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I recently have worked on compact convex sets in the context of time series and my question is related to that. If we have a set $$ C=\{\beta_1 X_1 +\beta_2 X_2 +\phi H_1 , |\beta_1|+|\beta_2|+|\phi|\leq K\} $$ where $X_1$ and $X_2$ are independent and $H_1$ consists of dependent columns(and correlated to $X_1$ and $X_2$). Also $X$ and $H_1$ are bounded. Then, under which conditions can we say that C is a compact and convex set?

TPArrow
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  • Almost never, I guess, because it looks like it is open in the finite dimensional subspace spanned by $X_1,X_2,H_1$. – Johannes Hahn May 28 '14 at 21:35
  • Dear @JohannesHahn, thank you for your reply. I know that in the situation where there are just $X_1$ and $X_2$, the set C is convex and compact. So what makes C so different if there exists H? – TPArrow May 28 '14 at 22:03
  • Alright, I was vague. Let me be more clear: The set you've defined is never compact in any hausdorff vector space topology. When you say you have proved it, you must have another topology. What topology is that? – Johannes Hahn May 29 '14 at 08:17
  • Thank you @JohannesHahn. I have not proved that, it is written in the paper: http://arxiv.org/pdf/1303.5817v4 on page 6 at the top of the page. – TPArrow May 29 '14 at 12:13
  • How about using weak inequality in your definition, instead of strict? Is that what you mean? What are $X_1,X_2$ and $H$? Vectors? – Per Alexandersson May 29 '14 at 18:04
  • Thank you @PerAlexandersson. $X_1$, $X_2$ are random variables. precisely $H$ consist of lags of response variable $Y$. Then the regression model is like $Y_t=f(X_t,Y_{t-1,...,t_n}$. Then $X_1$, $X_2$ and $H$ are vectors. – TPArrow May 29 '14 at 19:16
  • @HAMEDHM That is totally different! In that paper the inequality is not strict and of course that set is compact because it is the continuous image of a compact set. Here you are asking about a different set! – Johannes Hahn May 29 '14 at 19:38
  • Thank you @JohannesHahn I have corrected the mistake. Could you please explain it more? – TPArrow May 29 '14 at 19:58
  • There is really not much to explain... ${(\beta_1,\beta_2,\phi)\in\mathbb{R}^3 \mid |\beta_1|+|\beta_2|+|\phi|\leq K}$ is compact and your set is an image of this set under a continuous map. – Johannes Hahn May 29 '14 at 20:49
  • thank you @JohannesHahn. What about Convexity? – TPArrow May 29 '14 at 21:03
  • The set Johannes defined in the comment above is compact and convex. $C$ is an image of it under a linear transformation. Linear transformations preserve convexity and compactness. – Sasho Nikolov May 30 '14 at 02:51
  • Sorry, but I am confused. Firstly, is the form that @JohannesHahn has defined is equivalent to the form that I defined? secondly, is it compact and convex with respect to $\beta$'s and $\phi$ or with respect to $X$'s and $H$? third, if assume that $\phi$ consists of $\phi 1, \phi 2$ (and then H consists of two columns), then, is the results the same as above? fourth, I know it is very easy for you but is there any formal proof for convexity and compactness in current case? – TPArrow May 30 '14 at 06:56

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