Method of moment estimates for n Bernoulli trials Let $X_1, X_2 \dots X_N$ be the indicators of $n$ Bernoulli trials with probability of success $p$.
What is the method of moments estimate of $p$?
Exhibit method of moments estimates for $p \cdot (1 - p) / n$ using only the first moment and then using only the second moment of the population. How do we show that these estimates conincide?
Argue that in this, the indicators of $n$ Bernoulli trials case all frequency substitution estimates of $q(p)$ must agree with the mean of the indicators of $n$ Bernoulli trials.
Could I use the example , Let X1, X2, ..., Xn be normal random variables with mean μ and variance σ2. What are the method of moments estimators of the mean μ and variance σ2? in this URL  https://onlinecourses.science.psu.edu/stat414/node/193 ?
Also, Could I use the analysis of the variance of binomial distribution contained in the answers to this Mathematics Stack Exchange URL https://math.stackexchange.com/questions/240070/variance-of-binomial-distribution?
 A: Hint: "method of moments" means you set sample moments equal to population/theoretical moments.
For example, the first sample moment is $\bar{X} = n^{-1}\sum_{i=1}^n X_i$, and the second sample moment is $n^{-1}\sum_{i=1}^n X_i^2$. In general, the $k$th sample moment is $n^{-1}\sum_{i=1}^n X_i^k$, for some integer $k$.
The first population moment is $E[X] = \sum_x x P(X=x)$, and the second population moment is $E[X^2] = \sum_x x^2 P(X=x)$. In general the $k$th population moment is $E[X^k] = \sum_x x^k P(X=x)$. Or if your random variable $X$ is a continuous random variable, you would use integrals and density functions: $E[X] = \int x^k f(x) dx$.
In one of your cases, you would solve the following equation for the parameter of interest:
$$
E[X] = \bar{X} \tag{1}.
$$
With normal data, since you have two parameters ($\mu$ and $\sigma^2$), you need to solve two equations:
\begin{align*}
E[X] &= \bar{X}, \\
E[X^2] &= n^{-1}\sum_{i=1}^n X^2_i.
\end{align*}
What equations do I solve for Bernoulli data and one parameter to get the variance in terms of the one parameter?
A: After reading the analysis of the variance of binomial distribution contained in the answers to this Mathematics Stack Exchange URL https://math.stackexchange.com/questions/240070/variance-of-binomial-distribution, I believe that that the mean, variance and higher moments of the sum of n iid bernoulli variables with parameter p is binomial(p,n).So, the mean of S is np and the variance of S is np*(1 - p).
