Modifying the density of the real line
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I am having issues trying to do something that is fairly simple. Given the interval [0,1] I need to find a function from [0,1] to [0,1] that concentrates the point density around $x=0.5$.
So for example: [0,0.25,0.5,0.75,1] would become [0, 0.4, 0.5, 0.6, 1]
But I am not finding a function with this property that follows a normal distribution centered at 0.5.
general-topology functions real-numbers transformation
asked Nov 26 '18 at 17:14
MakoganMakogan
761217
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|
show 2 more comments
$begingroup$
I am having issues trying to do something that is fairly simple. Given the interval [0,1] I need to find a function from [0,1] to [0,1] that concentrates the point density around $x=0.5$.
So for example: [0,0.25,0.5,0.75,1] would become [0, 0.4, 0.5, 0.6, 1]
But I am not finding a function with this property that follows a normal distribution centered at 0.5.
general-topology functions real-numbers transformation
asked Nov 26 '18 at 17:14
MakoganMakogan
761217
$endgroup$
$begingroup$
What do you mean by "follows a normal distribution?"
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– Paul
Nov 26 '18 at 17:20
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@Paul I mean that numbers concentrate around 0.5 and density decreases as you move away from it. For example the straight line y=x can be said to have a uniform distribution.
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– Makogan
Nov 26 '18 at 17:32
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No function on $[0,1]$ will ever follow a normal distribution. For that you need a function on $(-infty,infty)$.
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– Ben W
Nov 26 '18 at 17:34
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That interval is homeomorphic to (0,1) so you can most definitely find an appropriate mapping. Regardless, the point is simply to concentrate the point density of that interval around 0.5 following a distribution that looks like a bell curve (most of the points are within an epsilon distance of 0.5)
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– Makogan
Nov 26 '18 at 17:45
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But that homeomorphic mapping will not take a normal distribution to a normal distribution, and it will not leave $0$ or $1$ fixed.
$endgroup$
– Paul
Nov 26 '18 at 17:55
|
show 2 more comments
$begingroup$
I am having issues trying to do something that is fairly simple. Given the interval [0,1] I need to find a function from [0,1] to [0,1] that concentrates the point density around $x=0.5$.
So for example: [0,0.25,0.5,0.75,1] would become [0, 0.4, 0.5, 0.6, 1]
But I am not finding a function with this property that follows a normal distribution centered at 0.5.
general-topology functions real-numbers transformation
asked Nov 26 '18 at 17:14
MakoganMakogan
761217
$endgroup$
I am having issues trying to do something that is fairly simple. Given the interval [0,1] I need to find a function from [0,1] to [0,1] that concentrates the point density around $x=0.5$.
So for example: [0,0.25,0.5,0.75,1] would become [0, 0.4, 0.5, 0.6, 1]
But I am not finding a function with this property that follows a normal distribution centered at 0.5.
general-topology functions real-numbers transformation
general-topology functions real-numbers transformation
asked Nov 26 '18 at 17:14
MakoganMakogan
761217
asked Nov 26 '18 at 17:14
MakoganMakogan
761217
asked Nov 26 '18 at 17:14
MakoganMakogan
761217
asked Nov 26 '18 at 17:14
MakoganMakogan
761217
asked Nov 26 '18 at 17:14
MakoganMakogan
761217
761217
$begingroup$
What do you mean by "follows a normal distribution?"
$endgroup$
– Paul
Nov 26 '18 at 17:20
$begingroup$
@Paul I mean that numbers concentrate around 0.5 and density decreases as you move away from it. For example the straight line y=x can be said to have a uniform distribution.
$endgroup$
– Makogan
Nov 26 '18 at 17:32
$begingroup$
No function on $[0,1]$ will ever follow a normal distribution. For that you need a function on $(-infty,infty)$.
$endgroup$
– Ben W
Nov 26 '18 at 17:34
$begingroup$
That interval is homeomorphic to (0,1) so you can most definitely find an appropriate mapping. Regardless, the point is simply to concentrate the point density of that interval around 0.5 following a distribution that looks like a bell curve (most of the points are within an epsilon distance of 0.5)
$endgroup$
– Makogan
Nov 26 '18 at 17:45
$begingroup$
But that homeomorphic mapping will not take a normal distribution to a normal distribution, and it will not leave $0$ or $1$ fixed.
$endgroup$
– Paul
Nov 26 '18 at 17:55
|
show 2 more comments
$begingroup$
What do you mean by "follows a normal distribution?"
$endgroup$
– Paul
Nov 26 '18 at 17:20
$begingroup$
@Paul I mean that numbers concentrate around 0.5 and density decreases as you move away from it. For example the straight line y=x can be said to have a uniform distribution.
$endgroup$
– Makogan
Nov 26 '18 at 17:32
$begingroup$
No function on $[0,1]$ will ever follow a normal distribution. For that you need a function on $(-infty,infty)$.
$endgroup$
– Ben W
Nov 26 '18 at 17:34
$begingroup$
That interval is homeomorphic to (0,1) so you can most definitely find an appropriate mapping. Regardless, the point is simply to concentrate the point density of that interval around 0.5 following a distribution that looks like a bell curve (most of the points are within an epsilon distance of 0.5)
$endgroup$
– Makogan
Nov 26 '18 at 17:45
$begingroup$
But that homeomorphic mapping will not take a normal distribution to a normal distribution, and it will not leave $0$ or $1$ fixed.
$endgroup$
– Paul
Nov 26 '18 at 17:55
$begingroup$
What do you mean by "follows a normal distribution?"
$endgroup$
– Paul
Nov 26 '18 at 17:20
$begingroup$
What do you mean by "follows a normal distribution?"
$endgroup$
– Paul
Nov 26 '18 at 17:20
$begingroup$
@Paul I mean that numbers concentrate around 0.5 and density decreases as you move away from it. For example the straight line y=x can be said to have a uniform distribution.
$endgroup$
– Makogan
Nov 26 '18 at 17:32
$begingroup$
@Paul I mean that numbers concentrate around 0.5 and density decreases as you move away from it. For example the straight line y=x can be said to have a uniform distribution.
$endgroup$
– Makogan
Nov 26 '18 at 17:32
$begingroup$
No function on $[0,1]$ will ever follow a normal distribution. For that you need a function on $(-infty,infty)$.
$endgroup$
– Ben W
Nov 26 '18 at 17:34
$begingroup$
No function on $[0,1]$ will ever follow a normal distribution. For that you need a function on $(-infty,infty)$.
$endgroup$
– Ben W
Nov 26 '18 at 17:34
$begingroup$
That interval is homeomorphic to (0,1) so you can most definitely find an appropriate mapping. Regardless, the point is simply to concentrate the point density of that interval around 0.5 following a distribution that looks like a bell curve (most of the points are within an epsilon distance of 0.5)
$endgroup$
– Makogan
Nov 26 '18 at 17:45
$begingroup$
That interval is homeomorphic to (0,1) so you can most definitely find an appropriate mapping. Regardless, the point is simply to concentrate the point density of that interval around 0.5 following a distribution that looks like a bell curve (most of the points are within an epsilon distance of 0.5)
$endgroup$
– Makogan
Nov 26 '18 at 17:45
$begingroup$
But that homeomorphic mapping will not take a normal distribution to a normal distribution, and it will not leave $0$ or $1$ fixed.
$endgroup$
– Paul
Nov 26 '18 at 17:55
$begingroup$
But that homeomorphic mapping will not take a normal distribution to a normal distribution, and it will not leave $0$ or $1$ fixed.
$endgroup$
– Paul
Nov 26 '18 at 17:55
|
show 2 more comments
1 Answer
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Map $[0,1]$ linearly to $[-1,1]$, apply $t mapsto t^{2n+1}$, which pushes numbers closer to $0$, the center point, and then map back to $[0,1]$, again linearly.
Explicitly $$x mapsto frac12((2x-1)^{2n+1} + 1)$$
As you make $n$ larger the values become more tightly concentrated around $0$.
answered Nov 26 '18 at 18:13
JonathanZJonathanZ
2,134613
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Map $[0,1]$ linearly to $[-1,1]$, apply $t mapsto t^{2n+1}$, which pushes numbers closer to $0$, the center point, and then map back to $[0,1]$, again linearly.
Explicitly $$x mapsto frac12((2x-1)^{2n+1} + 1)$$
As you make $n$ larger the values become more tightly concentrated around $0$.
answered Nov 26 '18 at 18:13
JonathanZJonathanZ
2,134613
$endgroup$
add a comment |
$begingroup$
Map $[0,1]$ linearly to $[-1,1]$, apply $t mapsto t^{2n+1}$, which pushes numbers closer to $0$, the center point, and then map back to $[0,1]$, again linearly.
Explicitly $$x mapsto frac12((2x-1)^{2n+1} + 1)$$
As you make $n$ larger the values become more tightly concentrated around $0$.
answered Nov 26 '18 at 18:13
JonathanZJonathanZ
2,134613
$endgroup$
add a comment |
$begingroup$
Map $[0,1]$ linearly to $[-1,1]$, apply $t mapsto t^{2n+1}$, which pushes numbers closer to $0$, the center point, and then map back to $[0,1]$, again linearly.
Explicitly $$x mapsto frac12((2x-1)^{2n+1} + 1)$$
As you make $n$ larger the values become more tightly concentrated around $0$.
answered Nov 26 '18 at 18:13
JonathanZJonathanZ
2,134613
$endgroup$
Map $[0,1]$ linearly to $[-1,1]$, apply $t mapsto t^{2n+1}$, which pushes numbers closer to $0$, the center point, and then map back to $[0,1]$, again linearly.
Explicitly $$x mapsto frac12((2x-1)^{2n+1} + 1)$$
As you make $n$ larger the values become more tightly concentrated around $0$.
answered Nov 26 '18 at 18:13
JonathanZJonathanZ
2,134613
edited Nov 26 '18 at 18:37
answered Nov 26 '18 at 18:13
JonathanZJonathanZ
2,134613
answered Nov 26 '18 at 18:13
JonathanZJonathanZ
2,134613
answered Nov 26 '18 at 18:13
JonathanZJonathanZ
2,134613
2,134613
add a comment |
add a comment |
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$begingroup$
What do you mean by "follows a normal distribution?"
$endgroup$
– Paul
Nov 26 '18 at 17:20
$begingroup$
@Paul I mean that numbers concentrate around 0.5 and density decreases as you move away from it. For example the straight line y=x can be said to have a uniform distribution.
$endgroup$
– Makogan
Nov 26 '18 at 17:32
$begingroup$
No function on $[0,1]$ will ever follow a normal distribution. For that you need a function on $(-infty,infty)$.
$endgroup$
– Ben W
Nov 26 '18 at 17:34
$begingroup$
That interval is homeomorphic to (0,1) so you can most definitely find an appropriate mapping. Regardless, the point is simply to concentrate the point density of that interval around 0.5 following a distribution that looks like a bell curve (most of the points are within an epsilon distance of 0.5)
$endgroup$
– Makogan
Nov 26 '18 at 17:45
$begingroup$
But that homeomorphic mapping will not take a normal distribution to a normal distribution, and it will not leave $0$ or $1$ fixed.
$endgroup$
– Paul
Nov 26 '18 at 17:55