Chapter 6: Problem 5
Let \(Y_{(1)}<\cdots
Short Answer
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chapter 6: Problem 5
Let \(Y_{(1)}<\cdots
These are the key concepts you need to understand to accurately answer the question.
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Get started for freeConsider a Poisson process of intensity \(\lambda\) in the plane. Find the distribution of the area of the largest disk centred on one point but containing no other points.
Classify the states of Markov chains with transition matrices $$ \left(\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ \frac{1}{2} & \frac{1}{2} & 0 \end{array}\right),\left(\begin{array}{llll} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right), \quad\left(\begin{array}{cccccc} \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 \\ \frac{1}{4} & \frac{3}{4} & 0 & 0 & 0 & 0 \\ \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & \frac{1}{4} & 0 & 0 \\ \frac{1}{4} & 0 & \frac{1}{4} & \frac{1}{4} & 0 & \frac{1}{4} \\ 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \end{array}\right). $$
Show that strict stationarity of a time series \(\left\\{Y_{j}\right\\}\) means that for any \(r\) we have $$ \operatorname{cum}\left(Y_{j_{1}}, \ldots, Y_{j_{r}}\right)=\operatorname{cum}\left(Y_{0}, \ldots, Y_{j_{r}-j_{1}}\right)=\kappa^{j_{2}-j_{1}, \ldots, j_{r}-j_{1}} $$ say. Suppose that \(\left\\{Y_{j}\right\\}\) is stationary with mean zero and that for each \(r\) it is true that \(\sum_{u}\left|\kappa^{u_{1}, \ldots, u_{r-1}}\right|=c_{r}<\infty\) The \(r\) th cumulant of \(T=n^{-1 / 2}\left(Y_{1}+\cdots+Y_{n}\right)\) is $$ \begin{aligned} \operatorname{cum}\left\\{n^{-1 / 2}\left(Y_{1}+\cdots+Y_{n}\right)\right\\} &=n^{-r / 2} \sum_{j_{1}, \ldots, j_{r}} \operatorname{cum}\left(Y_{j_{1}}, \ldots, Y_{j_{r}}\right) \\ &=n^{-r / 2} \sum_{j_{1}=1}^{n} \sum_{j_{2}, \ldots, j_{r}} \kappa^{j_{2}-j_{1}, \ldots, j_{r}-j_{1}} \\ &=n \times n^{-r / 2} \sum_{j_{2}, \ldots, j_{r}} \kappa^{j_{2}-j_{1}, \ldots, j_{r}-j_{1}} \\ & \leq n^{1-r / 2} \sum_{j_{2}, \ldots, j_{r}}\left|\kappa^{j_{2}-j_{1}, \ldots, j_{r}-j_{1}}\right| \leq n^{1-r / 2} c_{r} \end{aligned} $$ Justify this reasoning, and explain why it suggests that \(T\) has a limiting normal distribution as \(n \rightarrow \infty\), despite the dependence among the \(Y_{j}\). Obtain the cumulants of \(T\) for the MA(1) model, and convince yourself that your argument extends to the \(\mathrm{MA}(q)\) model. Can you extend the argument to arbitrary linear combinations of the \(Y_{j} ?\)
Consider two binary random variables with local characteristics $$ \begin{aligned} &\operatorname{Pr}\left(Y_{1}=1 \mid Y_{2}=0\right)=\operatorname{Pr}\left(Y_{1}=0 \mid Y_{2}=1\right)=1 \\ &\operatorname{Pr}\left(Y_{2}=0 \mid Y_{1}=0\right)=\operatorname{Pr}\left(Y_{2}=1 \mid Y_{1}=1\right)=1 \end{aligned} $$ Show that these do not determine a joint density for \(\left(Y_{1}, Y_{2}\right) .\) Is the positivity condition satisfied?
A Poisson process of rate \(\lambda(t)\) on the set \(\mathcal{S} \subset \mathbb{R}^{k}\) is a collection of random points with the following properties (among others): \- the number of points \(N_{\mathcal{A}}\) in a subset \(\mathcal{A}\) of \(\mathcal{S}\) has the Poisson distribution with mean \(\Lambda(\mathcal{A})=\int_{\mathcal{A}} \lambda(t) d t\) \- given \(N_{\mathcal{A}}=n\), the positions of the points are sampled randomly from the density \(\lambda(t) / \int_{\mathcal{A}} \lambda(s) d s, t \in \mathcal{A}\) (a) Assuming that you have reliable generators of \(U(0,1)\) and Poisson variables, show how to generate the points of a Poisson process of constant rate \(\lambda\) on the interval \(\left[0, t_{0}\right]\). (b) Let \(t=(x, y) \in \mathbb{R}^{2}, \eta, \xi \in \mathbb{R}, \tau>0, \lambda(x, y)=\tau^{-1}\\{1+\xi(y-\eta) / \tau\\}^{-1 / \xi-1}\). Give an algorithm to generate realisations from the Poisson process with rate \(\lambda(x, y)\) on $$ \mathcal{S}=\\{(x, y): 0 \leq x \leq 1, y \geq u, \lambda(x, y)>0\\}. $$ $$ \begin{array}{rrrrrrrrrrrrrrrrr} \hline 9 & 12 & 11 & 4 & 7 & 2 & 5 & 8 & 5 & 7 & 1 & 6 & 1 & 9 & 4 & 1 & 3 \\ 3 & 6 & 1 & 11 & 33 & 7 & 91 & 2 & 1 & 87 & 47 & 12 & 9 & 135 & 258 & 16 & 35 \\\ \hline \end{array} $$
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