Mean of $1/X$ for $X\sim Beta( \alpha,\beta)$

Question

Having a hard time calculating $\mathbb{E}[\frac{1}{X}]$ for $X\sim Beta( \alpha,\beta)$

I know that $$\mathbb{E}[\frac{1}{X}] = \int_0^1 x^{-1}\frac{x^{\alpha-1}(1-x)^{\beta-1 }}{B(\alpha,\beta)}dx = \int_0^1 \frac{x^{\alpha-2}(1-x)^{\beta-1 }}{B(\alpha,\beta)}dx $$

But I'm having a hard time getting past that integral. Also, since $\alpha>0$ by definition, does this mean the integral is undefined for $\alpha<1$?

Answer 1

The $\text{Beta}(\alpha, \beta)$ PDF is given by $$f(x) = \dfrac{1}{B(\alpha, \beta)}x^{\alpha - 1}(1-x)^{\beta - 1} = \dfrac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha - 1}(1-x)^{\beta - 1} $$ for $x \in (0, 1)$, where it is assumed that $\alpha, \beta > 0$.

Now $$\mathbb{E}\left[\dfrac{1}{X}\right]=\int_{0}^{1}\dfrac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-2}(1-x)^{\beta - 1}\text{ d}x$$ as you mention. The next step is to attempt to transform the integrand into a beta-distribution PDF. To make the parametrization clear, let $\alpha^{*} = \alpha - 1$, so that $$\mathbb{E}\left[\dfrac{1}{X}\right]=\int_{0}^{1}\dfrac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha^{*}-1}(1-x)^{\beta - 1}\text{ d}x$$ The integrand looks very similar to a $\text{Beta}(\alpha^{*}, \beta)$ pdf. How can we transform it as such?

Recall the Gamma function recursion property, i.e., $\Gamma(x + 1) = x\Gamma(x)$. Observe that $\alpha = \alpha^{*}+1$. Thus, it follows that $$\Gamma(\alpha) = \Gamma(\alpha^{*}+1)=\alpha^{*}\Gamma(\alpha^{*})\text{.}$$ Furthermore, $$\Gamma(\alpha+\beta)=\Gamma(\alpha^{*}+\beta+1)=(\alpha^{*}+\beta)\Gamma(\alpha^{*}+\beta)\text{.}$$ Thus, $$\begin{align} \mathbb{E}\left[\dfrac{1}{X}\right]&=\int_{0}^{1}\dfrac{(\alpha^{*}+\beta)\Gamma(\alpha^{*}+\beta)}{\alpha^{*}\Gamma(\alpha^{*})\Gamma(\beta)}x^{\alpha^{*}-1}(1-x)^{\beta - 1}\text{ d}x \\ &=\dfrac{\alpha^{*}+\beta}{\alpha^{*}}\int_{0}^{1}\dfrac{\Gamma(\alpha^{*})\Gamma(\beta)}{\Gamma(\alpha^{*}+\beta)}x^{\alpha^{*}-1}(1-x)^{\beta - 1}\text{ d}x \end{align}$$ and assuming $\alpha^{*} > 0$ (or $\alpha > 1$), the integral above is thus $1$. Thus, $$\mathbb{E}\left[\dfrac{1}{X}\right] = \dfrac{\alpha^{*}+\beta}{\alpha^{*}} = \dfrac{\alpha+\beta-1}{\alpha - 1}$$ as long as $\alpha > 1$ and $\beta > 0$.

Answer 2

Use the following characterization of the beta distribution, namely, $X\sim \text{Beta}(\alpha, \beta)$ if $$ X\stackrel{d}{=} \frac{Z}{Z+W}; \quad Z\sim\text{Gamma}(\alpha),W\sim \text{Gamma}(\beta)\tag{0} $$ and $Z\perp W$ (i.e. Z is independent of W). Further $X\perp (Z+W)$ and $Z+W\sim \text{Gamma}(\alpha +\beta)$. Here $Z\sim \text{Gamma}(\alpha)$ means that distribution of $Z$ is induced by the pdf $$ f_{Z}(z)=\frac{1}{\Gamma(\alpha)}z^{\alpha-1}e^{-z}\quad(z>0, \alpha>0)\tag{1}. $$ From the definition of the gamma function and (1) it is clear that $EZ^d=\Gamma(\alpha+d)/\Gamma(\alpha)$ provided that $\alpha>-d$. Since $X\perp(Z+W)$ it follows at once that $$ EX^{-1}=\frac{EZ^{-1}}{E(Z+W)^{-1}}=\frac{\Gamma(\alpha-1)/\Gamma(\alpha)}{\Gamma(\alpha+\beta-1)/\Gamma(\alpha+\beta)}=\frac{1/(\alpha-1)}{1/(\alpha+\beta-1)}=\frac{\alpha+\beta-1}{\alpha-1} $$ provided that $\alpha>1, \beta>0$ . For more info see this stats stackexchange link and this wikipedia link