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My rather basic question is related to an example from Spanos (1986, p.41), which I quote (verbatim) below.

Let $S$ be the real line $\mathbb{R} = \{x : -\infty<x<\infty\}$ and the set of events of interest be $$ J = \{B_{x} : x \in \mathbb{R}\} \quad \text{where } B_{x} = \{z : z \leqslant x\} = (\infty, -x]. $$

I am wondering if there is a typing error and that $B_{x}$ is, in fact, equal to $(-\infty, x]$?

My understanding is that $J$ is the set of all elements of the real numbers for which it is true that $z$ is less than or equal to the real number $x$.

So, my question is to ask for clarification around the notation $(\infty, -x]$. Is this correct or is it even legitimate interval notation? Perhaps it's a typo or that I am ignorant to some convention?

Thanks!

Reference:

Spanos, A (1986) Statistical foundations of econometric modelling. Cambridge University Press

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    $\begingroup$ Certainly a typo. $\endgroup$
    – Guest
    Dec 29, 2015 at 4:31
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    $\begingroup$ Yes, it's definitely a misplaced minus sign. $\endgroup$ Dec 29, 2015 at 5:38

1 Answer 1

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So, my question is to ask for clarification around the notation $(∞,−x]$. Is this correct or is it even legitimate interval notation? Perhaps it's a typo or that I am ignorant to some convention?

It is definitely just a misplaced minus sign.   It's not what was meant at all.

$B_x \mathop{\,:=\,} (-\infty; x]$ is the leftward ray with a maximum of $x$.   It is the set of real points less than or equal to $x$.   That is $B_x=\{z: -\infty < z\leq x\}$

My understanding is that $J$ is the set of all elements of the real numbers for which it is true that $z$ is less than or equal to the real number $x$.

Not quite.   $J \mathop{\,:=\,} \{B_x: x\in \Bbb R\}$ is the set of all such rays that can be generated from all real points.   It is not a set of points, but rather a set of sets of points.

$$J= \{\{ z : -\infty < z\leq x\}: -\infty< x< \infty\}$$

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  • $\begingroup$ Thanks very much for clarifying the typo and explaining the correct interpretation of the set $J$. Very helpful! Cheers, Graham. $\endgroup$ Dec 29, 2015 at 9:12

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