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5

You are looking at a truly tiny simulation. Here's one that went out to $n=2^{1000} \approx 1.07\times 10^{301}:$ (In order to plot it I thinned the walk to $999$ values equally spaced horizontally and connected them with line segments. The actual simulation is, of course, much more detailed than can be shown here :-).) Clearly this walk repeatedly hits (...


0

The following demonstrates that when $(X_t)$ converges in probability to zero, it is not necessarily the case that the running mean also converges in probability to zero. Set $X_t=0$ with probability $\frac{t-1}{t}$, and $X_t=2t$ with probability $\frac{1}{t}$. Clearly, $(X_t)$ converges in probability to zero. Let $Y_t=\frac{1}{t}\sum_{i=1}^tX_i$ be the ...


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Indeed, one may, for example, establish results of the following type: If $\{z_i\}$ is an uncorrelated sequence with expected value $\mu<\infty$ and \begin{equation} \sum_{i=1}^{\infty}i^{-2}\sigma_i^2<\infty, \end{equation} it holds that $\bar{z}\to_p\mu$. The result follows from Kronecker's Lemma, which says: If some positive real sequences $\{a_i\...


2

If $\mathbb{E}[X_i]=\mu<0$, with $\text{var}(X_i)=\sigma^2$, then \begin{align}\mathbb P(\bar X_n \ge 0)&=\mathbb P(\bar X_n \ge \mu-\sqrt{n}\mu\sigma/\sqrt{n}\sigma)\\ &=\mathbb P\left(\bar X_n \ge \mu-\frac{\sqrt{n}\mu}{\sigma}\underbrace{\frac{\sigma}{\sqrt{n}}}_{\text{sd}(\bar X_n)}\right)\\ &=\mathbb P\left(\frac{\sqrt{n}}{\sigma}\{\bar ...


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Yes, I would not trust the inferences (unless the higher R-hat were for the log-kernel, lp__, or perhaps for a generated quantity that does not go into the log-kernel). This question was considered in a paper by Jeff Gill whose abstract is Increasingly, political science researchers are turning to Markov chain Monte Carlo methods to solve inferential ...


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