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?

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    $\begingroup$ Do you mean the 95% confidence interval centered on the mean? $\endgroup$ – Jack M Aug 6 '18 at 11:00
  • $\begingroup$ 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. $\endgroup$ – BGM Aug 6 '18 at 12:47
  • $\begingroup$ @BGM can you expand on that a bit? Lets say I know the SD. Where would I go from there? $\endgroup$ – Munki Fisht Aug 6 '18 at 14:04
  • $\begingroup$ @JackM Yes. The CI is centered on the mean. $\endgroup$ – Munki Fisht Aug 6 '18 at 14:05

If you know the mean is $\mu$ and the standard deviation is $\sigma$, then the shape parameter of a Gamma distribution is $\dfrac{\mu^2}{\sigma^2}$ and the scale parameter is $\dfrac{\sigma^2}{\mu}$, making the corresponding rate parameter $\dfrac{\mu}{\sigma^2}$

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 $\frac{50-40}{2}=5$, 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 $k\theta$ and the variance $k\theta^2$, suggesting with these numbers that $\theta \approx \frac{25}{40} = 0.625$ (equivalent to a rate of $1.6$) and $k \approx \frac{40^2}{25}=64$

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 $40$


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