Method of moments estimator for distribution with density $p_{\theta}(x)=\theta(1+x)^{-(1+\theta)}$

Question

Let $X_{1},...,X_{n}$ be a sample from probability distribution with density $p_{\theta}(x)=\theta(1+x)^{-(1+\theta)}$ with $x>0$ and $0$ elsewhere, with $\theta>1$ unknown. Determine the method of moments estimator for $\theta$.

I let $\hat{\theta}_{MM}=E(p_{\theta}(x))$ Thus I need the find the expectation of the density function. Since it is continuous, I construct

$$E(p_{\theta}(x))=\int\limits_{0}^{\infty} x \theta(1+x)^{-(1+\theta)}dx,$$ but when I try to solve this by integration by parts I get stuck. I let $u=x$ and $dV=\theta(1+x)^{-(1+\theta)}$, which doesn't work.

Where do I go wrong? Any help, hints or suggestions would much be appreciated!

Answer 1

Integration by parts works. If $dv=\theta(1+x)^{-(1+\theta)}$, then $v=-(1+x)^{-\theta}$, and $$ \int_0^{\infty} x \theta(1+x)^{-(1+\theta)}dx=-\frac x{(1+x)^\theta}\,\Big|_0^\infty+\int_0^\infty(1+x)^{-\theta}\,dx\tag1 $$ Check that the first term on the RHS of (1) equals zero, while the second term equals $$ \frac{(1+x)^{-\theta+1}}{-\theta+1}\,\Big|_0^\infty=\frac1{\theta-1}. $$

Answer 2

The density $p_{\theta}$ is the pdf of a Beta distribution of the second kind with parameter $(1,\theta)$.

The population mean is calculated as

\begin{align} \mu&=\int_0^\infty \frac{x\theta}{(1+x)^{1+\theta}}\,dx \\\\&=\theta\int_0^\infty\frac{x^{\color{red}{2}-1}}{(1+x)^{\color{red}{2}+\color{blue}{\theta-1}}}\,dx \\\\&=\theta\,B(\color{red}{2},\color{blue}{\theta-1})\qquad\qquad,\,\text{ since }\theta>1 \\\\&=\theta\,\frac{\Gamma(2)\Gamma(\theta-1)}{\Gamma(2+\theta-1)} \\\\&=\theta\,\frac{\Gamma(\theta-1)}{\Gamma(\theta+1)} \\\\&=\frac{\theta\,\Gamma(\theta-1)}{\theta(\theta-1)\Gamma(\theta-1)} \\\\&=\frac{1}{\theta-1} \end{align}

So you have your method of moments estimator of $\theta$, given by

$$\hat\theta(X_1,X_2,\ldots,X_n)=\frac{1+\overline X}{\overline X}$$

, where $\overline X$ is the sample mean.

Answer 3

You don't need integration by parts, or knowledge of Beta distributions. If you write $x$ as $1+x-1$, you end up with the difference between two easy integrals. Then $\mu=\theta(\frac{1}{\theta-1}-\frac{1}{\theta})=\frac{1}{\theta-1}$.