Newest questions tagged ergodic - Cross Validated most recent 30 from stats.stackexchange.com 2019-11-22T15:19:31Z https://stats.stackexchange.com/feeds/tag?tagnames=ergodic&sort=newest https://creativecommons.org/licenses/by-sa/4.0/rdf https://stats.stackexchange.com/q/436296 0 Minimizing the rate of geometric ergodicity of a Metropolis-Hastings kernel 0xbadf00d https://stats.stackexchange.com/users/222528 2019-11-15T20:12:00Z 2019-11-18T09:29:31Z <p>Let <span class="math-container">$\kappa$</span> denote the transition kernel of the symmetric Metropolis-Hastings algorithm with proposal kernel <span class="math-container">$Q$</span> and target distribution <span class="math-container">$\mu$</span>.</p> <blockquote> <p>What would be a general guideline if we're aiming to show that (in a particular instance of the problem) <span class="math-container">$\kappa$</span> is geometric ergodic with a preferably small rate <span class="math-container">$\rho\in(0,1)$</span>, i.e. <span class="math-container">$$\left\|\kappa^n(x,\;\cdot\;)-\mu\right\|\le M(x)\rho^n\;\;\;\text{for all }n\in\mathbb N\tag1,$$</span> where <span class="math-container">$\left\|\;\cdot\;\right\|$</span> denotes the total variation norm and <span class="math-container">$M(x)&lt;\infty$</span>, for <span class="math-container">$\mu$</span>-a.e. <span class="math-container">$x\in E$</span>?</p> </blockquote> <p>Let <span class="math-container">$x\in E$</span>. We know that <span class="math-container">$$\left\|\kappa(x,\;\cdot\;)-\mu\right\|=\frac12\sup_{\left\|f\right\|_\infty\le1}|(\kappa f)(x)-\mu f|\tag2.$$</span> Maybe we can do something like this: Let <span class="math-container">$A_x:=\left\{\alpha(x,\;\cdot\;)=1\right\}$</span> and <span class="math-container">$R_x:=A_x^c$</span>. Now left <span class="math-container">$f:E\to\mathbb R$</span> be <span class="math-container">$\mathcal E$</span>-measurable with <span class="math-container">$\left\|f\right\|_\infty\le1$</span>. Then, <span class="math-container">\begin{equation}\begin{split}|(\kappa f)(x)-\mu f|&amp;=\left|\int\lambda({\rm d}y)(k(x,y)-p(y))(f(y)-f(x))\right|\\&amp;\le2\int\lambda({\rm d}y)|k(x,y)-p(y)|\end{split}\tag3\end{equation}</span> and <span class="math-container">\begin{equation}\begin{split}&amp;\int\lambda({\rm d}y)|k(x,y)-p(y)|\\&amp;\;\;\;\;\;\;\;\;\;\;\;\;=\int_{A_x\:\cap\:\{\:p\:=\:0\:\}}\lambda({\rm d}y)q(x,y)+\int_{A_x\:\cap\:\{\:p\:&gt;\:0\:\}}\mu({\rm d}y)\left|\frac{q(x,y)}{p(y)}-1\right|\\&amp;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+\int_{R_x}\mu({\rm d}y)\left|\frac{q(x,y)}{p(x)}-1\right|,\end{split}\tag4\end{equation}</span> where <span class="math-container">$k(x,y):=q(x,y)\alpha(x,y)$</span> for <span class="math-container">$x,y\in E$</span>.</p> <blockquote> <p>Maybe we can attempt to minimize <span class="math-container">$(4)$</span>. Is this a feasible approach? My goal would be to apply this in the particular instance of my other question: <a href="https://mathoverflow.net/q/346120/91890">https://mathoverflow.net/q/346120/91890</a>.</p> </blockquote> <p><em>Definitions</em>: Let</p> <ul> <li><span class="math-container">$(E,\mathcal E,\lambda)$</span> be a measure space;</li> <li><span class="math-container">$p$</span> be a probability density on <span class="math-container">$(E,\mathcal E,\lambda)$</span> and <span class="math-container">$\mu:=p\lambda$</span>;</li> <li><span class="math-container">$Q$</span> be a Markov kernel on <span class="math-container">$(E,\mathcal E)$</span> with <span class="math-container">$$Q(x,B)=\int_B\lambda({\rm d}y)q(x,y)\;\;\;\text{for all }(x,B)\in E\times\mathcal E\tag1$$</span> for some symmetric <span class="math-container">$\mathcal E^{\otimes2}$</span>-measurable <span class="math-container">$q:E^2\to[0,\infty)$</span>;</li> <li><span class="math-container">$$\alpha(x,y):=\left.\begin{cases}\displaystyle1\wedge\frac{p(y)}{p(x)}&amp;\text{, if }p(x)&gt;0\\0&amp;\text{, otherwise}\end{cases}\right\}\;\;\;\text{for }x,y\in E;$$</span></li> <li><span class="math-container">$$\tilde\kappa(x,B):=\int_BQ(x,{\rm d}y)\alpha(x,y)\;\;\;\text{for }(x,B)\in E\times\mathcal E$$</span> and <span class="math-container">$$\kappa(x,B):=\tilde\kappa(x,B)+\underbrace{\left(1-\tilde\kappa(x,B)\right)}_{=:\:r(x)}1_B(x)\;\;\;\text{for }(x,B)\in E\times\mathcal E$$</span></li> </ul> https://stats.stackexchange.com/q/436087 2 Is the Metropolis-Hastings kernel always aperiodic, irreducible and geometrically ergodic? 0xbadf00d https://stats.stackexchange.com/users/222528 2019-11-14T16:36:35Z 2019-11-15T09:28:58Z <p>Let <span class="math-container">$(E,\mathcal E,\lambda)$</span> be a measure space, <span class="math-container">$Q$</span> be a Markov kernel on <span class="math-container">$(E,\mathcal E)$</span> with <span class="math-container">$$Q(x,B)=\int_B\lambda({\rm d}y)q(x,y)\;\;\;\text{for all }(x,B)\in E\times\mathcal E$$</span> for some symmetric <span class="math-container">$q:E\times E\to[0,\infty)$</span>, <span class="math-container">$p$</span> be a probability density on <span class="math-container">$(E,\mathcal E,\lambda)$</span> and <span class="math-container">$\mu:=p\lambda$</span>.</p> <blockquote> <p>Let <span class="math-container">$\kappa$</span> denote the transition kernel of the chain <span class="math-container">$(X_n)_{n\in\mathbb N_0}$</span> generated by the Metropolis-Hastings algorithm with proposal kernel <span class="math-container">$Q$</span> and target distribution <span class="math-container">$\mu$</span>.</p> <p>I want to know under which additional assumptions (if any) <span class="math-container">$\kappa$</span> is geometrically or uniformly ergodic. Moreover, is <span class="math-container">$\kappa$</span> always <span class="math-container">$\phi$</span>-irreducible and aperiodic? In particular, I'd like to know whether we can apply Proposition 2.1 and Theorem 2.1 in the paper <a href="https://projecteuclid.org/download/pdf_1/euclid.ecp/1453832497" rel="nofollow noreferrer">Geometric ergodicity and hybrid Markov chains</a>.</p> </blockquote> <p>If necessary, assume that <span class="math-container">$(E,\mathcal E)$</span> is a <span class="math-container">$\mathbb R$</span>-vector space, <span class="math-container">$(E,\mathcal E,\lambda)$</span> is translation-invariant and <span class="math-container">$q(x,y)=\tilde q(x-y)$</span> for some <span class="math-container">$\mathcal E$</span>-measurable <span class="math-container">$\tilde q:E\to[0,\infty)$</span>. Moreover, you may assume that <span class="math-container">$\left\{p&gt;0\right\}\subseteq\left\{q(x,\;\cdot\;)&gt;0\right\}$</span> for all <span class="math-container">$x\in E$</span>.</p> https://stats.stackexchange.com/q/436084 1 Does a Metropolis-Hastings chain always obey a central limit theorem? 0xbadf00d https://stats.stackexchange.com/users/222528 2019-11-14T16:31:21Z 2019-11-15T09:28:40Z <p>Let <span class="math-container">$(E,\mathcal E,\lambda)$</span> be a measure space, <span class="math-container">$Q$</span> be a Markov kernel on <span class="math-container">$(E,\mathcal E)$</span> with <span class="math-container">$$Q(x,B)=\int_B\lambda({\rm d}y)q(x,y)\;\;\;\text{for all }(x,B)\in E\times\mathcal E$$</span> for some symmetric <span class="math-container">$q:E\times E\to[0,\infty)$</span>, <span class="math-container">$p$</span> be a probability density on <span class="math-container">$(E,\mathcal E,\lambda)$</span> and <span class="math-container">$\mu:=p\lambda$</span>.</p> <blockquote> <p>Let <span class="math-container">$\kappa$</span> denote the transition kernel of the chain <span class="math-container">$(X_n)_{n\in\mathbb N_0}$</span> generated by the Metropolis-Hastings algorithm with proposal kernel <span class="math-container">$Q$</span> and target distribution <span class="math-container">$\mu$</span>.</p> <p>I want to know what we need to assume (if anything) to ensure that for a given <span class="math-container">$f\in L_0^2(\mu):=\left\{h\in L^2(\mu):\mu h=0\right\}$</span>, there is a <span class="math-container">$\sigma^2(f)\ge0$</span> with <span class="math-container">$$\frac1{\sqrt n}\sum_{i=0}^{n-1}f(X_i)\xrightarrow{n\to\infty}\mathcal N(0,\sigma^2(f))\tag1.$$</span></p> </blockquote> <p>I know several conditions ensuring <span class="math-container">$(1)$</span>, e.g. that <span class="math-container">$f$</span> admits a solution <span class="math-container">$g\in L^2(\mu)$</span> of the Poisson equation <span class="math-container">$$f=(1-\kappa)g\tag2$$</span> or that <span class="math-container">$$\sum_{n=0}^\infty\left\|\kappa^nf\right\|_{L^2(\mu)}&lt;\infty\tag3$$</span> (in which case <span class="math-container">$g=\sum_{n=0}^\infty\kappa^nf$</span> is precisely the solution of <span class="math-container">$(2)$</span>.) For example, if <span class="math-container">$$c:=\sum_{n=0}^\infty\left\|\kappa^n\right\|_{\mathfrak L(L^2_0(\mu))}&lt;\infty,$$</span> then <span class="math-container">$(3)$</span> is satisfied (hence <span class="math-container">$(1)$</span> holds for all <span class="math-container">$f$</span>).</p> <blockquote> <p>I wonder, for example, whether <span class="math-container">$(2)$</span> or <span class="math-container">$(3)$</span> hold for all <span class="math-container">$f$</span> or if there are additional assumptions on <span class="math-container">$(E,\mathcal E,\lambda),p,q$</span> ensuring this.</p> </blockquote> <p>If necessary, assume that <span class="math-container">$(E,\mathcal E)$</span> is a <span class="math-container">$\mathbb R$</span>-vector space, <span class="math-container">$(E,\mathcal E,\lambda)$</span> is translation-invariant and <span class="math-container">$q(x,y)=\tilde q(x-y)$</span> for some <span class="math-container">$\mathcal E$</span>-measurable <span class="math-container">$\tilde q:E\to[0,\infty)$</span>. Moreover, you may assume that <span class="math-container">$\left\{p&gt;0\right\}\subseteq\left\{q(x,\;\cdot\;)&gt;0\right\}$</span> for all <span class="math-container">$x\in E$</span>.</p> <p><em>Remark</em>: Please take note of my related question: <a href="https://stats.stackexchange.com/q/436087/222528">Is the Metropolis-Hastings kernel always aperiodic, irreducible and geometrically ergodic?</a>.</p> https://stats.stackexchange.com/q/434689 0 Why is $\frac1n\sum_{i=0}^{n-1}\sum_{j=0}^i1_{\{\:X_i\:=\:Y_j\:\}}f(Y_j)$ an equivalent representation for the usual Metropolis-Hastings estimator? 0xbadf00d https://stats.stackexchange.com/users/222528 2019-11-05T16:18:07Z 2019-11-06T19:55:12Z <p>At the beginning of section 2 of the paper <a href="https://projecteuclid.org/download/pdfview_1/euclid.aos/1291388375" rel="nofollow noreferrer">A Vanilla Rao-Blackwellization of Metropolis-Hastings Algorithms</a>, the usual Metorpolis-Hastings estimator of <span class="math-container">$\int f$</span> given by the ergodic average <span class="math-container">$\frac1n\sum_{i=0}^{n-1}f(X_i)$</span> may equivalently be written as <span class="math-container">$\frac1n\sum_{i=0}^{n-1}\sum_{j=0}^i1_{\left\{\:X_i\:=\:Y_j\:\right\}}f(Y_j)$</span>, where <span class="math-container">$(X_n)_{n\in\mathbb N_0}$</span> is the chain generated by the algorithm, <span class="math-container">$(Y_n)_{n\in\mathbb N}$</span> is the corresponding sequence of proposals and <span class="math-container">$Y_0:=X_0$</span>.</p> <p>However, while the idea behind this equivalent representation is clear to me, I don't understand why it holds. Couldn't it be the case that, for a particular outcome <span class="math-container">$\omega$</span>, <span class="math-container">$X_i(\omega)=Y_j(\omega)=Y_k(\omega)$</span> and hence <span class="math-container">$f(Y_j(\omega))$</span> is mistakenly counted (at least) twice compared to the ergodic mean?</p> <hr> <p><strong>EDIT</strong>: Let's try to figure out if we can prove <span class="math-container">$$\operatorname P[\exists 1\le i&lt;j\le k:Y_{n_i}=Y_{n_j}]=0\;\;\;\text{for all }k\in\mathbb N\text{ and }1\le n_1&lt;\cdots&lt;n_k\tag1$$</span> as suggested in <a href="https://stats.stackexchange.com/a/434728/222528">Taylor's answer</a>. It would be sufficient to show that, given <span class="math-container">$1\le m&lt;n$</span>, <span class="math-container">$$\operatorname P\left[Y_m=Y_n\right]=0\tag2.$$</span> In order for this to make sense, we technically need to assume that <span class="math-container">$\Delta:=\left\{(x,x):x\in E\right\}\in\mathcal E^{\otimes2}$</span>, where <span class="math-container">$(E,\mathcal E)$</span> denotes the state space. We know that <span class="math-container">$$Z_k:=(X_{k-1},Y_k)\;\;\;\text{for }k\in\mathbb N$$</span> is a time-homogeneous Markov chain with transition kernel <span class="math-container">$$\kappa_{\text{aug}}((x,y),A\times B):=(1-\alpha(x,y))\delta_x(A)Q(x,B)+\delta_y(A)\alpha(x,y)Q(y,B)$$</span> for <span class="math-container">$x,y\in E$</span> and <span class="math-container">$A,B\in\mathcal E$</span>, where <span class="math-container">$\alpha$</span> denotes the acceptance function of the algorithm.. Thus, <span class="math-container">$$(Z_m,Z_n)\sim\mathcal L(Z_m)\otimes\kappa_{\text{aug}}\tag3$$</span> and <span class="math-container">$$\mathcal L(Z_m)\sim\mathcal L(X_{m-1})\otimes Q\tag4,$$</span> where <span class="math-container">$Q$</span> denotes the proposal kernel. Assume <span class="math-container">$Q$</span> and the target distribution <span class="math-container">$\mu$</span> have a density <span class="math-container">$q$</span> and <span class="math-container">$p$</span> with respect to a common reference measure <span class="math-container">$\lambda$</span>. For simplicity, let's focus on the case <span class="math-container">$n-m=1$</span>. Then, <span class="math-container">\begin{equation}\begin{split}&amp;\operatorname P\left[Y_{n-1}=Y_n\right]=\operatorname P\left[(Y_{n-1},Y_n)\in\Delta\right]\\&amp;\;\;\;\;=\operatorname P\left[X_{m-1}\in{\rm d}x_1\right]\int Q(x_1,{\rm d}y_1)\int\kappa_{\text{aug}}((x_1,y_1),{\rm d}(x_2,y_2))1_\Delta((y_1,y_2))\end{split}\tag5\end{equation}</span> and <span class="math-container">\begin{equation}\begin{split}&amp;\int\kappa_{\text{aug}}((x_1,y_1),{\rm d}(x_2,y_2))1_\Delta((y_1,y_2))\\&amp;\;\;\;\;=(1-\alpha(x_1,y_1))\int Q(x_1,{\rm d}y_2)1_\Delta((y_1,y_2))+\alpha(x_1,y_1)\int Q(y_1,{\rm d}y_2)1_\Delta((y_1,y_2))\end{split}\tag6\end{equation}</span> for all <span class="math-container">$x_1,y_1\in E$</span>.</p> <blockquote> <p>How can we show that <span class="math-container">$(5)$</span> (or maybe already <span class="math-container">$(6)$</span>) vanishes?</p> </blockquote> https://stats.stackexchange.com/q/421262 0 Is Autocorrelation of Posterior Samples always a problem in MCMC UmaN https://stats.stackexchange.com/users/28943 2019-08-08T14:24:19Z 2019-08-09T15:13:07Z <p>I am experimenting with MCMC methods and have implemented a basic Metropolis-Hastings algorithm.</p> <p>One potential issue with this is that MH posterior samples are autocorrelated. I could verify that mine were as well.</p> <p>This is often referred to as a known issue and various fixes are suggested such as "thinning" where only every N:th sample is used.</p> <p>My basic question is: why (or rather when) is this a problem?</p> <p>In my problem, I just want to get a good posterior distribution of a model parameter, e.g. density and moments.</p> <p>Even if the samples are autocorrelated, if I have enough of them, shouldn't I be fine?</p> <p>Let's say that I am not fine (as I am sure someone will explain):</p> <p>What if I run my chain until it converges, then grab 5000 samples and draw from them in a randomized fashion. Clearly the samples will then not be autocorrelated. Isn't this a better solution than "thinning"?</p> <p>In my experience when I used thinning I had to basically throw away 20 times the samples I actually used, causing excessive execution time.</p> <p>Any feedback is welcome.</p> https://stats.stackexchange.com/q/408331 1 Stationary Process Ergodicity Andrea https://stats.stackexchange.com/users/247895 2019-05-14T16:42:10Z 2019-05-14T17:39:36Z <p>Can you give me an example of a stationary nonergodic stochastic process that is time continuous? </p> https://stats.stackexchange.com/q/399356 0 What is Ergodic Variance KingJ https://stats.stackexchange.com/users/205608 2019-03-25T19:11:47Z 2019-03-26T12:22:17Z <p>I am curious as to the definition of ergodic variance in relation to an estimate of some parameter. It was mentioned to me by a teacher although I have not been able to find any references to it.</p> https://stats.stackexchange.com/q/397598 5 What does the distribution of samples from an MCMC method converge to without repeated samples? KiaSh https://stats.stackexchange.com/users/190214 2019-03-14T22:16:47Z 2019-03-15T10:14:51Z <p>Suppose I have an absolutely continuous distribution with density <span class="math-container">$f(x)$</span> and I use an mcmc sampler which has accept/reject step to sample from this distribution. In the final samples, there are some singular points (i.e., samples on top of each other). Does the distribution of the samples without these singular points converge to anything meaningful? </p> https://stats.stackexchange.com/q/391999 3 Do measurable maps preserve stationary ergodicity? J.Beck https://stats.stackexchange.com/users/235951 2019-02-11T20:39:25Z 2019-02-11T20:50:45Z <p>In a recent effort to establish stationary ergodicity for a certain stochastic process, I just happened to come across a statement, which I find to be little bit confounding.</p> <p>Given two measurable spaces <span class="math-container">$(\,E,\mathcal{E}\,)$</span> and <span class="math-container">$(\,\tilde{E},\tilde{\mathcal{E}}\,)$</span>, let <span class="math-container">$(v_t)_{\,t\in\mathbb{Z}}$</span> be a stationary ergodic sequence of <span class="math-container">$E$</span>-valued random elements and define a measurable function <span class="math-container">$f:{E}^\mathbb{N}\to \tilde{E}$</span>. Then the sequence <span class="math-container">$(\tilde{v}_t)_{\,t\in\mathbb{Z}}$</span> defined by <span class="math-container">$$\tilde{v}_t=f(v_t,v_{t-1},...)\qquad \text{for all }\quad t\in\mathbb{Z}$$</span> is stationary ergodic. (Straumann and Mikosch, 2006, p. 2455 - <a href="https://projecteuclid.org/euclid.aos/1169571804" rel="nofollow noreferrer">https://projecteuclid.org/euclid.aos/1169571804</a>)</p> <p>I do have to admit that I am somewhat irritated by this statement. In particular, let <span class="math-container">$f$</span> be prescribed by <span class="math-container">$$(v_t,v_{t-1},...)\,\mapsto\,\sum_{j=0}^\infty\rho^jv_{t-j}.$$</span> Then, I'd argue that <span class="math-container">$(\tilde{v}_t)_{\,t\in\mathbb{Z}}$</span> is not necessarily stationary (as would follow from the statement above). In fact, stationarity of <span class="math-container">$\tilde{v}_t$</span> would depend on whether <span class="math-container">$|\rho|&lt;1$</span> or not.</p> <p>I'd very much appreciate, if someone could tell me where I am going wrong, what I am missing, or whether the statement is not entirely accurate.</p> <p>For reference, the statement is supposed to build on proposition 4.3 (p. 26) in Ulrich Krengel's (1985) monograph <a href="https://books.google.com/books?hl=en&amp;lr=&amp;id=t4_BDh8gt2kC" rel="nofollow noreferrer"><em>Ergodic Theorems</em></a>.</p> https://stats.stackexchange.com/q/391108 0 Do we need ergodic-stationarity of the response variable in OLS spline regression? RScrlli https://stats.stackexchange.com/users/224433 2019-02-06T15:38:10Z 2019-02-06T15:43:41Z <p>I was wondering if we need the response variable to be ergodic stationarity when estimating an OLS spline regression. My intuition tells me that it's not needed but I would like to have a confirmation regarding it. Thanks in advance.</p> https://stats.stackexchange.com/q/386653 0 Prior/degree of belief/degree of lack-of-information/algorithms/complexity user70990 https://stats.stackexchange.com/users/0 2019-01-11T05:58:53Z 2019-01-11T07:16:29Z <p>For a long time I had a bit of difficulty understanding what "degree of belief" means.</p> <p>Recently I had some thoughts about it and I wonder if they make any sense, or is there some literature about this already.</p> <p>Basically I think about some algorithm that generates data. Our task is to figure out what is this algorithm given some data we observe. </p> <p>So prior is the information that we have so far have about the algorithm.</p> <p>Let's say that there are 100 different possible algorithms and one of them is the data generating algorithm. Our task is to figure out which algorithm has generated the data that we observed.</p> <p>From prior observations we can exclude 75 algorithms because they are not compatible with the data that has so far been observed.</p> <p>So in this sense, the prior is a representation of the already observed data. Sort of like an entropy/compression. If the prior is saying that we excluded 99 algorithms already then it means that the data that has been seen so far is compressed into the prior perfectly.</p> <p>This is a thought experiment, and I would like to read more about this interpretation of bayesian statistics. I wonder, is there some literature about this ? </p> <p>In other words, I would like to understand the bayesian view in terms of algorithms/information theory/compression/entropy/turing machines/ergodicity.</p> <p>I like the idea that entropy is a measure of the lack of information about a "system", but instead of system, I would rather say "algorithm" that generates data. </p> <p>There are two sources where uncertainty can come from:</p> <p>1) We cannot measure every degree of freedom (internal state) of the algorithm. </p> <p>2) The data we have observed is finite and it is not enough to pin down precisely which algorithm was generating the data we observed.</p> <p>So these are my intuitive understandings about bayesian view. Is there some literature about this already? Does this have some name? Where can I read about this interpretation more?</p> https://stats.stackexchange.com/q/385111 0 Stationary and ergodic r.v: relation between error and independent variable Albert https://stats.stackexchange.com/users/232535 2018-12-31T19:27:52Z 2018-12-31T19:27:52Z <p>In time series often hold the condition that a r.v. is stationary and ergodic, allowing the application of the law of large number. If in a model as: Y= a + bX + u where u is the error term, we assume that u is stationary and ergodic, this imply something about Y? also Y is stationary and ergodic? </p> https://stats.stackexchange.com/q/379989 2 Law of Large Numbers for Covariance Stationary Processes... Difference and Relationship between LLN and Ergodicity ColorStatistics https://stats.stackexchange.com/users/198058 2018-12-02T23:10:15Z 2019-05-26T12:35:57Z <p>We have a covariance stationary time series. We must assume that the time series was produced by an ergodic process if we are to make the bridge between the realization of the time series that we observe and the population that might have generated it. The assumption of an ergodic process amounts to assuming that the mean of our time series converges in probability to the population mean. </p> <p>The Law of Large Numbers for a covariance stationary process doesn't seem to add anything to the assumption of ergodicity we've made. Or does it?</p> <p>(1) <em>What does the Law of Large Numbers for a covariance stationary process give us that we did not already know, having assumed ergodicity?</em> (2) <em>Ergodicity is something untestable about the process that created the time series. The process is either ergodic or not, and we can't know it. Can we think of the LLN for a process as a statement about conditions under which LLN holds, hence a statement about conditions when we can expect a process to be ergodic?</em></p> https://stats.stackexchange.com/q/362279 1 Magnitude of non-ergodicity effect on the individual's risk of bankruptcy Antoni Parellada https://stats.stackexchange.com/users/67822 2018-08-15T05:53:42Z 2018-08-15T14:01:44Z <p><a href="https://youtu.be/f1vXAHGIpfc" rel="nofollow noreferrer">Dr. Ole Peters presents the concept of (non-)ergodicity with the following gambling example</a>:</p> <p>You're given $\$100$to play a game where you toss a coin once a minute. If it comes up heads, you win$50\%$of your wager, while you lose$40\%$if tails. Averaging over$1$million sequences (or players) results in the plot on the left below, which shows this is a favorable game:</p> <p><a href="https://i.stack.imgur.com/j8Dnd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/j8Dnd.png" alt="enter image description here"></a></p> <p>This is the ensemble perspective, or the idea of parallel worlds, which do not apply to the individual, who can go broke (<a href="https://en.wikipedia.org/wiki/Ergodicity#Markov_chains" rel="nofollow noreferrer">absorbing states</a>), and drop out of the game. Considering longer time sequences, the average over time ($1$year) will also get rid of the noise, and result in the plot to the right above.</p> <p>I am having problems reproducing these plots in R. The idea is there, but the drop is much more rapid in the time perspective plot:</p> <p><a href="https://i.stack.imgur.com/CxVOK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CxVOK.png" alt="enter image description here"></a></p> <blockquote> <p>Is there a much more rapid bankruptcy tendency for the individual player than led to believe by the plots on the talk? Or is it simply differences in the frequency and length of sampling in the original code from the lecture, which is not shared, as for example, less sampling points along the time line?</p> </blockquote> <pre><code># ENSEMBLE PATTERN: sam = 60 # Sixty samples (1 per minute) mat = matrix(0, sam, 10^6) # The experiment is repeated 1 million times for (i in 1:10^6){ mat[1,i] = 100 # Each time the money wagered is$100 for(j in 2:(sam-1)){ # H's wins 50% tails loses 40% mat[j,i] &lt;- mat[j - 1,i] * sample(c(1.5,0.6), 1) } } plot(rowMeans(mat), type='l', main = 'Ensemble perspective', ylab='Dollars', xlab='Trials', col = 3) # TIME PATTERN: set.seed(0) s = 100 # Initial money gambled samples = 60 * 24 * 30 * 12 # Samples at 1 per minute during 12 months vec = 0 # Empty vector vec = 100 # Starting at $100 for (i in 2:samples) vec[i] &lt;- vec[i - 1] * sample(c(1.5,0.6), 1) plot(vec, type='l', main = 'Time perspective', ylab='Dollars', xlab='Trials', col = 2) </code></pre> https://stats.stackexchange.com/q/349348 1 How do I create an iid Rademacher sequence? eBopBob https://stats.stackexchange.com/users/187201 2018-06-01T12:49:30Z 2018-06-01T15:38:14Z <p>The lecture notes say:</p> <p>Let$(\Omega,\mathcal{A},P) = ((0,1],\mathcal{B}((0,1]),\lambda)$where$\lambda$is the Lebesgue measure on the unit interval.</p> <p>Define$X(\omega) = 1$for$\omega &gt; 1/2$and$X(\omega) = - 1$for$\omega \le 1/2$. Then$T(\omega) = 2\omega - [2\omega]$.</p> <p>It can be shown that$X_{k} = \pm 1$with prob.$1/2$is an iid Rademacher sequence.</p> <p>Can someone please outline how this is possible?</p> https://stats.stackexchange.com/q/344937 5 Ergodicity explained in layman terms WhiteGlove https://stats.stackexchange.com/users/207374 2018-05-07T15:54:35Z 2018-05-09T22:16:09Z <p>I've been told that Ergodicity gives us a practical vision of processes WSS (Wise-sense stationary) and a bunch of integrals. For me, it is not enough to fully understand it.</p> <p><strong>Could someone explain me Ergodicity in a simple way?</strong></p> <p><strong>EDIT:</strong></p> <p>Thank you all for those interested in the question and answered, here i will share an example:</p> <p>y(t) is a random process where {i(t),q(t)} are two random stationary processes, incorrelated, null mean and autocorrelation Ri(z) = Rq(z).</p> <p><strong>y(t) = i(t)cos(2πf0t)−q(t)sin(2πf0t)</strong></p> <p>The exercise asks for mean and autocorrelation of y(t) and finally if that process is stationary or cyclostationary.</p> <p>I have resolved that already but, what about ergodicity. </p> <p><strong>Is this process ergodic?</strong> How could i demonstrate such thing?</p> https://stats.stackexchange.com/q/343563 4 Stationary Distribution of Multiplicative Autoregressive Model Shanks https://stats.stackexchange.com/users/160116 2018-04-30T09:53:46Z 2018-04-30T16:18:56Z <p>I know for the additive autoregressive model the stationary distribution of$\{X_t\}can be found, if it exists, in the following way: \begin{align} X_t &amp;= \alpha X_{t-1} + \epsilon_t\\ \Rightarrow X_t &amp;= (1-\alpha B)^{-1} \epsilon_t \\ &amp;= \epsilon_t + \alpha \epsilon_{t-1} + \cdots. [\text{if}~ -1&lt;\alpha&lt;1] \end{align} Now ife_t \stackrel{\text{iid}}{\thicksim} N(0, 1)$, say, one can easily find the distribution of$X_t$. </p> <p><strong>My question:</strong> What will be the stationary distribution, if it exists, of the multiplicative autoregressive model$X_t = \alpha X_{t-1}\epsilon_t$? What will be the condition(s) of$\alpha$in this case? It will also be helpful if some hints are suggested to how it can be generalized to the higher order models.</p> https://stats.stackexchange.com/q/325536 1 Is strict stationarity a sufficient condition for ergodicity? M. Hansen https://stats.stackexchange.com/users/193204 2018-01-28T23:31:30Z 2018-03-13T22:48:51Z <p>For a given time series, <strong>is strict stationarity a sufficient condition for ergodicity?</strong> I am wondering if it isn't also sufficient for a time series to be <strong>weakly stationary</strong> because then the mean is a constant. I can't see why the variance shouldn't be allowed to vary over time in order for the time series to be ergodic. </p> https://stats.stackexchange.com/q/325015 3 How are ergodicity and "weak dependence" related? shenflow https://stats.stackexchange.com/users/182258 2018-01-25T11:29:20Z 2019-03-27T02:34:37Z <p>I understand that <strong><em>weak dependence</em></strong> is a broad concept, the definition I am referring to is the one <em>Wooldridge</em> (2013) uses as an assumption that has to be fulfilled (amongst other assumptions) so that the estimators in a time series linear regression model are asymptotically consistent. That is:</p> <blockquote> <p>A process$\{X_t\}$is weakly dependent if the correlation between$\{X_t\}$and$\{X_{t+h}\}$goes to zero relatively quickly as$h\to \infty$.</p> </blockquote> <p>Instead of saying that stochastic processes have to be covariance stationary and weakly dependent, other authors say they have to be covariance stationary and <strong><em>ergodic</em></strong>. </p> <p>How are <strong><em>ergodicity</em></strong> and <strong><em>weak dependence</em></strong> related? Are they interchangable in the context of time series OLS assumptions?</p> <p>Thank you.</p> https://stats.stackexchange.com/q/319190 1 Wide-Sense Stationary but not ergodic Clarinetist https://stats.stackexchange.com/users/46427 2017-12-16T21:20:11Z 2017-12-16T22:04:18Z <p>This is based on example 2.2 from <em>Machine Learning: A Bayesian and Optimization Perspective</em> by Theodoridis. Please note that I'm not at all familiar with ergodic theory and I'm reading this with background from Casella and Berger.</p> <p>A wide-sense stationary process (WSS), by definition, is a stochastic process$(u_n)_{n \in \mathbb{Z}}$such that for each$k$,$\mathbb{E}[u_k] = \mu$for all$k$(i.e., the means are identical) and if we define $$r(n, m) = \mathbb{E}[u_nu_m]$$ we have$r(n, n-k) = r(k)$(i.e., the "autocorrelation" (what Theodoridis defines as such, technically the cross-product moment)) only depends on the difference between times.</p> <p>Consider a WSS$(u_n)_{n \in \mathbb{Z}}$for which the expected value is$\mu$at each time, and $$\mathbb{E}[u_nu_{n-k}] = r_u(k)\text{.}$$ Let$v_n = au_n$, where$a$is a random variable taking values in$\{0, 1\}$, each with probability$0.5$. Assuming independence of$a$and$u_n$, we obtain$\mathbb{E}[v_n] = 0.5\mu$and$\mathbb{E}[v_nv_{n-k}] = 0.5r_u(k)$.</p> <p>Theodoridis then remarks:</p> <blockquote> <p>Thus,$v_n$is WSS. However, it is not covariance-ergodic. Indeed, some of the realizations will be equal to zero (when$a = 0$), and the mean value and autocorrelation, which will result from them as time averages, will be zero, which is different from the ensemble averages.</p> </blockquote> <p><strong>I don't understand any of the above quote, other than that$v_n$is WSS</strong>. By "covariance-ergodic," the author talks about $$\text{cov}(k) = \lim_{N \to \infty}\dfrac{1}{2N+1}\sum_{n=-N}^{N}(u_n - \mu)(u_{n-k}-\mu)$$ with limits "in the mean-square sense" equaling$0$(I think?).</p> <p>Could someone enlighten me on this?</p> https://stats.stackexchange.com/q/311816 2 Examples of ergodic process JetLag https://stats.stackexchange.com/users/179171 2017-11-04T05:02:35Z 2017-11-04T09:55:33Z <p>An ergodic process is a process in which the structures of <em>inter-individual variation</em> and <em>intra-individual variation</em> are asymptotically equivalent <a href="http://www.tandfonline.com/doi/abs/10.1207/s15366359mea0204_1" rel="nofollow noreferrer">(Molenaar, 2004)</a>.</p> <p>In other words: A process is <strong>non-ergodic</strong> in case results of analysis of inter-individual variation do not generalize to the level of intra-individual change in time, and vice versa.</p> <p>Equivalently, a process is <strong>ergodic</strong> in case results of analysis of inter-individual variation validly generalize to the level of intra-individual change in time, and vice versa.</p> <p>I tried to google it but I couldn't find satisfactory and intuitive and easy to understand examples of this process. I'm wondering if anyone could come up with an example to contrast these two processes?</p> https://stats.stackexchange.com/q/311609 1 Is the matrix$\hat X_k\hat X^T_k$ergodic? ZHUANG https://stats.stackexchange.com/users/183283 2017-11-03T02:33:27Z 2017-11-05T06:08:48Z <p>Let$X_k = A_{k-1}X_{k-1} + \omega_{k-1}$, and$Y_k = H_{k} X_k +\eta_k$, where$\omega_k$and$\eta_k$are i.i.d. Gaussian random variables. We now can get unbiased estimate of the random variable$X_k$through$\hat X_k = (H_k)^{\dagger}Y_k$assuming$H_k$has full column rank, where$(H_k)^\dagger$represent a pseudo inverse. </p> <p>Q: Is the matrix$\hat X_k\hat X^T_k$ergodic?</p> <p>It feels like a no. but the simulations tell me the time averages are still good estimates of the ensemble averages.</p> https://stats.stackexchange.com/q/306681 2 Determine whether a time series is ergodic user179860 https://stats.stackexchange.com/users/179860 2017-10-06T22:19:04Z 2017-10-06T22:19:04Z <p>How to determine if this time series is ergodic? According to the ergodic theorem the mean must converge to the expected value starting point. How do I determine that? Hoverever I have plotted the mean observations for the last 500 days, and it seems that it doesn't converge so I can conclude it is not ergodic? <a href="https://i.stack.imgur.com/EQWWN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EQWWN.png" alt="enter image description here"></a></p> https://stats.stackexchange.com/q/301315 2 Markov chain which is also a projection Ted https://stats.stackexchange.com/users/103653 2017-09-04T11:14:41Z 2017-09-04T14:59:06Z <p>Let$P$be the transition matrix of a Markov chain, and assume that$P^2=P$. One immediate conclusion is that$P=P^\infty$.</p> <p>Furthermore, assume that there is a state$i$such as each state$j$(including$j=i$), is accessible from$i$:$i\rightarrow j$.</p> <p>I am fairly sure that this means that$P$is irreducible (and thus, has only one stationary distribution, and all the columns of its matrix are equal), but I can't find a quick argument why this is true.</p> <p>I think I can get a proof using the coefficients, to show that if there is a state$j$such as$j\not\rightarrow i$, then$P^\infty\neq P$, because the probability of being in state$j$will "grow over time", but I was wondering whether there was a more direct argument, perhaps making use of a standard result.</p> https://stats.stackexchange.com/q/283715 5 Why is ergodicity not a requirement for ARIMA models besides stationarity? JTicker https://stats.stackexchange.com/users/163387 2017-06-05T19:18:21Z 2017-06-06T12:33:46Z <p>I frequently read that ARIMA models must be fitted on stationary data. But stationarity does not ensure ergodicity, which I understand is necessary to deduce population parameters from a single time series sample. Why is ergodicity not a requirement for ARIMA modeling? Do we just assume it?</p> <p>Also, is there an example of a ergodic, but non-stationary process? Can you forecast these types of series?</p> https://stats.stackexchange.com/q/282206 0 Within-chain and between-chain average difference in MCMC simulation Will https://stats.stackexchange.com/users/155928 2017-05-28T19:15:17Z 2017-05-28T19:15:17Z <p>I have done 400 repetitions of a particular MCMC simulation (Metropolis–Hastings algorithm) to get a quantity of interest$N$. The simulation reaches its steady-state after ~$10^5$iterations. The typical correlation length is around ~$10^5$iterations too but to be on the safe side, I have used a burn-in time of$10^6$iterations and I let the simulation run until I reach$10^7$iterations.</p> <p>I have calculated the average value for$N$using two methods:</p> <ul> <li>By calculating the average value at the$10^{7\text{th}}$<strong>between</strong> the 400 independent chains.</li> <li>And by calculating the average value <strong>within</strong> each chain (all values between iteration$10^6$and iteration$10^7$) </li> </ul> <p><strong>Problem:</strong> I have performed a few checks (normality, t-tests, etc.) but the two values of$N$are still significantly different (one is 8 times bigger).</p> <p>Would someone know why this could be the case? I suppose this has to do with ergodicity, but I am clearly not sure of it because I cannot see why.</p> https://stats.stackexchange.com/q/276583 2 Calculate of Lyapunov Exponents of a sequence of random matrices Diego Fonseca https://stats.stackexchange.com/users/150865 2017-04-29T03:58:26Z 2017-04-29T03:58:26Z <p>Let$(\Omega,\mathcal{F},\mathbb{P}):=(M^{\mathbb{N}_{0}},\mathcal{M}^{\mathbb{N}_{0}},\mathbb{P})$be a probability space where$M=\left\{0,1,2,3,4\right\}$,$\mathcal{M}^{\mathbb{N}_{0}}$is product$\sigma$-algebra. Note that if we consider the cylinder$I=\left[x_{0},x_{1},\ldots,x_{k}\right]$defined by \begin{equation}I=\left[x_{0},x_{1},\ldots,x_{n}\right]:=\left\{\mathbf{y}=(y_{i})_{i\in\mathbb{N}}\in M^{\mathbb{N}_{0}} \left|y_{i}=x_{i} \mbox{ para }i=0,1,\ldots,n \right. \right\}.\end{equation} Then <strong>the collection of all cylinders form a semi-algebra that we called$\mathcal{S}$</strong>, futhermore, by measure theory we know that \begin{equation}\mathcal{M}^{\mathbb{N}_{0}}=\sigma(\mathcal{S}).\end{equation} In this sense, given a fixed transition matrix$\Pi$of size$5\times 5$and your invariant measure$\pi$we define$\mathbb{P}$in$S$as $$\mathbb{P}(I):=\pi(x_{0})\Pi(x_{0},x_{1})\Pi(x_{1},x_{2})\cdots \Pi(x_{k-1},x_{k}).$$ </p> <p><strong>The question:</strong> In this context, let the stationary sequence$(Y_{n})_{n\in\mathbb{N}}$defined by$Y_{n}:=Y_{0}\circ \varphi^{n} $where$\varphi$is the 1-left shift (i.e.$\varphi(x_{0},x_{1},x_{2},x_{3},\ldots)= (x_{1},x_{2},x_{3},x_{4},\ldots)$) and$Y_{0}:=\mathrm{Proy}_{0}$(i.e.$\mathrm{Proy}_{0}(x_{0},x_{1},x_{2},x_{3},\ldots)=x_{0}$). We define the random matrix $$B_{n}:=\left(\begin{array}{cc} Y_{n} &amp; 0 \\ 0&amp; 2\end{array}\right).$$ and $$A_{n}:=B_{n}B_{n-1}\cdots B_{1}B_{0}.$$ Determine the Lyapunov exponents$\lambda_{1}$and$\lambda_{2}$of$(A_{n})_{n\in\mathbb{N}}$. [Note: Consider$Y_{0}\sim \pi $]</p> <p><strong>My attempt:</strong> Considering the <strong>Theorem 6.15</strong> (Furstenberg-Kesten) in this <a href="https://www.mathematik.hu-berlin.de/~fiebig/paper_Imkeller/stochastic_dynamics.pdf" rel="nofollow noreferrer">link</a> we should calculate$\widehat{\Omega}$, I think that $$\widehat{\Omega}:=\left\{\mathbf{y}=(y_{n})_{n\in\mathbb{N}}\in M^{\mathbb{N}} | y_{i}=0 \mbox{ for some }i\right\}.\tag{1}$$ In this sense, we know that the singular values of$A_{n}$are $$\delta_{1}(A_{n})=2^{n+1} \qquad \mbox{ and }\qquad \delta_{2}(A_{n})=Y_{n}Y_{n-1}\cdots Y_{1}Y_{0}.$$ Then, for any$\omega\in \widehat{\Omega}$the Lyapunov exponents are $$\Lambda_{1}(\omega)=\lim_{n\rightarrow \infty} \frac{1}{n} \log\delta_{1}(A_{n})=\log(2)$$ and for$n$large we have</p> <p>$$\Lambda_{2}(\omega)=\lim_{n\rightarrow \infty} \frac{1}{n} \log\delta_{2}(A_{n})=\lim_{n\rightarrow \infty} \frac{1}{n} \log\left(Y_{n}(\omega)Y_{n-1}(\omega)\cdots Y_{1}(\omega)Y_{0}(\omega)\right)=\lim_{n\rightarrow \infty} \frac{1}{n} \log\left(0\right)=-\infty.$$</p> <p><strong>The problem is show that$\widehat{\Omega}$in (1) is the given in Theorem 6.15.</strong></p> https://stats.stackexchange.com/q/270819 4 Metropolis Hastings with estimated posterior Till Hoffmann https://stats.stackexchange.com/users/17643 2017-03-30T14:52:15Z 2017-04-01T12:52:27Z <p>I am interested in samples of$\theta$from the posterior distribution</p> <p>$$P(\theta|x) = \int d\phi P(\theta|\phi)P(\phi|x)$$</p> <p>where$x$are data and$\phi$are nuisance parameters. In principle, I can use a Metropolis Hastings sampler to sample from$\theta$and$\phi$and discard the samples of the nuisance parameter.</p> <p>In this case, I can sample from$P(\phi|x)$directly such that I can approximate the marginal posterior by Monte Carlo integration</p> <p>$$P(\theta|x)\approx\frac{1}{n}\sum_{i=1}^n P(\theta|\phi_i)\equiv \hat P,$$</p> <p>where$\phi_i$are samples from$P(\phi|x)$. Of course, the approximation$\hat P$is not deterministic because of the sampling error. I seem to remember that running the following algorithm samples from the posterior in this case but I can no longer find the reference. Is this correct? If so, do you have a reference?</p> <pre><code># Sampler in pseudo-python theta = initial_value for step in range(num_steps): # Sample phi phi = sample_phi_given_x(x) # Evaluate current posterior posterior_estimate = mean(posterior(theta, phi)) # Sampled from proposal and evaluate posterior at proposal candidate = sample_from_proposal(theta) posterior_candidate_estimate = mean(posterior(candidate, phi)) # Accept or reject the proposal if random_uniform() &lt; posterior_candidate_estimate / posterior_estimate: theta = candidate </code></pre> https://stats.stackexchange.com/q/269497 3 Robert Casella Independence Sampler Result user2379888 https://stats.stackexchange.com/users/45729 2017-03-24T03:03:41Z 2017-03-26T15:14:40Z <p>In the second edition of the Robert &amp; Casella book (<a href="http://amzn.to/2lQDmJR" rel="nofollow noreferrer">Monte Carlo Statistical Methods</a>), the authors have a result, Theorem 7.8, on the independent Metropolis-Hastings sampler: Letting$f$be the density of the target measure an$g$the density of the proposal, if a minimization condition is satisfied$f(x)\leq M g(x)$for all$x$in the support of$f$, then the chain is uniformly ergodic and $$\| K^n(x,\cdot) - f\|_{TV} \leq 2 (1-1/M)^n$$</p> <p>The proof begins with an odd result (equation 7.9), \begin{equation} \|K(x,\cdot) -f\|_{TV} = 2 \sup_{A} |\int_{A} K(x,y) - f(y) dy|. \end{equation} Where is this factor of 2 coming from? Ignoring the factor of 2 for a moment, the authors then obtain \begin{equation} \|K(x,\cdot) -f\|_{TV} \leq 2 (1- 1/M), \end{equation} which I agree with, except, again, for the factor of 2. They then say that, by induction, one can obtain the result.</p> <p>The best that I can obtain is $$(2\|K^n(x, \cdot) - f\|_{TV})\leq (2\|K(x, \cdot) - f\|_{TV})^n \leq (2(1-1/M))^n,$$ which is based on a computation in Roberts &amp; Rosenthal (2004) in Probability Surveys.</p> <p>This calculation has been driving me nuts.</p> https://stats.stackexchange.com/q/254761 4 Ergodic theorem Rhafi Sheikh https://stats.stackexchange.com/users/86190 2017-01-05T18:44:56Z 2017-01-06T07:33:44Z <p>We know that,</p> <p>If$p_{jj}$is the transition probability of staying in state j at n-th step and j at (n-1)-th step then the state j is said to be recurrent if,</p> <p>$\sum_{n=0}^\infty p_{jj}^n = \infty$</p> <p>and transient if,</p> <p>$\sum_{n=0}^\infty p_{jj}^n \lt \infty$.</p> <p>My problem of understanding is that if j is a recurrent state then the probability of returning to state j from state j in one step is$p_{jj}^1$,in two step is$p_{jj}^2$and .......so on. So the probability of ever returning to the state j is the sum of the transition probabilities of returning to state j from j where n=1,2,......$\infty\$.These probabilities are the probabilities of mutually events. So ,here how the probability of occurring an event is infinity?</p> <p>I know the proof of the theorem but i don't understand the intuition behind the theorem and why the sum of probability is equal to infinity,if the state is recurrent?.</p> <p>Can someone explain the theorem intuitively? </p>