From the low-discrepancy wiki I have:

$$\displaystyle \prod_{i=1}^s[a_i, b_i) = \{𝑥 \in \mathbb{R}^s : a_i \le x_i \le b_i\}$$ where $$0 \le a_i < b_i \le 1$$

How can I read it? Should I first make do a cartesian product of $[a_i,b_i)$? For example for $s=3$ I would have: $$ [a_1, b_1) \times [a_2, b_2) \times [a_3, b_3) = \{ \{a_1, a_2, a_3\}, \{a_1, a_2, b_3\}, \{a_1, b_2, a_3\}, \{a_1, b_2, b_3\}, \{b_1, a_2, a_3\}, \{b_1, a_2, b_3\}, \{b_1, b_2, a_3\}, \{b_1, b_2, b_3\}\} $$

then I would put it into set-builder notation.. but for first $\{a_1, a_2, a_3\}$ I don't have $b_i$ value.

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    $\begingroup$ What is your question? $\endgroup$
    – lisyarus
    Apr 23 '19 at 9:50
  • $\begingroup$ @lisyarus I am sorry. Stackexchenge posted my question when I was trying to add a link. I have edited it. $\endgroup$ Apr 23 '19 at 10:06

You should read, $\displaystyle \prod_{i=1}^s[a_i,b_i)$ as, "the product of all intervals closed $a_i$, open $b_i$ from $i=1$ to $s$". And $\displaystyle \{\tilde x\in \Bbb R^s:\tilde x=(x_1,x_2,\dots x_s) \text{ where }x_i\in[a_i,b_i),\forall i=1,2\dots s\}$ as, "the collection of all $\tilde x$ ($x$ tilde) in $\Bbb R^s$ such that $x$ tilde is of the form of $s$-tuple $(x_1,x_2,\dots x_s)$ where $x_i$ belongs to $[a_i,b_i)$ for all $i=1,2,... s$".

Hope it works.

Further, $\displaystyle \prod_{i=1}^s[a_i,b_i]=[a_1,b_1]\times [a_2,b_2]\times ... \times [a_s,b_s]$ where $\times$ is the cartesian product defined here.

  • $\begingroup$ After I compute $\displaystyle \prod_{i=1}^s[a_i, b_i) = \{𝑥 \in \mathbb{R}^s : a_i \le x_i \le b_i\}$ I will get collection of s-tuples: $\{ s-tuple, s-tuple...\}$. So, for $s = 1$ I will get $\{a_1, a_1 + v, a_1 + 2*v, ...a_1 + k*v\}$ where $v$ is some "small value" and $a_1 + k*v$ is smaller than $b_1$. So, for $v=0.4$, $a_1=0$ and $b_1=1$ I will get $\{0,0.4,0.8\}$. For $s=2$ I need to compute the Cartesian product: $\{0,0.4,0.8\}\times\{0,0.4,0.8| = \{\{0,0\},\{0,0.4\},\{0,0.8\},\{0.4,0\},\{0.4,0.4\},\{0.4,0.8\},\{0.8,0\},\{0.8,0.4\},\{0.8,0.8\}\}$. Is this correct? How do I find $v$? $\endgroup$ Apr 29 '19 at 11:40
  • $\begingroup$ I think you have confused with math.stackexchange.com/questions/2115752/… this one. Further for $s=1$ you simply get an interval which cannot be written in your form $\{a1,a1+v...\}$ etc. $\endgroup$ Apr 30 '19 at 2:33
  • $\begingroup$ As for link: instead of $\{\{0,0\},...\}$ I should write $\{(0,0)\}$ and $\{0, 0.4, 0.8\}$ as $\{(0), (0.4), (0.8)\}$. That is what you mean? As for $s=1$ do you mean that I have forgotten parenthesis ($()$)? $\endgroup$ May 9 '19 at 16:06
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    $\begingroup$ Definitely yes. Thats what I mean. $\endgroup$ May 10 '19 at 1:59

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