How to find which percentile is a value in a skewed normal distribution
$begingroup$
I have a skewed normal distribution for which I know the average, standard deviation, skewness & kurtosis (which is different from zero).
Given a number $X,$ how can estimate which percentile corresponds to that value? (I'm ok with getting an approximate value of this percentile.)
I used z-score tables in the past (before having skewed distributions), but they seem to apply only to non-skewed distributions.
Thanks for your help.
statistics
edited Jan 7 at 14:57
Namaste
1
asked Jan 7 at 13:37
NisalonNisalon
162
$endgroup$
add a comment |
$begingroup$
I have a skewed normal distribution for which I know the average, standard deviation, skewness & kurtosis (which is different from zero).
Given a number $X,$ how can estimate which percentile corresponds to that value? (I'm ok with getting an approximate value of this percentile.)
I used z-score tables in the past (before having skewed distributions), but they seem to apply only to non-skewed distributions.
Thanks for your help.
statistics
edited Jan 7 at 14:57
Namaste
1
asked Jan 7 at 13:37
NisalonNisalon
162
$endgroup$
$begingroup$
Assuming you mean a skewed normal distribution (as defined at en.wikipedia.org/wiki/Skew_normal_distribution) you could use the method of moments to estimate the parameters of the skewed normal distribution and then estimate any desired function of that distribution. The problem is that you won't know how good your estimate of any percentile will be unless maybe you also have the sample size.
$endgroup$
– JimB
Jan 7 at 15:03
$begingroup$
I have the sample size
$endgroup$
– Nisalon
Jan 7 at 16:16
$begingroup$
If you have the sample size, you could sample from the estimated distribution to get some idea as to how good the parameter estimates might be.
$endgroup$
– JimB
Jan 7 at 16:59
add a comment |
$begingroup$
I have a skewed normal distribution for which I know the average, standard deviation, skewness & kurtosis (which is different from zero).
Given a number $X,$ how can estimate which percentile corresponds to that value? (I'm ok with getting an approximate value of this percentile.)
I used z-score tables in the past (before having skewed distributions), but they seem to apply only to non-skewed distributions.
Thanks for your help.
statistics
edited Jan 7 at 14:57
Namaste
1
asked Jan 7 at 13:37
NisalonNisalon
162
$endgroup$
I have a skewed normal distribution for which I know the average, standard deviation, skewness & kurtosis (which is different from zero).
Given a number $X,$ how can estimate which percentile corresponds to that value? (I'm ok with getting an approximate value of this percentile.)
I used z-score tables in the past (before having skewed distributions), but they seem to apply only to non-skewed distributions.
Thanks for your help.
statistics
statistics
edited Jan 7 at 14:57
Namaste
1
asked Jan 7 at 13:37
NisalonNisalon
162
edited Jan 7 at 14:57
Namaste
1
asked Jan 7 at 13:37
NisalonNisalon
162
edited Jan 7 at 14:57
Namaste
1
edited Jan 7 at 14:57
Namaste
1
edited Jan 7 at 14:57
Namaste
1
1
asked Jan 7 at 13:37
NisalonNisalon
162
asked Jan 7 at 13:37
NisalonNisalon
162
asked Jan 7 at 13:37
NisalonNisalon
162
162
$begingroup$
Assuming you mean a skewed normal distribution (as defined at en.wikipedia.org/wiki/Skew_normal_distribution) you could use the method of moments to estimate the parameters of the skewed normal distribution and then estimate any desired function of that distribution. The problem is that you won't know how good your estimate of any percentile will be unless maybe you also have the sample size.
$endgroup$
– JimB
Jan 7 at 15:03
$begingroup$
I have the sample size
$endgroup$
– Nisalon
Jan 7 at 16:16
$begingroup$
If you have the sample size, you could sample from the estimated distribution to get some idea as to how good the parameter estimates might be.
$endgroup$
– JimB
Jan 7 at 16:59
add a comment |
$begingroup$
Assuming you mean a skewed normal distribution (as defined at en.wikipedia.org/wiki/Skew_normal_distribution) you could use the method of moments to estimate the parameters of the skewed normal distribution and then estimate any desired function of that distribution. The problem is that you won't know how good your estimate of any percentile will be unless maybe you also have the sample size.
$endgroup$
– JimB
Jan 7 at 15:03
$begingroup$
I have the sample size
$endgroup$
– Nisalon
Jan 7 at 16:16
$begingroup$
If you have the sample size, you could sample from the estimated distribution to get some idea as to how good the parameter estimates might be.
$endgroup$
– JimB
Jan 7 at 16:59
$begingroup$
Assuming you mean a skewed normal distribution (as defined at en.wikipedia.org/wiki/Skew_normal_distribution) you could use the method of moments to estimate the parameters of the skewed normal distribution and then estimate any desired function of that distribution. The problem is that you won't know how good your estimate of any percentile will be unless maybe you also have the sample size.
$endgroup$
– JimB
Jan 7 at 15:03
$begingroup$
Assuming you mean a skewed normal distribution (as defined at en.wikipedia.org/wiki/Skew_normal_distribution) you could use the method of moments to estimate the parameters of the skewed normal distribution and then estimate any desired function of that distribution. The problem is that you won't know how good your estimate of any percentile will be unless maybe you also have the sample size.
$endgroup$
– JimB
Jan 7 at 15:03
$begingroup$
I have the sample size
$endgroup$
– Nisalon
Jan 7 at 16:16
$begingroup$
I have the sample size
$endgroup$
– Nisalon
Jan 7 at 16:16
$begingroup$
If you have the sample size, you could sample from the estimated distribution to get some idea as to how good the parameter estimates might be.
$endgroup$
– JimB
Jan 7 at 16:59
$begingroup$
If you have the sample size, you could sample from the estimated distribution to get some idea as to how good the parameter estimates might be.
$endgroup$
– JimB
Jan 7 at 16:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The Wiki page Skew normal distribution provides the information to estimate the parameters using the sample mean ($bar{x}$), standard deviation ($s$), and skewness ($hat{gamma}$). The 3 parameters to be estimated are $mu$, $sigma$, and $alpha$.
If $|hat{gamma}|<1$, then $hat{alpha}$ is found in two steps:
$$delta =sqrt{frac{pi left| hat{gamma} right| ^{2/3}}{2 left(left| hat{gamma} right| ^{2/3}+left(frac{4-pi }{2}right)^{2/3}right)}}$$
$$hat{alpha} = text{sgn}(hat{gamma})sqrt{frac{delta }{1-delta ^2}}$$
Otherwise $hat{alpha}$ is the solution to
$$hat{gamma} =frac{sqrt{2} (4-pi ) hat{alpha} ^3}{left((pi -2) hat{alpha} ^2+pi right)^{3/2}}$$
which needs to be performed numerically. Then $hat{mu}$ and $hat{sigma}$ are
$$hat{sigma} =frac{s}{sqrt{1-frac{2 hat{alpha} ^2}{pi left(hat{alpha} ^2+1right)}}}$$
$$hat{mu} =bar{x}-frac{sqrt{frac{2}{pi }} hat{alpha} hat{sigma} }{sqrt{hat{alpha} ^2+1}}$$
Now armed with the parameter estimates, then one can estimate the cumulative distribution function:
$$Pr(X le x)=Phileft(frac{x, -hat{mu} }{hat{sigma} }right)-2 Tleft(frac{x, -hat{mu} }{hat{sigma} },hat{alpha} right)$$
where $T$ is the Owen's T function: $T(x,a)=frac{int_0^a frac{exp left(-left(left(t^2+1right) x^2right)right)}{2 left(t^2+1right)} , dt}{2 pi }$ .
Here is an implementation using Mathematica:
I know code should be given as text but in this case because it is unlikely that you have Mathematica (and it would look much messier as text), it should be instructive as to the process.
To estimate the percentage of the distribution no larger than a specified value you'll need to use the cumulative distribution function (CDF) described on the Wiki page. Using Mathematica for values of $X$ being 7.5 and 5.2:
If you have access to the statistical package R, then the sn package will perform these calculations.
answered Jan 7 at 16:57
JimBJimB
61547
$endgroup$
$begingroup$
Sorry, I'm not sure you actually answer my question : how can I get the percentile corresponding to one given number ?
$endgroup$
– Nisalon
Jan 7 at 17:15
$begingroup$
I've added the Mathematica code for that and mentioned thesnpackage in R.
$endgroup$
– JimB
Jan 7 at 17:57
$begingroup$
I don't want code (as this code will be integrated in PHP), I want to understand the process to be able to implement it
$endgroup$
– Nisalon
Jan 7 at 20:40
$begingroup$
Added a description of the process.
$endgroup$
– JimB
Jan 7 at 21:39
add a comment |
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1 Answer
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$begingroup$
The Wiki page Skew normal distribution provides the information to estimate the parameters using the sample mean ($bar{x}$), standard deviation ($s$), and skewness ($hat{gamma}$). The 3 parameters to be estimated are $mu$, $sigma$, and $alpha$.
If $|hat{gamma}|<1$, then $hat{alpha}$ is found in two steps:
$$delta =sqrt{frac{pi left| hat{gamma} right| ^{2/3}}{2 left(left| hat{gamma} right| ^{2/3}+left(frac{4-pi }{2}right)^{2/3}right)}}$$
$$hat{alpha} = text{sgn}(hat{gamma})sqrt{frac{delta }{1-delta ^2}}$$
Otherwise $hat{alpha}$ is the solution to
$$hat{gamma} =frac{sqrt{2} (4-pi ) hat{alpha} ^3}{left((pi -2) hat{alpha} ^2+pi right)^{3/2}}$$
which needs to be performed numerically. Then $hat{mu}$ and $hat{sigma}$ are
$$hat{sigma} =frac{s}{sqrt{1-frac{2 hat{alpha} ^2}{pi left(hat{alpha} ^2+1right)}}}$$
$$hat{mu} =bar{x}-frac{sqrt{frac{2}{pi }} hat{alpha} hat{sigma} }{sqrt{hat{alpha} ^2+1}}$$
Now armed with the parameter estimates, then one can estimate the cumulative distribution function:
$$Pr(X le x)=Phileft(frac{x, -hat{mu} }{hat{sigma} }right)-2 Tleft(frac{x, -hat{mu} }{hat{sigma} },hat{alpha} right)$$
where $T$ is the Owen's T function: $T(x,a)=frac{int_0^a frac{exp left(-left(left(t^2+1right) x^2right)right)}{2 left(t^2+1right)} , dt}{2 pi }$ .
Here is an implementation using Mathematica:
I know code should be given as text but in this case because it is unlikely that you have Mathematica (and it would look much messier as text), it should be instructive as to the process.
To estimate the percentage of the distribution no larger than a specified value you'll need to use the cumulative distribution function (CDF) described on the Wiki page. Using Mathematica for values of $X$ being 7.5 and 5.2:
If you have access to the statistical package R, then the sn package will perform these calculations.
answered Jan 7 at 16:57
JimBJimB
61547
$endgroup$
$begingroup$
Sorry, I'm not sure you actually answer my question : how can I get the percentile corresponding to one given number ?
$endgroup$
– Nisalon
Jan 7 at 17:15
$begingroup$
I've added the Mathematica code for that and mentioned thesnpackage in R.
$endgroup$
– JimB
Jan 7 at 17:57
$begingroup$
I don't want code (as this code will be integrated in PHP), I want to understand the process to be able to implement it
$endgroup$
– Nisalon
Jan 7 at 20:40
$begingroup$
Added a description of the process.
$endgroup$
– JimB
Jan 7 at 21:39
add a comment |
$begingroup$
The Wiki page Skew normal distribution provides the information to estimate the parameters using the sample mean ($bar{x}$), standard deviation ($s$), and skewness ($hat{gamma}$). The 3 parameters to be estimated are $mu$, $sigma$, and $alpha$.
If $|hat{gamma}|<1$, then $hat{alpha}$ is found in two steps:
$$delta =sqrt{frac{pi left| hat{gamma} right| ^{2/3}}{2 left(left| hat{gamma} right| ^{2/3}+left(frac{4-pi }{2}right)^{2/3}right)}}$$
$$hat{alpha} = text{sgn}(hat{gamma})sqrt{frac{delta }{1-delta ^2}}$$
Otherwise $hat{alpha}$ is the solution to
$$hat{gamma} =frac{sqrt{2} (4-pi ) hat{alpha} ^3}{left((pi -2) hat{alpha} ^2+pi right)^{3/2}}$$
which needs to be performed numerically. Then $hat{mu}$ and $hat{sigma}$ are
$$hat{sigma} =frac{s}{sqrt{1-frac{2 hat{alpha} ^2}{pi left(hat{alpha} ^2+1right)}}}$$
$$hat{mu} =bar{x}-frac{sqrt{frac{2}{pi }} hat{alpha} hat{sigma} }{sqrt{hat{alpha} ^2+1}}$$
Now armed with the parameter estimates, then one can estimate the cumulative distribution function:
$$Pr(X le x)=Phileft(frac{x, -hat{mu} }{hat{sigma} }right)-2 Tleft(frac{x, -hat{mu} }{hat{sigma} },hat{alpha} right)$$
where $T$ is the Owen's T function: $T(x,a)=frac{int_0^a frac{exp left(-left(left(t^2+1right) x^2right)right)}{2 left(t^2+1right)} , dt}{2 pi }$ .
Here is an implementation using Mathematica:
I know code should be given as text but in this case because it is unlikely that you have Mathematica (and it would look much messier as text), it should be instructive as to the process.
To estimate the percentage of the distribution no larger than a specified value you'll need to use the cumulative distribution function (CDF) described on the Wiki page. Using Mathematica for values of $X$ being 7.5 and 5.2:
If you have access to the statistical package R, then the sn package will perform these calculations.
answered Jan 7 at 16:57
JimBJimB
61547
$endgroup$
$begingroup$
Sorry, I'm not sure you actually answer my question : how can I get the percentile corresponding to one given number ?
$endgroup$
– Nisalon
Jan 7 at 17:15
$begingroup$
I've added the Mathematica code for that and mentioned thesnpackage in R.
$endgroup$
– JimB
Jan 7 at 17:57
$begingroup$
I don't want code (as this code will be integrated in PHP), I want to understand the process to be able to implement it
$endgroup$
– Nisalon
Jan 7 at 20:40
$begingroup$
Added a description of the process.
$endgroup$
– JimB
Jan 7 at 21:39
add a comment |
$begingroup$
The Wiki page Skew normal distribution provides the information to estimate the parameters using the sample mean ($bar{x}$), standard deviation ($s$), and skewness ($hat{gamma}$). The 3 parameters to be estimated are $mu$, $sigma$, and $alpha$.
If $|hat{gamma}|<1$, then $hat{alpha}$ is found in two steps:
$$delta =sqrt{frac{pi left| hat{gamma} right| ^{2/3}}{2 left(left| hat{gamma} right| ^{2/3}+left(frac{4-pi }{2}right)^{2/3}right)}}$$
$$hat{alpha} = text{sgn}(hat{gamma})sqrt{frac{delta }{1-delta ^2}}$$
Otherwise $hat{alpha}$ is the solution to
$$hat{gamma} =frac{sqrt{2} (4-pi ) hat{alpha} ^3}{left((pi -2) hat{alpha} ^2+pi right)^{3/2}}$$
which needs to be performed numerically. Then $hat{mu}$ and $hat{sigma}$ are
$$hat{sigma} =frac{s}{sqrt{1-frac{2 hat{alpha} ^2}{pi left(hat{alpha} ^2+1right)}}}$$
$$hat{mu} =bar{x}-frac{sqrt{frac{2}{pi }} hat{alpha} hat{sigma} }{sqrt{hat{alpha} ^2+1}}$$
Now armed with the parameter estimates, then one can estimate the cumulative distribution function:
$$Pr(X le x)=Phileft(frac{x, -hat{mu} }{hat{sigma} }right)-2 Tleft(frac{x, -hat{mu} }{hat{sigma} },hat{alpha} right)$$
where $T$ is the Owen's T function: $T(x,a)=frac{int_0^a frac{exp left(-left(left(t^2+1right) x^2right)right)}{2 left(t^2+1right)} , dt}{2 pi }$ .
Here is an implementation using Mathematica:
I know code should be given as text but in this case because it is unlikely that you have Mathematica (and it would look much messier as text), it should be instructive as to the process.
To estimate the percentage of the distribution no larger than a specified value you'll need to use the cumulative distribution function (CDF) described on the Wiki page. Using Mathematica for values of $X$ being 7.5 and 5.2:
If you have access to the statistical package R, then the sn package will perform these calculations.
answered Jan 7 at 16:57
JimBJimB
61547
$endgroup$
The Wiki page Skew normal distribution provides the information to estimate the parameters using the sample mean ($bar{x}$), standard deviation ($s$), and skewness ($hat{gamma}$). The 3 parameters to be estimated are $mu$, $sigma$, and $alpha$.
If $|hat{gamma}|<1$, then $hat{alpha}$ is found in two steps:
$$delta =sqrt{frac{pi left| hat{gamma} right| ^{2/3}}{2 left(left| hat{gamma} right| ^{2/3}+left(frac{4-pi }{2}right)^{2/3}right)}}$$
$$hat{alpha} = text{sgn}(hat{gamma})sqrt{frac{delta }{1-delta ^2}}$$
Otherwise $hat{alpha}$ is the solution to
$$hat{gamma} =frac{sqrt{2} (4-pi ) hat{alpha} ^3}{left((pi -2) hat{alpha} ^2+pi right)^{3/2}}$$
which needs to be performed numerically. Then $hat{mu}$ and $hat{sigma}$ are
$$hat{sigma} =frac{s}{sqrt{1-frac{2 hat{alpha} ^2}{pi left(hat{alpha} ^2+1right)}}}$$
$$hat{mu} =bar{x}-frac{sqrt{frac{2}{pi }} hat{alpha} hat{sigma} }{sqrt{hat{alpha} ^2+1}}$$
Now armed with the parameter estimates, then one can estimate the cumulative distribution function:
$$Pr(X le x)=Phileft(frac{x, -hat{mu} }{hat{sigma} }right)-2 Tleft(frac{x, -hat{mu} }{hat{sigma} },hat{alpha} right)$$
where $T$ is the Owen's T function: $T(x,a)=frac{int_0^a frac{exp left(-left(left(t^2+1right) x^2right)right)}{2 left(t^2+1right)} , dt}{2 pi }$ .
Here is an implementation using Mathematica:
I know code should be given as text but in this case because it is unlikely that you have Mathematica (and it would look much messier as text), it should be instructive as to the process.
To estimate the percentage of the distribution no larger than a specified value you'll need to use the cumulative distribution function (CDF) described on the Wiki page. Using Mathematica for values of $X$ being 7.5 and 5.2:
If you have access to the statistical package R, then the sn package will perform these calculations.
answered Jan 7 at 16:57
JimBJimB
61547
edited Jan 7 at 21:39
answered Jan 7 at 16:57
JimBJimB
61547
answered Jan 7 at 16:57
JimBJimB
61547
answered Jan 7 at 16:57
JimBJimB
61547
61547
$begingroup$
Sorry, I'm not sure you actually answer my question : how can I get the percentile corresponding to one given number ?
$endgroup$
– Nisalon
Jan 7 at 17:15
$begingroup$
I've added the Mathematica code for that and mentioned thesnpackage in R.
$endgroup$
– JimB
Jan 7 at 17:57
$begingroup$
I don't want code (as this code will be integrated in PHP), I want to understand the process to be able to implement it
$endgroup$
– Nisalon
Jan 7 at 20:40
$begingroup$
Added a description of the process.
$endgroup$
– JimB
Jan 7 at 21:39
add a comment |
$begingroup$
Sorry, I'm not sure you actually answer my question : how can I get the percentile corresponding to one given number ?
$endgroup$
– Nisalon
Jan 7 at 17:15
$begingroup$
I've added the Mathematica code for that and mentioned thesnpackage in R.
$endgroup$
– JimB
Jan 7 at 17:57
$begingroup$
I don't want code (as this code will be integrated in PHP), I want to understand the process to be able to implement it
$endgroup$
– Nisalon
Jan 7 at 20:40
$begingroup$
Added a description of the process.
$endgroup$
– JimB
Jan 7 at 21:39
$begingroup$
Sorry, I'm not sure you actually answer my question : how can I get the percentile corresponding to one given number ?
$endgroup$
– Nisalon
Jan 7 at 17:15
$begingroup$
Sorry, I'm not sure you actually answer my question : how can I get the percentile corresponding to one given number ?
$endgroup$
– Nisalon
Jan 7 at 17:15
$begingroup$
I've added the Mathematica code for that and mentioned the
sn package in R.$endgroup$
– JimB
Jan 7 at 17:57
$begingroup$
I've added the Mathematica code for that and mentioned the
sn package in R.$endgroup$
– JimB
Jan 7 at 17:57
$begingroup$
I don't want code (as this code will be integrated in PHP), I want to understand the process to be able to implement it
$endgroup$
– Nisalon
Jan 7 at 20:40
$begingroup$
I don't want code (as this code will be integrated in PHP), I want to understand the process to be able to implement it
$endgroup$
– Nisalon
Jan 7 at 20:40
$begingroup$
Added a description of the process.
$endgroup$
– JimB
Jan 7 at 21:39
$begingroup$
Added a description of the process.
$endgroup$
– JimB
Jan 7 at 21:39
add a comment |
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$begingroup$
Assuming you mean a skewed normal distribution (as defined at en.wikipedia.org/wiki/Skew_normal_distribution) you could use the method of moments to estimate the parameters of the skewed normal distribution and then estimate any desired function of that distribution. The problem is that you won't know how good your estimate of any percentile will be unless maybe you also have the sample size.
$endgroup$
– JimB
Jan 7 at 15:03
$begingroup$
I have the sample size
$endgroup$
– Nisalon
Jan 7 at 16:16
$begingroup$
If you have the sample size, you could sample from the estimated distribution to get some idea as to how good the parameter estimates might be.
$endgroup$
– JimB
Jan 7 at 16:59