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The Inverse Gamma Distribution has p.d.f. defined as...Hi,
I am trying to solve the integral of the Inverse Gamma Distribution.
Does this equate to 1 as it is a pdf?
Thanks
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
$\chi$ $\sigma$
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
$\chi$ $\sigma$
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 $\alpha>0$...On reflection I was thinking - Is the same true if I am integrating over a definite integral from 0 to a constant?
Your last explanation was very helpful.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
$\chi$ $\sigma$