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How can I solve for $t$ in $$S = 1 - \gamma(\alpha/p,(\lambda t)^p) / \Gamma (\alpha / p)$$ where $$ \gamma(s,x)=\int_0^x t^{s-1}e^{-t} dt$$

$S$ is the survival function for a generalized gamma distribution. Thank you in advance!

Milad D
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    Search “Inverse Gamma Regularized” in WolframAlpha – Тyma Gaidash Aug 26 '22 at 21:20
  • Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Aug 26 '22 at 21:25

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Define the upper incomplete Gamma function as $$\Gamma(a,z)=\int_z^\infty t^{a-1}\mathrm e^{-t}\mathrm dt \\ =\Gamma(a)-\gamma(a,z)$$ Hence $$S=1-\frac{\Gamma(a/p)-\Gamma\big(a/p,(\lambda t)^p\big)}{\Gamma(a/p)}$$ Define the regularized incomplete Gamma function, $$Q(a,z)=\frac{\Gamma(a,z)}{\Gamma(a)}$$ So $$S=Q\big(a/p,(\lambda t)^p\big)$$ Define the inverse w.r.t $z$, $$Q^{-1}(a,s)~~\text{is the solution for }z~\text{in} \\ s=Q(a,z)$$ Therefore $$Q^{-1}(a/p,S)=(\lambda t)^p \\ \frac{\big(Q^{-1}(a/p,S)\big)^{1/p}}{\lambda}=t$$ $Q^{-1}$ is implemented in the Wolfram Language as InverseGammaRegularized.

K.defaoite
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