Which estimator is preferred for a random sample from $P_\theta(X=x)=\theta^x(1-\theta)^{1-x}, x=0,1; 0 \le \theta \le \frac{1}{2}$? Let $X_1,\cdots,X_n$ be an i.i.d sample from $P_\theta(X=x)=\theta^x(1-\theta)^{1-x}, x=0,1; 0 \le \theta \le \frac{1}{2}$. Its the method of moments estimator of the MLE better? Why?
My work:
I found the following two estimators:
$\hat{\theta}_{MoM}=\bar{X}$ with MSE: $\frac{\theta(1-\theta)}{n}$
$\hat{\theta}_{MLE}=min(\bar{X},1/2)$ with piece-wise MSE: $\frac{\theta(1-\theta)}{n}, \bar{X} < 1/2$ and $\theta^2-\theta+1/4, \bar{X} > 1/2$
How do I show from here which estimator is better?
Updated Work:
When $\bar{X} \le \frac{1}{2}, MSE(\hat{\theta}_{MLE})=MSE(\hat{\theta}_{MM})$
When $\bar{X} > \frac{1}{2}$, let $g(\theta)=MSE(\hat{\theta}_{MLE}) - MSE(\hat{\theta}_{MM}) = (1 + \frac{1}{n})\theta^2 - (1+\frac{1}{n})\theta + \frac{1}{4} \stackrel{\text{set}}{<} 0$
which is a positive quadratic equation, decreasing until $\theta = 1/2$. 
By solving the quadratic equation, we get $\theta < \frac{1}{2}-\frac{1}{2\sqrt{n+1}} < \frac{1}{2}$.
So, $g(\theta) < 0$ for $\theta < \frac{1}{2}$. Therefore, we prefer MLE over method of moments estimator.
 A: If your sample is large and your data is independent and identically distributed, you may assume that all the assymptotic properties of the maximum likelihood estimator are valid: (1) Consistency; (2) Functional invariance; (3) Efficiency; (4) Assymptotic mean square error. See Wikipedia for details or Introduction to the Mathematical andStatistical Foundations of Econometrics - Bierens.
Unfortunatelly, the maximum likelihood estimator has no finite sample properties. I am not sure if it is possble to derive finite sample properties for the method of moments in the case, but usually if the sample is small, you can compare both estimators using Monte Carlo simulations. 
I did a Monte Carlo simulation to compare the results. I basically found that if $\theta$ is small, the error of both estimators is the same. On the other hand, if $\theta$ is close to 0.5, then the ML works better since $\theta$ is limited by 0.5 and the value of the parameter is never larger than 0.5.
import matplotlib.pyplot as plt
import numpy as np



def evalMSE(estimatedParameter,realParameter):
    mse=0
    for i in range(len(estimatedParameter)):
        mse=mse+(estimatedParameter[i]-realParameter)**2
    return mse    

def generateData(theta,sampleSize):
    sample=np.empty([sampleSize])
    for i in range(sampleSize):
        x=np.random.uniform(0,1)
        if(x<1-theta):
            sample[i]=0
        else:
            sample[i]=1
    return sample


def MM(data):
    return np.mean(data)

def ML(data):
    return np.min([np.mean(data),0.5])

if __name__=="__main__":
    numberSamples=1000
    sampleSize=100

    vectorML=np.empty([numberSamples])
    vectorMM=np.empty([numberSamples])

    theta=0.01
    for i in range(0,numberSamples):
        data=generateData(theta,sampleSize)
        vectorMM[i]=MM(data)
        vectorML[i]=ML(data)
    mseMM=evalMSE(vectorMM,theta)    
    mseML=evalMSE(vectorML,theta)
    plt.figure(num=1)        
    plt.hist(vectorMM,bins=30,density=True)
    plt.figure(num=2)
    plt.hist(vectorML,bins=30,density=True)
    print(mseMM)
    print(mseML)

For $\theta=0.4$
We get 
mseMM
Out[5]: 2.466999999999995

mseML
Out[6]: 2.3957999999999933

The distributions of $\theta$ are shown below:
$\theta$ distribution by MM:

$\theta$ distribution by ML:

A: MLE of $\theta$ is more precisely given by
$$\hat\theta_{MLE}=\overline XI_{0\le \overline X\le \frac12}+\frac12I_{\overline X>\frac12}=\begin{cases}\overline X&,\text{ if }0\le\overline X\le \frac12 
\\ \frac12&,\text{ if }\overline X>\frac12
\end{cases}$$
Method of moments estimator of $\theta$ is as you say $\hat\theta_{MOM}=\overline X$.
Therefore for $\theta\in[0,\frac12]$, 
\begin{align}
\operatorname{MSE}_{\theta}(\hat\theta_{MLE})&=\mathbb E_{\theta}(\hat\theta_{MLE}-\theta)^2
\\&=\sum_{0\le j\le \frac12}(j-\theta)^2\mathbb P_{\theta}(\overline X=j)+\sum_{j>\frac12}\left(\frac12-\theta\right)^2\mathbb P_{\theta}(\overline X=j)
\end{align}
And
\begin{align}
\operatorname{MSE}_{\theta}(\hat\theta_{MOM})&=\mathbb E_{\theta}(\overline X-\theta)^2
\\&=\sum_{0\le j\le \frac12}(j-\theta)^2\mathbb P_{\theta}(\overline X=j)+\sum_{j>\frac12}(j-\theta)^2\mathbb P_{\theta}(\overline X=j)
\end{align}
So for every $\theta\in[0,\frac12]$,
\begin{align}
\operatorname{MSE}_{\theta}(\hat\theta_{MLE})-\operatorname{MSE}_{\theta}(\hat\theta_{MOM})&=\sum_{j>\frac12}\left[\left(\frac12-\theta\right)^2-(j-\theta)^2\right]\mathbb P_{\theta}(\overline X=j)
\end{align}
You just have to check the sign of this expression to conclude.
Similar question: How to show that $E[(\hat\theta -\theta)^2]<Var(\bar X)=\dfrac{1}{n}$?
