time takes "a univariate or multivariate time-series, or a vector or matrix." How about time(airline.ts)?

From wikipedia: "The goal of Gibbs Sampling here is to approximate the distribution of $P(\mathbf{Z}|\mathbf{W};\alpha,\beta)$" Notation can be found on the wiki site or from the original paper here. ...

This is all from here: http://cims.nyu.edu/~holmes/teaching/asa15/Lecture2.pdf TLDR: your still using the spectral decomposition theorem; you just have to find the right symmetric matrix. Detailed ...

The normal conditional density's normalizing constant is not free of lambda, so Wolfram is integrating the wrong thing. $$(2\pi)^{-1}\int_0^\infty \underbrace{(2\pi\tau^2\lambda^2)^{-1/2}}_{\text{... View answer Accepted answer 3 votes Multiplying X by W gives you transformed data. Multiplying X by P gives you X again (P is the identity matrix because the eigenvectors are orthogonal). Observe that$$ \text{Var}\left[W^...

Hint: if $X \sim \text{HalfNormal}(\sigma^2)$ and $Y \sim \text{Normal}(0,\sigma^2)$, then for any symmetric and integrable $f$ $$E[f(X)] = E[f(|Y|)] = E[f(Y)|Y \ge 0] = 2 E[f(Y) 1(Y \ge 0)].$$ I ...

You're mistaken about a few things (and that's okay!). In this case the optimal proposal density...is [available]. I believe this is only true if $f$, the state transition is Gaussian. It can be ...

Actually, that formula above is not for the modified Bessel function of the first kind, it is for the (non-modified) Bessel function of the first kind! Furthermore, the Encyclopedia of Mathematics has ...

If $$\sqrt{n}[g(\hat{\theta}) - g(\theta)] \rightarrow_{d} N(0, \sigma^2[g'(\theta)]^2)$$ (note the typo you made in the variance term) then for a "large" $n$ it is approximately true that \... View answer 3 votes As I mention here, the interarrival times T_1, T_2, \ldots are distributed as \text{Exponential}(\lambda) random variables. You seem to be interested in T=\min (T_1,...,T_n) somehow, but as @... View answer Accepted answer 3 votes You are correct that Z\mid \mu_x,\mu_y,\sigma^2_x,\sigma^2_y \sim \mathcal{N}(\mu_x + \mu_y, \sigma_x^2 + \sigma_y^2). First note that this likelihood is not identifiable. This is because \mu_x,\... View answer Accepted answer 3 votes Here's a few things that might help. I left the derivation of the prior predictive distribution left for you because it follows some of the same tricks I used towards the end. First, the expectation ... View answer Accepted answer 3 votes Yes, if you're drawing one x_{T+1} for every sample joint posterior sample x_{1:T},\theta. To be more clear, you are correct if you do the following steps for i=1,2,\ldots: -1. Draw \theta[i]... View answer Accepted answer 3 votes First, notice that, if you use some properties of the trace operator, \begin{align*} p(\mu, \Sigma) &\propto \lvert \Sigma\rvert^{-((\nu_0+d)/2+1)}\exp\Big(-\frac{1}{2}\text{tr}(\Lambda_0\Sigma^{-... View answer 3 votes Yes, you are correct that there is a lack of identifiability unless one of the coefficent vectors is fixed. There are some reasons that don't mention this. I can't speak to why they omit this detail, ... View answer Accepted answer 3 votes Note, I am treating this as if it has been tagged with a self-study tag. Regarding (a), yes, the density must integrate to 1, and it must be nonnegative everywhere. Integrate over that unit square ... View answer Accepted answer 3 votes To quote Brockwell and Davis: "[t]he summability of |\gamma(\cdot)| implies that the series converges absolutely..." When you look at the definition of the spectral density f(\lambda) = \frac{1}{...

Main idea: Let $y \in \{1,2\}$ be the output, and $x \in \mathbb{R}^2$ be the input. Using Bayes' theorem: $p(y=1 \mid x) = \frac{p(x \mid y=1)\pi_1}{p(x)}$ $p(y=2 \mid x) = \frac{p(x \mid y=2)\pi_2}... View answer Accepted answer 3 votes In a multiple Bayesian linear regression model, do all variables (dependent and predictors) get prior distributions? All parameters get a prior. Random variables/data might get a prior if you're ... View answer 3 votes All particle filters are for state state space models, but not all state space models need a particle filter. Below is a quick explanation. You need a few probability distributions to define a state ... View answer Accepted answer 3 votes Yes, the variance of the sum is the sum of the variances, in this case. By the law of the unconscious statistician, for$i\neq j: \begin{align*} E\left[\frac{X_i}{Y_i}\frac{X_j}{Y_j} \right] &= \... View answer Accepted answer 3 votes Hint: the likelihood forT$is $$f_{T}(t ; N) = N(1-e^{-t})^{N-1}e^{-t}.$$ To verify this, first find the cdf of$T, and then differentiate. \begin{align*} F_T(t) &= P(T \le t)\\ &= [F_{... View answer 3 votes What do you mean by "find?" I can tell you\pi(\theta \mid x)is proportional to $$f(x \mid \theta) \pi(\theta) \propto \exp\left[-\frac{(\theta - x)^2}{2} - \log(1 + \theta^2) \right],$$ but I ... View answer 3 votes Regarding your edit, that's not right. You also need the product rule: \begin{align*} \frac{d^2\log p(y | \phi)}{d\phi^2} &= \frac{d}{d\phi} \left( \frac{d \log p(y|\theta(\phi))}{d \theta} \... View answer Accepted answer 3 votes The formula in likelihood is correct, but the bigger your data set gets, the more problems you're going to have with numerical underflow. That's why so many of the points in your graph are being ... View answer Accepted answer 3 votes Letz = 2m$, then$\frac{\text{d}z}{\text{d}m} = 2$and your integral equals $$\int_0^\infty m^{x+1}e^{-2m}\text{d}m = 2^{-x-2}\int_0^{\infty} z^{x+1} e^{-z}\text{d}z = 2^{-x-2} \Gamma(x+2).$$ ... View answer 3 votes First, you have the wrong log-likelihood. You might want to go back and check that part. Second, this likelihood, when thought of as a function in$\theta$, is only nonzero whenever$\theta$is less ... View answer 3 votes No, there is no way to explain this discrepancy because it does not make sense to sum together an AR and MA model in the way you have, because$X_t$cannot be both an AR model and an MA model at the ... View answer Accepted answer 3 votes A discrete Markov chain$\theta_1, \theta_2, \ldots$with transition pmf$k(\theta_{t+1} \mid \theta_t)$is reversible if joint pmfs for two consecutive time points are "symmetric", or in other words$...
Is it necessarily true that $\theta_2 | Y$ is the mean of $\theta_1 |Y$? No, they just have the same "centers." The mean of $\theta_1 | Y$ is  E[\theta_1 \mid Y] = E[ E(\theta_1 \mid \theta_2) \mid ...