distribution of the estimator) by looking at the variation (standard deviation) you see in the value of the estimator across boostrapped datasets.
Your SE should not be going to zero (and thus, confidence interval width going to zero and ...
3
Is an estimate of the performance of a (Bayesian or Frequentist) prediction model a frequency guarantee of that prediction model?
No. Performance estimates do not guarantee performance; they estimate it. Also, most performance guarantees are asymptotic. Do you have an infinite amount of data?
Unbiased estimators are guaranteed to be perfectly accurate. ...
2
I imagine that people could come up with a justification for the two step approach, but to me, it seems like a bit of a waste. Most critical from my perspective, if you run everything in a single model,
$logit(p(y_{it}=1))=\beta_0 + u_{0i} + (\beta_1 + u_{1i})x$
then the $u_{0i}$ and $u_{1i}$ remain latent variables, thus reducing measurement error that is ...
2
It should be emphasized that you don't need to sort in order to sample. The method given in the tutorial works, but it is extremely inefficient. It basically does $\Theta(n \log n)$ operations for what can be done in $\Theta(1)$.
If you can sample a random floating point number from 0 to 1, you can sample a random integer from 1 to n. And Excel can give you ...
1
That inequality is an application of the Cauchy–Schwarz inequality:
$$|\langle \mathbf{u},\mathbf{v}\rangle| ^2 \leq \langle \mathbf{u},\mathbf{u}\rangle \cdot \langle \mathbf{v},\mathbf{v}\rangle,$$
where $\langle\cdot,\cdot\rangle$ is the inner product.
For random variables $Y_1$ and $Y_2$, the expected value of their product is an inner product:
$$\langle ...
1
Thanks to whuber for the hint that broke my mental block on this one.
The transformation to avoid would be
y <- 152.5 - 1.5 * x
This is a bad idea because it means x and y are negatively correlated. The example refers to test scores, so this would obviously be a bad way to rescale them!
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Example: If $X \sim \mathsf{Beta}(2,1)$, which has density function $f_X(x) = 2x,$ for $0 < x < 1$ and $0$ elsewhere. [See Wikipedia on beta distributions.]
Then
$$\mu_X = E(X) = \int_0^1 xf_X(x)\,dx = \int_0^1 x(2x)\,dx\\
=\int_0^1 2x^2\,dx = 2/3.$$
Also,
$$\sigma_X^2=Var(X) = E(X^2) - \mu_X^2 = \int_0^1 x^2f_X(x)\,dx - (2/3)^2\\ = \int_0^1 2x^3\, dx -...
1
This Question has two interesting and somewhat similar parts. The first question is in the title
and the second is the question about the data table you provided.
Title.
Suppose the data are normal in particular, or symmetrical in general (provided that the population mean exists). Then it is
possible to get a reasonable estimate of $\mu$ as the midrange, ...
1
Only if you estimate the concatenated variance with degrees of freedom equals to $2$:
np.var(np.concatenate((x1-np.mean(x1), x2-np.mean(x2))), ddof=2)
Because, $(n_i-1)S_i^2=\sum_{n=1}^{n_i} (x_{i,n}-\hat\mu_i)^2$ and when summed over $i$, this is going to be equal to the unnormalized variance of the concatenated array. And you need to normalize it with $...
1
One of the libraries in R has a runs.test procedure, which you can explore. My purpose here here is to give
an idea how looking at runs can help you decide whether
your observations are randomly sampled from the same population.
To begin we look specifically at sequences of Bernoulli
trials, as mentioned in your Question. (Randomness tests
for other ...
1
0. Your question suggests (by use of the word "the") that there is only one basis for the natural cubic splines. There are in fact many, and this is only one particular basis for them.
1. It sounds like you want an intuitive way to interpret $d_k(x)$. Nice, intuitive ways of understanding things don't always exist, and I doubt such an ...
1
Symplectic model of Statistical Physics and Information Geometry is given by Souriau model of "Lie groups Thermodynamics":
Lie Group Cohomology and (Multi)Symplectic Integrators: New Geometric Tools for Lie Group Machine Learning Based on Souriau Geometric Statistical Mechanics,
Lie Group Statistics and Lie Group Machine Learning Based on Souriau ...
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