Is it possible to determine shape and scale for a gamma distribution from a mean and confidence interval?

Question

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?

Answer 1

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 $40$