# Is it possible to calculate the scale param θ for the Gamma distribution, given shape param K and a quantile value Q?

Stats-ophils,

I am running into a problem, in which I'd like to generate a Gamma distribution (in Julia) and I know the value of the quantile Q(0.9) = 130 as well as the shape parameter k=2.

Is it possible to come up with a function to calculate the shape parameter θ?

Best A stats-inclined

Yes it's possible.

Suppose that $$X\sim \mathrm{Gamma}(\alpha, \beta)$$ where $$\alpha$$ is the shape parameter and $$\beta$$ the scale parameter.

Define the incomplete gamma function as $$\Gamma(a, z)=\int_{z}^{\infty}t^{a-1}\mathrm{e}^{-t}\;\mathrm{d}t$$

and the generalized incomplete gamma function as $$\Gamma(a, z_0, z_1)=\int_{z_0}^{z_1}t^{a-1}\mathrm{e}^{-t}\;\mathrm{d}t = \Gamma(a,z_0)-\Gamma(a,z_1).$$

Further, define the generalized regularized/normalized incomplete gamma function as $$Q(a,z_0,z_1)=\frac{\Gamma(a, z_0, z_1)}{\Gamma(a)}$$ Finally, the inverse gamma regularized/normalized function is the solution for $$z_1$$ in $$s = Q(a, z_0, z_1)$$

The $$q$$-quantile of a Gamma distribution is given by $$\beta\;\mathrm{InverseGammaRegularized(\alpha, 0, q)}$$. Assuming that the $$q$$-quantile is $$k$$, solving for $$\beta$$ we have $$\beta = \frac{k}{\mathrm{InverseGammaRegularized(\alpha, 0, q)}}$$ For your example $$\beta = \frac{130}{\mathrm{InverseGammaRegularized(2, 0, 0.9)}}=33.4214$$

I'm not sure if this function is implemented in Julia.

• Thank you! This helps a lot, understanding-wise. I have meanwhile built a solution based on a linear model fit. I fear that the Inverse Gamma Regularized function is not implemented in Julia.. – dmeu Jul 14 at 9:56