The integral of the Inverse Gamma Distribution's probability density function (pdf) is confirmed to equal 1, as demonstrated through the substitution method involving the Gamma function. The pdf is defined as f(x;α,β) = (β^α / Γ(α)) x^(-(1+α)) e^(-β/x) for x > 0. The integral, when evaluated, simplifies to Γ(α)/Γ(α) = 1. For definite integrals, the probability P{X > γ} can be computed using the derived formula involving the Gamma function, confirming that integration over a range is feasible for α > 0.
PREREQUISITESStatisticians, data scientists, and researchers involved in probability theory and statistical modeling will benefit from this discussion, particularly those working with the Inverse Gamma Distribution and related statistical computations.
statty said:Hi,
I am trying to solve the integral of the Inverse Gamma Distribution.
Does this equate to 1 as it is a pdf?
Thanks
chisigma said:The Inverse Gamma Distribution has p.d.f. defined as...
$\displaystyle f(x;\alpha,\beta)= \frac{\beta^{\alpha}}{\Gamma(\alpha)}\ x^{-(1+\alpha)}\ e^{- \frac{\beta}{x}},\ x>0$ (1)
... where $\alpha$ and $\beta$ are constant. The integral of (1), operating the substitution $\displaystyle \frac{\beta}{x}=\xi$ is...
$\displaystyle \frac{\beta^{\alpha}}{\Gamma(\alpha)}\ \int_{0}^{\infty} x^{-(1+\alpha)}\ e^{- \frac{\beta}{x}}\ dx = \frac{\beta^{\alpha}}{\Gamma(\alpha)}\ \beta^{- \alpha} \int_{0}^{\infty} \xi^{\alpha-1}\ e^{- \xi}\ d \xi = \frac {\Gamma(\alpha)}{\Gamma(\alpha)}=1 $
Kind regards
$\chi$ $\sigma$
chisigma said:The Inverse Gamma Distribution has p.d.f. defined as...
$\displaystyle f(x;\alpha,\beta)= \frac{\beta^{\alpha}}{\Gamma(\alpha)}\ x^{-(1+\alpha)}\ e^{- \frac{\beta}{x}},\ x>0$ (1)
... where $\alpha$ and $\beta$ are constant. The integral of (1), operating the substitution $\displaystyle \frac{\beta}{x}=\xi$ is...
$\displaystyle \frac{\beta^{\alpha}}{\Gamma(\alpha)}\ \int_{0}^{\infty} x^{-(1+\alpha)}\ e^{- \frac{\beta}{x}}\ dx = \frac{\beta^{\alpha}}{\Gamma(\alpha)}\ \beta^{- \alpha} \int_{0}^{\infty} \xi^{\alpha-1}\ e^{- \xi}\ d \xi = \frac {\Gamma(\alpha)}{\Gamma(\alpha)}=1 $
Kind regards
$\chi$ $\sigma$
statty said:On reflection I was thinking - Is the same true if I am integrating over a definite integral from 0 to a constant?
chisigma said:I think You have in mind to compute, given a r.v. X which is 'Inverse Gamma', the probability $P\{ X> \gamma\}$ or something like that. This is perfectly possible for $\alpha>0$...
$\displaystyle P\{X>\gamma\}= \frac{\beta^{\alpha}}{\Gamma(\alpha)}\ \int_{\gamma}^{\infty} x^{-(1+\alpha)}\ e^{-\frac{\beta}{x}}\ dx = \frac{1}{\Gamma(\alpha)}\ \int_{0}^{\frac{\beta}{\gamma}} \xi^{\alpha-1}\ e^{- \xi}\ d \xi =$
$\displaystyle = \frac{\beta^{\alpha}}{\gamma^{\alpha}\ \Gamma(\alpha)}\ \sum_{n=0}^{\infty} (-1)^{n}\ \frac{(\frac{\beta}{\gamma})^{n}}{(n+\alpha)\ n!}$ (1)
Kind regards
$\chi$ $\sigma$