Konstantin
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## 18 Answers

Accepted answer
6 votes

Answer: Your code seems to be in line with the cited paper, and the incosistency is an artefact of non-standard definition of principal components used in scikit-learn of versions < 0.22 (this is ...

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6 votes

Answer: Whatever the distribution of $X_1,...,X_n$, $$\mathbb{E} Y_2 \geq \mathbb{E} Y_1.$$ Details: For any $n$ numbers $X_1,..., X_n$ it is true that $$\sum_i |X_i| \geq |\sum_i X_i|$$ and ...

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Accepted answer
5 votes

Answer: There is a mistake in the formula for $\theta$. The correct computation must align autocovariances of the MA components of two representations. The correct formula is $$\theta = \frac{\... View answer 5 votes Answer: Posterior of \sigma^2|Y_1,..., Y_n is an instance of inverse gamma distribution with the probability density$$ p(\sigma^2|Y_1,...,Y_n) = \frac{\beta^\alpha}{\Gamma(\alpha)} (\sigma^2)^{-\...

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Accepted answer
4 votes

Answer: Correct, $\Sigma^{-1}(I-P)$, positive definite, would also be a kernel. Correct, $\Sigma^{-1}$, positive definite is a valid kernel. $\Sigma^{-1} K = \frac{1}{\sigma^2_1}I-\frac{\sigma^2_2}{\... View answer Accepted answer 4 votes The answer: $$P_{X_1 > X_2} = \frac{B(\frac{1}{2};a_2,a_1)}{B(a_1,a_2)},$$ where$B(\alpha,\beta) = \int_0^1 t^{\alpha-1}(1-t)^{\beta-1}dt$is the Beta function and$...

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Accepted answer
3 votes

Short answer: Indeed, when the same customer may be approached at most $n$ times, it is optimal to start with offer $y_1=\frac{n-1}{n}x$ and decrease the price by $\frac{x}{n}$ with every refusal. ...

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Accepted answer
3 votes

Answer: Although it does not resemble any distribution I know, it is possible to obtain a compact expression for the probability density. Denote $W=p_1+p_2$ and $Z=p_1+p_3$ then their joint density ...

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Accepted answer
2 votes

The answer : \begin{align} p(y_1=1|y_1^o = 1, y_{-1}^o = \mathbf{0} ) = \frac{ p(y_1^o=1|y_1=1) \cdot \sum_{y_2=0}^1 \sum_{y_3=0}^1 \sum_{y_4=0}^1 \prod_{i=2}^4 p(y_i^o=0|y_i) \phi_{1,y_2} \phi_{...

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Accepted answer
2 votes

You are on the right track. Substitute expressions for the densities of normals into the Bayes law (and don't keep track of constants !!) \begin{align} f(\theta|X) &\propto \exp(-\frac{1}{2}(X-\...

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Accepted answer
1 votes

Just follow the instructions: A horizontal move represents deletion A vertical move represents insertion A diagonal move represents match 'deletion', 'insertion' and 'match' refer to the ...

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1 votes

Answer: Every unit-root state-space model has an equivalent ARIMA(0,1,1) representation, but not every ARIMA(0,1,1) model has an equivalent state-space model representation. The following holds true ...

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1 votes

Answer: A decent first pass would be a generalized ARMA model followed by the analysis of impulse response functions (IRFs). Interesting IRF measurements: peak, delay until peak, halftime of peak ...

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1 votes

Try numpy.random.dirichlet in Python. Once you install Python and the numpy package (to install both in the easiest way start here), you can generate your samples in a new file by the code import ...

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1 votes

Intuitively, quadrature approximation of an integral is replacing the function $f(x)$ under the integral by a close enough step-function $\hat f (x)$ (piecewise constant with a finite number of jumps)....

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0 votes

Answer: there is no way to do it, the premise is wrong. For an AR(1) process $x_t = \rho x_{t-1} + e_t$ where $e_t$ is a white-noise process, the equation $$\mathbb{V}x_t = \rho^2 \mathbb{V}x_t + \... View answer 0 votes Your source has Theorem 2', which says that Z SOSD X if for every concave function u$$ \mathbb{E}u(Z) \geq \mathbb{E}u(X), whatever the sign of the first derivative, u'. Now, \begin{align} ... View answer Accepted answer 0 votes According to the Bayes law the joint probability of a sample observation (x,y) can be decomposed into the familiar product: P(x,y) = P(x|y)P(y). In other words, we can always partition the set ...

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