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For a gamma probability distribution function of the form: $\frac{\beta ^{\alpha }}{\Gamma\left(\alpha \right)}x^{\alpha -1}e^{-\beta x}$,

is it sill a valid pdf if instead of $x$ we have $1/x$?

More specifically, does the density function $\frac{\beta ^{\alpha }}{\Gamma\left(\alpha \right)}(1/x)^{\alpha -1}e^{-\beta /x}$ integrate to 1 over the interval (0, $\infty$) and can we still say that E(x) = $\frac{\alpha}{\beta}$?

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  • $\begingroup$ I am not sure how to format the equation so that it shows up how I would like it to. $\endgroup$
    – Guest
    Apr 22, 2018 at 17:44
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    $\begingroup$ Put dollar signs around the whole thing; I'll edit it so you can see. Also note the gamma function is indicated by "\Gamma". $\endgroup$
    – jbowman
    Apr 22, 2018 at 17:47

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No, it's not. When you are working with probability density functions, you are really working with the derivative of the associated cumulative density function. If we have a 1-1 function, let us say $y = 1/x$, we can make the straightforward substitution $1/y$ for $x$ everywhere in the CDF, and that works, but not so with the PDF - you need to account for the change of variable by multiplying by the Jacobian of the transform, otherwise, although the variable has been transformed, the derivative (i.e., the PDF) hasn't been. The first slide of this presentation explains why very well.

Let's work through this as an example. We'll define $y = g(x) = 1/x$. The transform from $y$ back to $x$ is $g^{-1}(y)$ and $= 1/y$. The Jacobian of the transform $g^{-1}(y)$ is $|dx/dy| = 1/y^2$. Substituting $1/y$ for $x$ in the expression for the Gamma density function and multiplying by the Jacobian gives us:

$$f(y) = y^{-2}{\beta^{\alpha} \over \Gamma(\alpha)}y^{1-\alpha}e^{-\beta/y} ={\beta^{\alpha} \over \Gamma(\alpha)}y^{-1-\alpha}e^{-\beta/y} $$

which is the density function of the inverted Gamma distribution. As the link shows, the mean of $y$ is $\beta / (\alpha - 1)$ but exists only when $\alpha > 0$.

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    $\begingroup$ So, I would be able to say, in theory, that if I arrived at a density function of the form of f(y) above, then I arrived at the density for an inverted Gamma distribution and could conclude that the integral of that function over its support is 1 since it is a valid pdf? $\endgroup$
    – Guest
    Apr 22, 2018 at 19:00
  • $\begingroup$ That's correct. $\endgroup$
    – jbowman
    Apr 22, 2018 at 19:03
  • $\begingroup$ Thank you so much for explaining this! You have cleared up a lot. $\endgroup$
    – Guest
    Apr 22, 2018 at 19:06
  • $\begingroup$ If you think that basically answered your question, please accept the answer - it'll help inform others that the question has been answered acceptably well. Thanks! $\endgroup$
    – jbowman
    Apr 22, 2018 at 19:25
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    $\begingroup$ Do I just accept by clicking the green check mark? $\endgroup$
    – Guest
    Apr 22, 2018 at 19:31

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