Timeline for Decision boundaries and Gaussian density functions
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| May 27, 2022 at 13:00 | comment | added | Benjamin Wang | @whuber cool visualisation | |
| Jan 8, 2015 at 3:31 | comment | added | whuber♦ | Re the visualization: Imagine the intersection of a spherical bowl with a bowl that is so long and shallow (but with steep sides) that it looks almost like a folded piece of paper. That is what things look like when $b=2/a$ and $c=1$. The steep sides of the paper rise above the bowl's sides, but the valley along the center of the paper falls below the bowl's sides. Thus, the region where the paper is higher consists of two separate areas delimited by the arcs of the hyperbola. | |
| Mar 17, 2013 at 22:49 | history | edited | Dilip Sarwate | CC BY-SA 3.0 |
corrected speculations in last paragraph
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| Mar 17, 2013 at 16:06 | comment | added | whuber♦ | Good beginning! I fixed a subtraction error in the final formula. As a result, you might want to revisit your speculations in the last paragraph. To get started, note that the equation is linear if and only if all the quadratic coefficients are zero, which is equivalent to $a=1$, $b=1$, and $c=0$. (There is another special case where the equation represents a pair of lines.) Whether the locus is elliptical or hyperbolic depends on the sign of the determinant $(a-1)(b-1)-c^2$; both signs are possible. | |
| Mar 17, 2013 at 16:02 | history | edited | whuber♦ | CC BY-SA 3.0 |
added 13 characters in body
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| Oct 18, 2012 at 12:44 | history | edited | Dilip Sarwate | CC BY-SA 3.0 |
cleaned up the peroration
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| Oct 17, 2012 at 21:31 | history | answered | Dilip Sarwate | CC BY-SA 3.0 |