Suppose you have $X_1,\ldots,X_{20}$ $f(x|\theta) = 0.1(1+\theta x)$ for $-1
Attempt: Isn't the moment estimator of theta just the first moment which is the mean?
Suppose you have $X_1,\ldots,X_{20}$ $f(x|\theta) = 0.1(1+\theta x)$ for $-1
Attempt: Isn't the moment estimator of theta just the first moment which is the mean?
$ \int_{-1}^1 1+\theta x\;dx = 2, $ so you need $f(x|\theta) = \frac 12 (1+\theta x)$.
If you mean what I think you mean, I would call it a method-of-moments estimator of $\theta$ rather than a "moment estimator", since the latter term might be mistaken for "estimator of the moment(s)".
The first moment of this distribution is $ \int_{-1}^1 x f(x \mid \theta) \; dx, $ which by my reckoning is $\theta/3$. The first moment of the sample is $(X_1+\cdots+X_{20})/20$. You need to equate the first moment of the distribution with the first moment of the sample and then solve for $\theta$.
The method-of-moments estimator of $\theta$ would be equal to the sample mean only if $\theta$ were the population mean.