Let $x_1=2, x_2 = 1, x_3 = \sqrt5, x_4 = \sqrt2$ be the observed values of a random sample of size 4 from a uniform distribution $U(-\theta, \theta)$ where $\theta>0$. Then the method of moments estimate of $\theta$ is?
a) 1 b) 2 c) 3 d) 4
The way I approached this question is: The first method of moment is $\bar{X}$ = $\sum X_i/n$ = ($2+1+\sqrt5+\sqrt2)/4 = 1.66$. Expected value of $U(-\theta, \theta)$ = ($-\theta+\theta)/2 = 0$. I am unable to interpret further.
The mean of this distribution is always zero so it does not depend on the parameter $\theta$. For that reason, if you were using MOM estimation you could use the second raw moment, which is:
$$\mathbb{E}(X^2) = \frac{\theta^2}{3}.$$
Equating this to the sample moment $R_n^2 \equiv \tfrac{1}{n} \sum x_i^2$ gives the MOM estimator:
$$\hat{\theta}_n = \sqrt{3} \cdot R_n.$$
With your particular data $\mathbf{x}_4 = (2,1,\sqrt{5},\sqrt{2})$ you have $R_4^2 = 3$ so you get the estimated parameter value $\hat{\theta}_4 = \sqrt{3} \cdot R_4 = \sqrt{3} \cdot \sqrt{3} = 3$.
An alternative method: Pursuing the suggestion in the comment from whuber below, another option would be to form the MOM for $|X| \sim \text{U}[0, \theta)$. In this case we have the first raw moment (the mean):
$$\mathbb{E}(|X|) = \frac{\theta}{2}.$$
Equating this to the sample moment $A_n \equiv \tfrac{1}{n} \sum |x_i|$ gives the MOM estimator:
$$\hat{\theta}_n = 2 \cdot A_n.$$
With your data you have $A_4 = \tfrac{1}{4} (3+\sqrt{5}+\sqrt{2}) \approx 1.66257$ so you get the estimated parameter value $\hat{\theta}_4 \approx 3.32514$. This is similar to the result for the previous MOM estimate.
