Find the Method Moment Estimator of parameter $\theta$ Find the MME of parameter $\theta$ in the distribution with the density $f(x,\theta)=(\theta +1)x^{-(\theta+2)}$, for $x>1$ and $\theta >0$.
So far I think I have a basic understanding of the MME process, but I am confused about the the execution.
$E[x] =\int x\ f(x,\theta)\ dx$
$=\int_x^{\infty}\ t (1+\theta) \ t^{-(\theta+2)} \ dt$
$=\int_x^{\infty} (1+\theta) \ t^{-(\theta+1)} \ dt$
$=(\theta+1)\int_x^{\infty}t^{-(\theta+1)}dt=?...?= \dfrac{1+\theta}{\theta}, according \ to \ WolframAlpha $
So I have a few concerns... I am not sure if the bounds on my integral are correct, and even if they are not, I am unsure about how I would calculate $\int t^{-(\theta+1)}dt$.
Am I on the right track?
Thanks for helping!
 A: We recently covered the method of moments in my stats class too.  You want to calculate $E[X]$; however, there's no need to make it a function of $t$.  I worked through that integral and  presumably this is where your ellipsis and question marks came in  I got something different from that Wolfram answer:  The integral evaluated to $(1+\theta)/(\theta x)$.  As it should:  If $x$ is a limit of integration, it doesn't get a specific value that can cancel out somewhere else.
That said, the integral that Wolfram gave you is the one you want; if you calculate
$$\int_{-\infty}^{\infty}x\cdot f(x)\,dx = \int_{1}^{\infty}(1+\theta)x^{-(1+\theta)}\, dx$$
that's the answer you'll get.  At this point, you're almost finished:  We know that $\mu = E[X] = (1 + \theta)/\theta$, and so because we're using $\mu$ to estimate $\theta$, all we have to do now is write $\theta$ in terms of $\mu$.  The one difference is that, since these are estimates, based presumably on observed data, your answer should be giving $\widehat{\theta}$ in terms of $\bar{x}$.
From here, what answer do you get?
Almost forgot:  As for integrating $t^{-(\theta + 1)}$ or $x$ to the same power:  You're integrating with respect to $x$ (or $t$), so $\theta$ is just a constant.  Is that what you were asking?
