Mixing
Is it possible to determine shape and scale for a gamma distribution from a mean and confidence interval?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Having the 95% confidence interval and mean for a distribution and knowing nothing else (other than the data is skewed and will likely follow a gamma distribution) is there any way to determine the shape and scale of that gamma distribution? If not, what are the minimum data you would need to determine these?
probability probability-distributions gamma-distribution
asked Aug 6 at 10:47
Munki Fisht
111
add a comment |Â
up vote
2
down vote
favorite
Having the 95% confidence interval and mean for a distribution and knowing nothing else (other than the data is skewed and will likely follow a gamma distribution) is there any way to determine the shape and scale of that gamma distribution? If not, what are the minimum data you would need to determine these?
probability probability-distributions gamma-distribution
asked Aug 6 at 10:47
Munki Fisht
111
2
Do you mean the 95% confidence interval centered on the mean?
– Jack M
Aug 6 at 11:00
In general, if you have two unknowns, you need two independent equations to form a system to solve them. In your case, if you have functions of point estimatiors (the confidence limits) and equate to the realizations, you can solve the point estimates of the parameters out.
– BGM
Aug 6 at 12:47
@BGM can you expand on that a bit? Lets say I know the SD. Where would I go from there?
– Munki Fisht
Aug 6 at 14:04
@JackM Yes. The CI is centered on the mean.
– Munki Fisht
Aug 6 at 14:05
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Having the 95% confidence interval and mean for a distribution and knowing nothing else (other than the data is skewed and will likely follow a gamma distribution) is there any way to determine the shape and scale of that gamma distribution? If not, what are the minimum data you would need to determine these?
probability probability-distributions gamma-distribution
asked Aug 6 at 10:47
Munki Fisht
111
Having the 95% confidence interval and mean for a distribution and knowing nothing else (other than the data is skewed and will likely follow a gamma distribution) is there any way to determine the shape and scale of that gamma distribution? If not, what are the minimum data you would need to determine these?
probability probability-distributions gamma-distribution
asked Aug 6 at 10:47
Munki Fisht
111
asked Aug 6 at 10:47
Munki Fisht
111
asked Aug 6 at 10:47
Munki Fisht
111
asked Aug 6 at 10:47
Munki Fisht
111
111
2
Do you mean the 95% confidence interval centered on the mean?
– Jack M
Aug 6 at 11:00
In general, if you have two unknowns, you need two independent equations to form a system to solve them. In your case, if you have functions of point estimatiors (the confidence limits) and equate to the realizations, you can solve the point estimates of the parameters out.
– BGM
Aug 6 at 12:47
@BGM can you expand on that a bit? Lets say I know the SD. Where would I go from there?
– Munki Fisht
Aug 6 at 14:04
@JackM Yes. The CI is centered on the mean.
– Munki Fisht
Aug 6 at 14:05
add a comment |Â
2
Do you mean the 95% confidence interval centered on the mean?
– Jack M
Aug 6 at 11:00
In general, if you have two unknowns, you need two independent equations to form a system to solve them. In your case, if you have functions of point estimatiors (the confidence limits) and equate to the realizations, you can solve the point estimates of the parameters out.
– BGM
Aug 6 at 12:47
@BGM can you expand on that a bit? Lets say I know the SD. Where would I go from there?
– Munki Fisht
Aug 6 at 14:04
@JackM Yes. The CI is centered on the mean.
– Munki Fisht
Aug 6 at 14:05
2
2
Do you mean the 95% confidence interval centered on the mean?
– Jack M
Aug 6 at 11:00
Do you mean the 95% confidence interval centered on the mean?
– Jack M
Aug 6 at 11:00
In general, if you have two unknowns, you need two independent equations to form a system to solve them. In your case, if you have functions of point estimatiors (the confidence limits) and equate to the realizations, you can solve the point estimates of the parameters out.
– BGM
Aug 6 at 12:47
In general, if you have two unknowns, you need two independent equations to form a system to solve them. In your case, if you have functions of point estimatiors (the confidence limits) and equate to the realizations, you can solve the point estimates of the parameters out.
– BGM
Aug 6 at 12:47
@BGM can you expand on that a bit? Lets say I know the SD. Where would I go from there?
– Munki Fisht
Aug 6 at 14:04
@BGM can you expand on that a bit? Lets say I know the SD. Where would I go from there?
– Munki Fisht
Aug 6 at 14:04
@JackM Yes. The CI is centered on the mean.
– Munki Fisht
Aug 6 at 14:05
@JackM Yes. The CI is centered on the mean.
– Munki Fisht
Aug 6 at 14:05
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
If you know the mean is $mu$ and the standard deviation is $sigma$, then the shape parameter of a Gamma distribution is $dfracmu^2sigma^2$ and the scale parameter is $dfracsigma^2mu$, making the corresponding rate parameter $dfracmusigma^2$
As an illustration of what is possible, suppose you knew that the mean is $40$ and you had an interval of $[30,50]$ representing about $2$ standard deviations either side of the mean
Then the standard deviation is about $frac50-402=5$, and the variance is therefore about $5^2=25$
For a Gamma distribution with shape parameter $k$ and scale parameter $theta$, the mean would be $ktheta$ and the variance $ktheta^2$, suggesting with these numbers that $theta approx frac2540 = 0.625$ (equivalent to a rate of $1.6$) and $k approx frac40^225=64$
As a check, we can look at the corresponding interval for these parameters in R
> pgamma(50,shape=64,scale=0.625) - pgamma(30,shape=64,scale=0.625)
[1] 0.9553145
> c(qgamma(0.025,shape=64,scale=0.625),qgamma(0.975,shape=64,scale=0.625))
[1] 30.80487 50.37773
which shows this approach is not exact, but is not that far away.
$k=59.3749$ and $theta=0.66312$ would get you closer to the confidence interval with $2.5%$ each side but at the cost (due to the asymmetry of the Gamma distribution) of a corresponding mean of $39.372$ rather than $40$
answered Aug 6 at 14:47
Henry
93.1k469147
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If you know the mean is $mu$ and the standard deviation is $sigma$, then the shape parameter of a Gamma distribution is $dfracmu^2sigma^2$ and the scale parameter is $dfracsigma^2mu$, making the corresponding rate parameter $dfracmusigma^2$
As an illustration of what is possible, suppose you knew that the mean is $40$ and you had an interval of $[30,50]$ representing about $2$ standard deviations either side of the mean
Then the standard deviation is about $frac50-402=5$, and the variance is therefore about $5^2=25$
For a Gamma distribution with shape parameter $k$ and scale parameter $theta$, the mean would be $ktheta$ and the variance $ktheta^2$, suggesting with these numbers that $theta approx frac2540 = 0.625$ (equivalent to a rate of $1.6$) and $k approx frac40^225=64$
As a check, we can look at the corresponding interval for these parameters in R
> pgamma(50,shape=64,scale=0.625) - pgamma(30,shape=64,scale=0.625)
[1] 0.9553145
> c(qgamma(0.025,shape=64,scale=0.625),qgamma(0.975,shape=64,scale=0.625))
[1] 30.80487 50.37773
which shows this approach is not exact, but is not that far away.
$k=59.3749$ and $theta=0.66312$ would get you closer to the confidence interval with $2.5%$ each side but at the cost (due to the asymmetry of the Gamma distribution) of a corresponding mean of $39.372$ rather than $40$
answered Aug 6 at 14:47
Henry
93.1k469147
add a comment |Â
up vote
0
down vote
If you know the mean is $mu$ and the standard deviation is $sigma$, then the shape parameter of a Gamma distribution is $dfracmu^2sigma^2$ and the scale parameter is $dfracsigma^2mu$, making the corresponding rate parameter $dfracmusigma^2$
As an illustration of what is possible, suppose you knew that the mean is $40$ and you had an interval of $[30,50]$ representing about $2$ standard deviations either side of the mean
Then the standard deviation is about $frac50-402=5$, and the variance is therefore about $5^2=25$
For a Gamma distribution with shape parameter $k$ and scale parameter $theta$, the mean would be $ktheta$ and the variance $ktheta^2$, suggesting with these numbers that $theta approx frac2540 = 0.625$ (equivalent to a rate of $1.6$) and $k approx frac40^225=64$
As a check, we can look at the corresponding interval for these parameters in R
> pgamma(50,shape=64,scale=0.625) - pgamma(30,shape=64,scale=0.625)
[1] 0.9553145
> c(qgamma(0.025,shape=64,scale=0.625),qgamma(0.975,shape=64,scale=0.625))
[1] 30.80487 50.37773
which shows this approach is not exact, but is not that far away.
$k=59.3749$ and $theta=0.66312$ would get you closer to the confidence interval with $2.5%$ each side but at the cost (due to the asymmetry of the Gamma distribution) of a corresponding mean of $39.372$ rather than $40$
answered Aug 6 at 14:47
Henry
93.1k469147
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If you know the mean is $mu$ and the standard deviation is $sigma$, then the shape parameter of a Gamma distribution is $dfracmu^2sigma^2$ and the scale parameter is $dfracsigma^2mu$, making the corresponding rate parameter $dfracmusigma^2$
As an illustration of what is possible, suppose you knew that the mean is $40$ and you had an interval of $[30,50]$ representing about $2$ standard deviations either side of the mean
Then the standard deviation is about $frac50-402=5$, and the variance is therefore about $5^2=25$
For a Gamma distribution with shape parameter $k$ and scale parameter $theta$, the mean would be $ktheta$ and the variance $ktheta^2$, suggesting with these numbers that $theta approx frac2540 = 0.625$ (equivalent to a rate of $1.6$) and $k approx frac40^225=64$
As a check, we can look at the corresponding interval for these parameters in R
> pgamma(50,shape=64,scale=0.625) - pgamma(30,shape=64,scale=0.625)
[1] 0.9553145
> c(qgamma(0.025,shape=64,scale=0.625),qgamma(0.975,shape=64,scale=0.625))
[1] 30.80487 50.37773
which shows this approach is not exact, but is not that far away.
$k=59.3749$ and $theta=0.66312$ would get you closer to the confidence interval with $2.5%$ each side but at the cost (due to the asymmetry of the Gamma distribution) of a corresponding mean of $39.372$ rather than $40$
answered Aug 6 at 14:47
Henry
93.1k469147
If you know the mean is $mu$ and the standard deviation is $sigma$, then the shape parameter of a Gamma distribution is $dfracmu^2sigma^2$ and the scale parameter is $dfracsigma^2mu$, making the corresponding rate parameter $dfracmusigma^2$
As an illustration of what is possible, suppose you knew that the mean is $40$ and you had an interval of $[30,50]$ representing about $2$ standard deviations either side of the mean
Then the standard deviation is about $frac50-402=5$, and the variance is therefore about $5^2=25$
For a Gamma distribution with shape parameter $k$ and scale parameter $theta$, the mean would be $ktheta$ and the variance $ktheta^2$, suggesting with these numbers that $theta approx frac2540 = 0.625$ (equivalent to a rate of $1.6$) and $k approx frac40^225=64$
As a check, we can look at the corresponding interval for these parameters in R
> pgamma(50,shape=64,scale=0.625) - pgamma(30,shape=64,scale=0.625)
[1] 0.9553145
> c(qgamma(0.025,shape=64,scale=0.625),qgamma(0.975,shape=64,scale=0.625))
[1] 30.80487 50.37773
which shows this approach is not exact, but is not that far away.
$k=59.3749$ and $theta=0.66312$ would get you closer to the confidence interval with $2.5%$ each side but at the cost (due to the asymmetry of the Gamma distribution) of a corresponding mean of $39.372$ rather than $40$
answered Aug 6 at 14:47
Henry
93.1k469147
answered Aug 6 at 14:47
Henry
93.1k469147
answered Aug 6 at 14:47
Henry
93.1k469147
answered Aug 6 at 14:47
Henry
93.1k469147
93.1k469147
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873763%2fis-it-possible-to-determine-shape-and-scale-for-a-gamma-distribution-from-a-mean%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
Do you mean the 95% confidence interval centered on the mean?
– Jack M
Aug 6 at 11:00
In general, if you have two unknowns, you need two independent equations to form a system to solve them. In your case, if you have functions of point estimatiors (the confidence limits) and equate to the realizations, you can solve the point estimates of the parameters out.
– BGM
Aug 6 at 12:47
@BGM can you expand on that a bit? Lets say I know the SD. Where would I go from there?
– Munki Fisht
Aug 6 at 14:04
@JackM Yes. The CI is centered on the mean.
– Munki Fisht
Aug 6 at 14:05