What is the correct process of finding the point estimators for the following situation?

Question

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 How $\frac{1}{n}\sum_{i=1}^n X_i^2 - \bar X^2 = \frac{\sum_{i=1}^n (X_i - \bar X)^2}{n}$

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?

Answer 1

The method of moments is based on writing the unknown parameters as a function of the moments, and using the "standard" estimators for the moments ($\widehat{m_r}=\sum x^r/n$). If it's the case that the unkown parameters are precisely moments, then it's trivial that the estimators are determined by the procedure, and does not depend on the density.

In concrete, when (only when) we use MOM, the estimator of the mean of any distribution, will always be $\widehat = \sum x/n$, and the estimator of the variance ($\sigma^2=m_2-m_1^2$) will always be $\widehat\sigma^2 = \sum x^2/n- (\sum x/n)^2$, regardless of the distribution.

Answer 2

There is nothing illogical about your result. It just turns out that the way your problem is constructed, the first and second moments written in terms of the $\mu$ and $\sigma$ parameters of the above uniform distribution have a form identical to the moments of a Gaussian random variable with mean $\mu$ and standard deviation $\sigma$. Hence you obtain point estimators of a similar form under the MOM.

It is always possible for the point estimators of some parameters of one distribution to have a form identical to the point estimators of some other parameters of a different distribution.

Note that you do take the PDF of the uniform distribution into consideration when writing the moments in terms of $\mu$ and $\sigma$.