MHB Integral of the Inverse Gamma Distribution

AI Thread Summary
The discussion centers around the integral of the Inverse Gamma Distribution's probability density function (pdf) and its implications. It confirms that the integral of the pdf equals 1, validating it as a proper distribution. The conversation then shifts to calculating probabilities for a random variable following the Inverse Gamma Distribution, specifically for values greater than or less than a constant. A method is proposed for finding the cumulative distribution function (cdf) by relating it to the Gamma distribution, as the cdf of the Inverse Gamma does not have a closed form. The participants exchange ideas on the validity and approach of these calculations.
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Hi,
I am trying to solve the integral of the Inverse Gamma Distribution.
Does this equate to 1 as it is a pdf?
Thanks
 
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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

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

$\chi$ $\sigma$
 
Thank you very much!
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

$\chi$ $\sigma$
 
On reflection I was thinking - Is the same true if I am integrating over a definite integral from 0 to a constant?

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?

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

$\chi$ $\sigma$
 
Last edited:
Thank you for your reply, that is most helpful
 
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

$\chi$ $\sigma$

Your last explanation was very helpful.
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?
 
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