Let $X_1,X_2,...,X_n$ be a random sample from a uniform distribution on $(\mu-\sqrt 3\sigma,\mu+\sqrt3\sigma)$.
Here the unknown parameters are two, namely $\mu$ and $\sigma$, which are the population mean and standard deviation.
Find the point estimator of $\mu$ and $\sigma$.
I have tried to do that by the Method-of-Moments(MOM). The procedure is,
$M\prime_j=\mu\prime_j(\theta_1,\theta_2,...,\theta_k); j=1,2,...,k$
where $M\prime_j$ is the $j^{th}$ sample moment about zero & $M\prime_j=\frac{1}{n}\sum_{i=1}^n X_i^j$
& $\mu\prime_j$ is the $j^{th}$ moment about zero ,ie, $j^{th}$ raw moment.
Now,
$M\prime_1=\mu\prime_1=\mu\prime_1(\mu,\sigma)=\mu$
And
$M\prime_1=\frac{1}{n}\sum_{i=1}^n X_i=\bar X$
Again,
$M\prime_2=\mu\prime_2=\mu\prime_2(\mu,\sigma)=\sigma^2+\mu^2$
$\Rightarrow M\prime_2=\sigma^2+\mu^2$
$\Rightarrow \sigma^2=M\prime_2-\mu^2$
$\Rightarrow \sigma=\sqrt{\frac{1}{n}\sum_{i=1}^n(X_i-\bar X^2)}$
see https://math.stackexchange.com/questions/416581
Hence Method-of-Moment estimators are $\bar X$ for $\mu$
and $\sqrt{\frac{1}{n}\sum_{i=1}^n(X_i-\bar X^2)}$ for $\sigma$.
But the procedure seems illogical to me for the following reason:
$\bullet$ I haven't considered the pdf of uniform density. so this procedure is also applicable for normal density. Then where is the difference?
What is the correct process of finding the point estimators for the above situation?