Timeline for Probability of one point being closer to fixed $x \in \mathbf{R}^2$ than another
Current License: CC BY-SA 4.0
13 events
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| Oct 8, 2020 at 10:29 | vote | accept | Inter Veridium | ||
| Oct 7, 2020 at 22:23 | answer | added | Tyrel Stokes | timeline score: 2 | |
| Oct 5, 2020 at 19:01 | history | edited | Inter Veridium | CC BY-SA 4.0 |
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| Oct 5, 2020 at 18:49 | history | edited | Inter Veridium | CC BY-SA 4.0 |
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| Oct 5, 2020 at 18:37 | comment | added | Inter Veridium | @whuber Indeed, but we've also got a fixed point $x = (0.32, 0)$. So, e.g. assume that $x^{(0)} = (0, 0.5)$, $x^{(1)} = (1, 0.5)$. $x^{(0)}$ is closer to point $x$, as $d(x^{(0)}, x) < d(x^{(1)}, x)$. What is the probability that the fixed point $x = (0.32, 0)$ will be closer to $x^{(1)}$ than to $x^{(0)}$? Should I rephrase the question? | |
| Oct 5, 2020 at 18:07 | comment | added | whuber♦ | Because that is an unusual meaning of "added noise," please edit your post to explain it. (Most, if not all, readers would understand "adding noise" to mean that $x^{(0)}$ is a number to which a random, zero-expectation value has been added, producing another number, not a point in the plane.) I still cannot make sense of your question with this new interpretation, though: the point $(0,r)$ always has a distance at least $1$ from all points of the form $(1,y)$ and a distance no more than $1$ from $(0,0),$ so isn't the answer trivially zero? | |
| Oct 5, 2020 at 15:16 | comment | added | Inter Veridium | @whuber Sorry. I understood it as if we've got $x^{(0)} = (0, 0)$, and by adding noise we set it as $x^{(0)} = (0, r_1)$, where $r_1 \sim U[-1, 1]$, same for $x^{(1)} = (1, r_2)$, $r_2 \sim U[-1, 1]$. | |
| Oct 5, 2020 at 14:55 | comment | added | whuber♦ | Could you please explain what you mean by the "Euclidean distance" between a point $x$ in the plane and a point like $x^{(0)}$ or $x^{(1)}$ in the line? | |
| Oct 5, 2020 at 7:30 | comment | added | Inter Veridium | @TyrelStokes Two independent errors, adding the first to $x^{(0)}$ and second to $x^{(1)}$. | |
| Oct 5, 2020 at 3:22 | comment | added | Tyrel Stokes | When you add the error to $x^{(0)}$ and $x^{(1)}$, do you draw one value from U[-1,1] and add this value to both or do you draw two independent errors adding the first to $x^{(0)}$ and the second to $x^{(1)}$? | |
| Oct 4, 2020 at 23:48 | history | edited | Inter Veridium | CC BY-SA 4.0 |
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| Oct 4, 2020 at 12:25 | review | First posts | |||
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| Oct 4, 2020 at 12:20 | history | asked | Inter Veridium | CC BY-SA 4.0 |