Discretization of normal distribution over a finite range
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If I only have data about the mean and standard deviation of a distribution over a finite discrete range (e.g. integers 1 to 5). How do I properly reconstruct the distribution (= a distribution that has the same mean and standard deviation, assuming it is close to normal)?
normal-distribution
edited May 9 at 21:39
T J. Kim
314217
asked Aug 19 '14 at 8:10
kutschkem
291310
add a comment |
up vote
0
down vote
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If I only have data about the mean and standard deviation of a distribution over a finite discrete range (e.g. integers 1 to 5). How do I properly reconstruct the distribution (= a distribution that has the same mean and standard deviation, assuming it is close to normal)?
normal-distribution
edited May 9 at 21:39
T J. Kim
314217
asked Aug 19 '14 at 8:10
kutschkem
291310
Maybe match the two moments with the hypergeometric distribution? (In general, any distribution with 2 unknown parameters would work.) How proper the result is going to be? I have no idea...
– Tunococ
Aug 19 '14 at 8:16
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If I only have data about the mean and standard deviation of a distribution over a finite discrete range (e.g. integers 1 to 5). How do I properly reconstruct the distribution (= a distribution that has the same mean and standard deviation, assuming it is close to normal)?
normal-distribution
edited May 9 at 21:39
T J. Kim
314217
asked Aug 19 '14 at 8:10
kutschkem
291310
If I only have data about the mean and standard deviation of a distribution over a finite discrete range (e.g. integers 1 to 5). How do I properly reconstruct the distribution (= a distribution that has the same mean and standard deviation, assuming it is close to normal)?
normal-distribution
normal-distribution
edited May 9 at 21:39
T J. Kim
314217
asked Aug 19 '14 at 8:10
kutschkem
291310
edited May 9 at 21:39
T J. Kim
314217
asked Aug 19 '14 at 8:10
kutschkem
291310
edited May 9 at 21:39
T J. Kim
314217
edited May 9 at 21:39
T J. Kim
314217
edited May 9 at 21:39
T J. Kim
314217
314217
asked Aug 19 '14 at 8:10
kutschkem
291310
asked Aug 19 '14 at 8:10
kutschkem
291310
asked Aug 19 '14 at 8:10
kutschkem
291310
291310
Maybe match the two moments with the hypergeometric distribution? (In general, any distribution with 2 unknown parameters would work.) How proper the result is going to be? I have no idea...
– Tunococ
Aug 19 '14 at 8:16
add a comment |
Maybe match the two moments with the hypergeometric distribution? (In general, any distribution with 2 unknown parameters would work.) How proper the result is going to be? I have no idea...
– Tunococ
Aug 19 '14 at 8:16
Maybe match the two moments with the hypergeometric distribution? (In general, any distribution with 2 unknown parameters would work.) How proper the result is going to be? I have no idea...
– Tunococ
Aug 19 '14 at 8:16
Maybe match the two moments with the hypergeometric distribution? (In general, any distribution with 2 unknown parameters would work.) How proper the result is going to be? I have no idea...
– Tunococ
Aug 19 '14 at 8:16
add a comment |
2 Answers
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If $p_i$ is the probability of outcome $i$ and you want to achieve a certain $mu$ and $sigma$, you need to obey the constraints
$$ begin{align}sum p_i&=1\
sum ip_i &= mu\
sum i^2p_i &= sigma^2+mu^2end{align}$$
These are three equations in five unknowns, so there are many solutions (let us ignore the possibility that the solutions conflict with the conditions $p_ige 0$) and we can impose additional conditions "at will". Since you want "close to normal", I suggest we take the conditions from the next few non-central moments of the normal distribution:
$$ begin{align}sum i^3p_i &= mu^3+3musigma^2\
sum i^4p_i &= mu^4 + 6mu^2sigma^2 + 3sigma^4.end{align}$$
So now you have five linear equations in five unknowns - and I hope it turns out that all $p_i$ are $ge0$ in your specific case.
answered Aug 19 '14 at 8:29
Hagen von Eitzen
275k21268495
add a comment |
up vote
0
down vote
If there are just five values (possible), I fail to see the point of trying to fit to some "standard distribution" or even a continuous one (like normal). The (relative) frequency table says it all, in a simpler way, and it's even easy to visualize (e.g. like a histogram).
I know the thread is very old; I just couldn't resist.
answered Jan 8 '17 at 9:17
Vladimír Čunát
1011
add a comment |
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2 Answers
2
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2 Answers
2
active
oldest
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active
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active
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up vote
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down vote
If $p_i$ is the probability of outcome $i$ and you want to achieve a certain $mu$ and $sigma$, you need to obey the constraints
$$ begin{align}sum p_i&=1\
sum ip_i &= mu\
sum i^2p_i &= sigma^2+mu^2end{align}$$
These are three equations in five unknowns, so there are many solutions (let us ignore the possibility that the solutions conflict with the conditions $p_ige 0$) and we can impose additional conditions "at will". Since you want "close to normal", I suggest we take the conditions from the next few non-central moments of the normal distribution:
$$ begin{align}sum i^3p_i &= mu^3+3musigma^2\
sum i^4p_i &= mu^4 + 6mu^2sigma^2 + 3sigma^4.end{align}$$
So now you have five linear equations in five unknowns - and I hope it turns out that all $p_i$ are $ge0$ in your specific case.
answered Aug 19 '14 at 8:29
Hagen von Eitzen
275k21268495
add a comment |
up vote
0
down vote
If $p_i$ is the probability of outcome $i$ and you want to achieve a certain $mu$ and $sigma$, you need to obey the constraints
$$ begin{align}sum p_i&=1\
sum ip_i &= mu\
sum i^2p_i &= sigma^2+mu^2end{align}$$
These are three equations in five unknowns, so there are many solutions (let us ignore the possibility that the solutions conflict with the conditions $p_ige 0$) and we can impose additional conditions "at will". Since you want "close to normal", I suggest we take the conditions from the next few non-central moments of the normal distribution:
$$ begin{align}sum i^3p_i &= mu^3+3musigma^2\
sum i^4p_i &= mu^4 + 6mu^2sigma^2 + 3sigma^4.end{align}$$
So now you have five linear equations in five unknowns - and I hope it turns out that all $p_i$ are $ge0$ in your specific case.
answered Aug 19 '14 at 8:29
Hagen von Eitzen
275k21268495
add a comment |
up vote
0
down vote
up vote
0
down vote
If $p_i$ is the probability of outcome $i$ and you want to achieve a certain $mu$ and $sigma$, you need to obey the constraints
$$ begin{align}sum p_i&=1\
sum ip_i &= mu\
sum i^2p_i &= sigma^2+mu^2end{align}$$
These are three equations in five unknowns, so there are many solutions (let us ignore the possibility that the solutions conflict with the conditions $p_ige 0$) and we can impose additional conditions "at will". Since you want "close to normal", I suggest we take the conditions from the next few non-central moments of the normal distribution:
$$ begin{align}sum i^3p_i &= mu^3+3musigma^2\
sum i^4p_i &= mu^4 + 6mu^2sigma^2 + 3sigma^4.end{align}$$
So now you have five linear equations in five unknowns - and I hope it turns out that all $p_i$ are $ge0$ in your specific case.
answered Aug 19 '14 at 8:29
Hagen von Eitzen
275k21268495
If $p_i$ is the probability of outcome $i$ and you want to achieve a certain $mu$ and $sigma$, you need to obey the constraints
$$ begin{align}sum p_i&=1\
sum ip_i &= mu\
sum i^2p_i &= sigma^2+mu^2end{align}$$
These are three equations in five unknowns, so there are many solutions (let us ignore the possibility that the solutions conflict with the conditions $p_ige 0$) and we can impose additional conditions "at will". Since you want "close to normal", I suggest we take the conditions from the next few non-central moments of the normal distribution:
$$ begin{align}sum i^3p_i &= mu^3+3musigma^2\
sum i^4p_i &= mu^4 + 6mu^2sigma^2 + 3sigma^4.end{align}$$
So now you have five linear equations in five unknowns - and I hope it turns out that all $p_i$ are $ge0$ in your specific case.
answered Aug 19 '14 at 8:29
Hagen von Eitzen
275k21268495
answered Aug 19 '14 at 8:29
Hagen von Eitzen
275k21268495
answered Aug 19 '14 at 8:29
Hagen von Eitzen
275k21268495
answered Aug 19 '14 at 8:29
Hagen von Eitzen
275k21268495
275k21268495
add a comment |
add a comment |
up vote
0
down vote
If there are just five values (possible), I fail to see the point of trying to fit to some "standard distribution" or even a continuous one (like normal). The (relative) frequency table says it all, in a simpler way, and it's even easy to visualize (e.g. like a histogram).
I know the thread is very old; I just couldn't resist.
answered Jan 8 '17 at 9:17
Vladimír Čunát
1011
add a comment |
up vote
0
down vote
If there are just five values (possible), I fail to see the point of trying to fit to some "standard distribution" or even a continuous one (like normal). The (relative) frequency table says it all, in a simpler way, and it's even easy to visualize (e.g. like a histogram).
I know the thread is very old; I just couldn't resist.
answered Jan 8 '17 at 9:17
Vladimír Čunát
1011
add a comment |
up vote
0
down vote
up vote
0
down vote
If there are just five values (possible), I fail to see the point of trying to fit to some "standard distribution" or even a continuous one (like normal). The (relative) frequency table says it all, in a simpler way, and it's even easy to visualize (e.g. like a histogram).
I know the thread is very old; I just couldn't resist.
answered Jan 8 '17 at 9:17
Vladimír Čunát
1011
If there are just five values (possible), I fail to see the point of trying to fit to some "standard distribution" or even a continuous one (like normal). The (relative) frequency table says it all, in a simpler way, and it's even easy to visualize (e.g. like a histogram).
I know the thread is very old; I just couldn't resist.
answered Jan 8 '17 at 9:17
Vladimír Čunát
1011
answered Jan 8 '17 at 9:17
Vladimír Čunát
1011
answered Jan 8 '17 at 9:17
Vladimír Čunát
1011
answered Jan 8 '17 at 9:17
Vladimír Čunát
1011
1011
add a comment |
add a comment |
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Maybe match the two moments with the hypergeometric distribution? (In general, any distribution with 2 unknown parameters would work.) How proper the result is going to be? I have no idea...
– Tunococ
Aug 19 '14 at 8:16