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- Feb 13, 2012

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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

$\displaystyle f(x;\alpha,\beta)= \frac{\beta^{\alpha}}{\Gamma(\alpha)}\ x^{-(1+\alpha)}\ e^{- \frac{\beta}{x}},\ x>0$ (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

$\chi$ $\sigma$

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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

$\chi$ $\sigma$

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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

$\chi$ $\sigma$

- Feb 13, 2012

- 1,704

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?

$\displaystyle P\{X>\gamma\}= \frac{\beta^{\alpha}}{\Gamma(\alpha)}\ \int_{\gamma}^{\infty} x^{-(1+\alpha)}\ e^{-\frac{\beta}{x}}\ dx = \frac{1}{\Gamma(\alpha)}\ \int_{0}^{\frac{\beta}{\gamma}} \xi^{\alpha-1}\ e^{- \xi}\ d \xi =$

$\displaystyle = \frac{\beta^{\alpha}}{\gamma^{\alpha}\ \Gamma(\alpha)}\ \sum_{n=0}^{\infty} (-1)^{n}\ \frac{(\frac{\beta}{\gamma})^{n}}{(n+\alpha)\ n!}$ (1)

Kind regards

$\chi$ $\sigma$

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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

$\chi$ $\sigma$

I would like however to compute the probability $P\{ X< \gamma\}$

Can I follow an approach similar to yours above?

I had been thinking of following the following approach:

P1=integral(A(x)) over [0,x] where A(x) is the inverse gamma distribution function.

Integrating over [0,x] will get the cdf however this does not exist in closed form.

Hence, to compute this I can use the Gamma distribution cdf and a transformation. So if B has the Gamma distribution then C=1/B has the inverse Gamma distribution.

F(x)= P(C<=x)=P(1/B <=x)

=P(1/x<=B)

=1-P(B<1/x)

=1-F(1/x)

Hence I am finding the Gamma cdf and subtracting it from 1.

Any thoughts?