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Suppose $X\sim \text{Gamma}(\alpha, \beta). $ Determine the distribution of $Y = 1/X.$

This is what I reached as a solution: the distribution of $Y = 1/X$ is a gamma distribution with shape parameter $\alpha+1$ and scale parameter $1/(\beta\cdot y).$

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    $\begingroup$ Hint: stats.stackexchange.com/questions/277567/… $\endgroup$
    – kjetil b halvorsen ♦
    Commented Dec 26, 2022 at 20:11
  • $\begingroup$ Note that your scale parameter is a function of $y$, so it isn't a parameter at all! $\endgroup$
    – jbowman
    Commented Dec 27, 2022 at 2:14
  • $\begingroup$ I have added the self-study tag. From future, please do the same for homework problems by yourself. Also, whatever the attempt is, you should show so that the community may identify the mistake other wise this question is meaningless. $\endgroup$
    – User1865345
    Commented Dec 27, 2022 at 3:13

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With $$ Y = 1/X \overset{X >0}{\iff} X = 1/Y, \\ \frac{\mathrm d \, 1/y}{\mathrm d \, y} = -\frac{1}{y^2}, $$ and using the shape-rate parameterization, a change of variables yields $$ \begin{align} \mathop{f_Y}\left(y\right) &= \mathop{\text{Gamma}}\left(\frac 1 y;\alpha, \beta\right) \cdot \left| \frac{\mathrm d \, 1/y}{\mathrm d \, y} \right| \\ &= \frac{\beta^\alpha}{\mathop{\Gamma}\left(\alpha\right)} \left(\frac 1 y\right)^{\alpha -1} \exp\left(-\beta \cdot \frac 1 y\right) \cdot \left|-\frac{1}{y^2}\right| \\ &= \frac{\beta^\alpha}{\mathop{\Gamma}\left(\alpha\right)} y^{-\alpha - 1} \exp\left(-\frac \beta y\right) \;\; \forall \, y \in \mathbb{R}_{>0} \end{align} $$ for the probability density function $\mathop{f_Y}$ of $Y$.

What distribution does $\mathop{f_Y}$ correspond to?

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  • $\begingroup$ Added a minor format change, if you don't mind. +1. $\endgroup$
    – User1865345
    Commented Dec 27, 2022 at 9:25
  • $\begingroup$ @User1865345 Why didn't you like the version with $1/y$ in the numerator? $\endgroup$
    – statmerkur
    Commented Dec 27, 2022 at 9:33
  • $\begingroup$ I neither disliked it, nor it was wrong - just that the latter version is more clear in what the differential operator is working on. But again you can rollback if you want :-) $\endgroup$
    – User1865345
    Commented Dec 27, 2022 at 9:35
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    $\begingroup$ @User1865345 I think in this case it's really a matter of taste, so rolled back. But I appreciate the effort you put into formatting so many old and new posts! $\endgroup$
    – statmerkur
    Commented Dec 27, 2022 at 9:44
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    $\begingroup$ Not a problem! Appreciate your feedback and posts. $\endgroup$
    – User1865345
    Commented Dec 27, 2022 at 9:46

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