Inverse gamma distribution definition

Question

Wikipedia says the pdf for the gamma function is:

\[ X \sim \operatorname{Gamma}(\alpha,\beta) \implies \Pr(X=x) \propto x^{\alpha-1}e^{-\beta x} \]

If $Y = 1/X$, then

\[ \Pr(Y=y) = \Pr(X=1/y) \propto (1\mathbin/y)^{\alpha-1}e^{-\beta/y} = y^{-(\alpha-1)}e^{-\beta\mathbin/y} = y^{-\alpha+1}e^{-\beta\mathbin/y} \]

However, wikipedia says of the inverse gamma distribution that

\[ \Pr(Y=y) \propto y^{-\alpha\mathbf{-1}}e^{-\beta/y} \]

Eg, -1, not +1.

Is this just a parameterization difference, or a mistake in wikipedia, or a mistake in my derivation?

Answer 1

You have made a mistake assuming that you can derive the pdf of $1/X$ the way you would do it for the discrete case. But remember this is the continuous case so solving for the variable and substituting back is not enough. You still have to multiplty by the Jacobian. And also some people might take issue with the equality sign within the probability as in the continuous setting the probability of any value is zero. But you can say something like

$$P\left( y<Y <y +\Delta y\right) = f_Y(y)$$

for sufficiently small $\Delta y$.

Answer 2

To obtain the density for a transformed continuous random variable, you need to take account of the rate of change of the original random variable with respect to the new random variable. Since $y = 1/x$ we also have $x = 1/y$ so the absolute value of the Jacobian of the transformation is:

$$\Bigg| \frac{dx}{dy} (y) \Bigg| = \Bigg| \frac{d}{dy} \frac{1}{y} \Bigg| = \Bigg| - \frac{1}{y^2} \Bigg| = \frac{1}{y^2}.$$

Thus, using the rules densities of transformed random variables you should then have:

$$\begin{aligned} p_Y(y) &= p_X(1/y) \cdot \Bigg| \frac{dx}{dy} (y) \Bigg| \\[6pt] &\propto (1/y)^{\alpha-1} e^{-\beta /y} \cdot \frac{1}{y^2} \\[6pt] &= y^{-\alpha+1} e^{-\beta /y} \cdot y^{-2} \\[6pt] &= y^{-\alpha-1} e^{-\beta /y} \\[6pt] &\propto \text{InvGamma}(y|\alpha, \beta). \\[6pt] \end{aligned}$$