# Tag Info

## Hot answers tagged markov-process

13

Thinning has nothing to do with Bayesian inference, but everything to do with computer-based pseudo-random simulation. The whole point in generating a Markov chain $(\theta_t)$ via MCMC algorithms is to achieve more easily simulations from the posterior distribution, $\pi(\cdot)$. However, the penalty for doing so is creating correlation between the ...

6

Thinning does not really work. On the contrary, without thinning you end up with more samples, and so, more precise estimates (Link and Eaton, 2012). Adding to great answer by Xi'an, nowadays thinning is rarely recommend as it doesn't help that much. You could use thinning if you want to reduce disk, or memory, usage by storing less samples, but that seems ...

2

Assuming a fixed initial value $y_0\in\mathbb R$, we have $y_t = \rho^t y_0 + \sum_{i=1}^t\rho^{i-1}\varepsilon_t$ for $t=1,2,\ldots,T$. Note that the linear combination of independent normal random variables is again normal, so $\sum_{i=1}^t\rho^{i-1} \varepsilon_t$ has $N\left(0,\frac{1-\rho^{2(t-1)}}{1-\rho^2}\right)$ distribution. For any $t\in\{1,2,\... 2 Because this kind of matrix appears frequently in Markov chain theory, let's generalize a little. The matrix$\mathbb P\$ is an instance of the family $$\mathbb{P}_d = \pmatrix{0 & \frac{1}{d-1} & \frac{1}{d-1} & \cdots & \frac{1}{d-1} \\ \frac{1}{d-1} & 0 & \frac{1}{d-1} & \cdots & \frac{1}{d-1} \\ ... 1 They're just saying that the probability of ending in state j, given that you start in state i is the element in the ith row and jth column of the matrix. For example, if you start in state 3, the probability of transitioning to state 7 is the element in the 3rd row, and 7th column of the matrix: p_{37}. 1 The set \{ H^A < \infty \}  is \{ \omega, H^A(\omega) < \infty \} = \{\omega, \inf(n, X_n(w) \in A) < \infty\} where it is assumed that \inf ( \emptyset) = \infty. Thus \{H^A < \infty \} is the set of all the possible outcomes (the \omega) for which there is at least of n \in \mathbb N such that X_n(\omega) \in A 1 Showing it is a Markov process means showing that$$P(X_t = x_t | X_1 = x_1,\dots,X_{t-1} = x_{t-1}) = P(X_t=x_t | X_{t-1} = x_{t-1})In your case, you have \begin{align*} P(X_t = x_t | X_1 = x_1,\dots,X_{t-1} = x_{t-1}) & = P(X_{t-1} + w_t = x_t|X_1 = x_1,\dots,X_{t-1} = x_{t-1})\\ & = P(x_{t-1} + w_t = x_t|X_1 = x_1,\dots,X_{t-1} = x_{t-1})\\ &... 1 I think you are somehow mixing independence and conditional independence. The idea is not that X_{n+1} is independent from X_0 ... X_{n-1}, but that it is conditionally independent given X_n. This simply means that, once you know the value of X_n, the previous values X_0...X_{n-1} are irrelevant for the distribution of X_{n+1}. Taking your ... 1 There are two questions. Why is p_{11}^0 = 1? This concerns powers of the matrix (which I will write as P). Because matrix powers may be unfamiliar, here is a quick sketch of the theory. For powers to be useful, they need to have some basic properties, one of which is that P^1=P and another is that for any natural number n, P^{n+1} = P\,P^n = P^n\, ... 1 There is a major distinction between the data being used to determine the prior (bad!) and the data being used to determine the Metropolis proposal (good!), as the former is for inference (double use of the data) while the later is for simulation (manageable approximation of the posterior). Hence, there is nothing wrong in using a Normal approximation to the ... 1 The generator matrix is given by G = \begin{pmatrix} -(\lambda_1+\lambda_2)&\lambda_2&\lambda_1&0\\ \mu_2& -(\lambda_1+\mu_2)& 0 & \lambda_1\\ \mu_1& 0 & -(\lambda_2+\mu_1)& \lambda_2\\ 0 & \mu_1 & \mu_2 & -(\mu_1+\mu_2), \end{pmatrix} $$and the transition matrix for the embedded Markov chain is given by$$ ...

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