Inverse gamma distribution
If \(X\sim\text{Gamma}(\alpha,\,\beta)\), then the new random variable \(Y=1/X\) has the inverse gamma distribution. We denote this \(Y\sim\text{IG}(\alpha,\,\beta)\). You derived this on Problem Set 6 #6d.
Basic properties
| Notation | \(X\sim\text{IG}(\alpha,\,\beta)\) |
| Range | \((0,\,\infty)\) |
| Parameter space | \(\alpha,\,\beta>0\) |
| \(f(x)=\begin{cases}\frac{\beta^\alpha}{\Gamma(\alpha)}x^{-\alpha-1}e^{-\beta /x} &x> 0\\0 & x\leq0.\end{cases}\) | |
| Expectation | \(\frac{\beta}{\alpha-1}\) if \(\alpha>1\) |
| Variance | \(\frac{\beta^2}{(\alpha-1)^2(\alpha-2)}\) if \(\alpha>2\) |
Derivations
Let \(X\sim\text{Gamma}(\alpha,\,\beta)\) and consider \(Y=1/X\). So the forward transformation is \(g(x) = 1/ x\), and the inverse is \(g^{-1}(y) = 1/ y\). These are strictly decreasing, and they map positive numbers to positive numbers. So \(\text{Range}(Y)=(0,\, \infty)\). Fix \(y>0\). The change-of-variables formula says:
\[ \begin{aligned} f_Y(y) &= f_X \left( g^{-1}(y) \right) \left| \frac{d}{dy} g^{-1}(y) \right| \\[6pt] &= \frac{\beta^\alpha}{\Gamma(\alpha)} \left( \frac{1}{y} \right)^{\alpha-1} \exp \left( \frac{-\beta}{y} \right) \left|-\frac{1}{y^2}\right| \\[6pt] &= \frac{\beta^\alpha}{\Gamma(\alpha)} \left( \frac{1}{y} \right)^{\alpha-1} \exp \left( \frac{-\beta}{y} \right) \frac{1}{y^2} \\[6pt] &= \frac{\beta^\alpha}{\Gamma(\alpha)} \left( \frac{1}{y} \right)^{\alpha+1} \exp \left( \frac{-\beta}{y} \right) \\[6pt] &= \frac{\beta^\alpha}{\Gamma(\alpha)} y^{-\alpha-1} \exp \left( \frac{-\beta}{y} \right),&& y>0. \end{aligned} \]
Let \(X\sim\text{Gamma}(\alpha,\,\beta)\) and \(Y=1/X\sim\text{IG}(\alpha,\,\beta)\). Furthermore, assume \(\alpha > k\) for some \(k\in\mathbb{N}\). Then the \(k\)th raw moment of the inverse gamma variate \(Y\) is
\[ \begin{aligned} E(Y^k) &= E\left(\frac{1}{X^k}\right) \\ &= \int_0^\infty \frac{1}{x^k} f_X(x) \,\text{d}x \\ &= \int_0^\infty \frac{1}{x^k} \frac{\beta^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x} \,\text{d}x \\ &= \frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty \underbrace{x^{\alpha-k-1}e^{-\beta x}}_{\text{kernel of Gamma}(\alpha-k,\,\beta)} \,\text{d}x \\ &= \frac{\beta^\alpha}{\Gamma(\alpha)} \frac{\Gamma(\alpha-k)}{\beta^{\alpha-k}} \\ &= \beta^k\frac{\Gamma(\alpha-k)}{\Gamma(\alpha)}. \end{aligned} \]
So
\[ \begin{aligned} E(Y) &= \beta\frac{\Gamma(\alpha-1)}{\Gamma(\alpha)}=\beta\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-1+1)}=\beta\frac{\Gamma(\alpha-1)}{(\alpha-1)\Gamma(\alpha-1)}=\frac{\beta}{\alpha-1} \\ \\ E(Y^2) &= \beta^2\frac{\Gamma(\alpha-2)}{\Gamma(\alpha)} \\ &= \beta^2\frac{\Gamma(\alpha-2)}{\Gamma(\alpha-1+1)} \\ &= \beta^2\frac{\Gamma(\alpha-2)}{(\alpha-1)\Gamma(\alpha-1)} \\ &= \beta^2\frac{\Gamma(\alpha-2)}{(\alpha-1)\Gamma(\alpha-2+1)} \\ &= \beta^2\frac{\Gamma(\alpha-2)}{(\alpha-1)(\alpha-2)\Gamma(\alpha-2)} \\ &= \frac{\beta^2}{(\alpha-1)(\alpha-2)} \\ \\ \text{var}(Y) &= E(Y^2)-E(Y)^2 \\ &= \frac{\beta^2}{(\alpha-1)(\alpha-2)} - \frac{\beta^2}{(\alpha-1)^2} \\ &= \beta^2 \left[ \frac{1}{(\alpha-1)(\alpha-2)} - \frac{1}{(\alpha-1)^2} \right] \\ &= \beta^2 \frac{\alpha-1-(\alpha-2)}{(\alpha-1)^2(\alpha-2)} \\ &= \beta^2 \frac{\alpha-1-\alpha+2}{(\alpha-1)^2(\alpha-2)} \\ &= \frac{\beta^2}{(\alpha-1)^2(\alpha-2)} . \end{aligned} \]