$\frac{1}{n}\sum X^2_i -\bar{X^2}$ equals $\frac{1}{n}\sum(X_i-\bar{X})^2$?

Question

I have the following question. Mostly a clarification:

I know that $Var(X)=E(X-EX)^2 =E(X^2)-(EX)^2$

I have the following problem: $X_1,...X_n $ are idd $n(\mu,\sigma^2)$. Then I am using the Method of Moments and end up with the expressions:

1) $\bar{X}=\mu$

2) $\hat{\sigma^2}=(1/n)\sum X^2_i -\bar{X^2}=(1/n) \sum(X_i-\bar{X})^2$

When I first saw that expression I just thought about the expression above. First since $E(X_i)=\mu=\bar{X}$ then I thought $(1/n)$ as a pmf and that somehow $(1/n)\sum X^2_i=E(X^2)$. But I realized this is mistaken, since the $EX^2$ assumes that X follows a normal distribution.

My question is how do I go from here: $(1/n)\sum X^2_i -\bar{X^2}$ to here $ (1/n) \sum(X_i-\bar{X})^2$

MY ATTEMPT is the following:

Apply the Expectation to the whole expression 2, and put the Xbar inside the summation: $E\hat{\sigma^2}=E[(1/n)\sum X^2_i -\bar{X^2}]=E[(1/n)\sum (X^2_i -n\bar{X^2})]$

$(1/n)[\sum E(X^2_i) -nE(\bar{X^2})]=(1/n)[\sum E(X^2_i) -n\mu^2)]$

Now this is a summation of variances so I can just use the regular formula presented above, and I will substitute $\mu^2=\bar{X^2}$, then:

$(1/n)[\sum E(X_i-\bar{X})^2]= E \hat{\sigma^2}$

Since I have expectations in both sides, I can just remove them. Is this the right approach? It seems like a long thing to just be left out.

Thanks!!

Answer 1

First, you need to be more careful with some of your statements above because you seem to be confusing expected values and sample means. A statement like $\bar{X} = \mu$ is not true in any meaningful way since $\bar{X}$ is a random variable and $\mu$ a constant that $\bar{X}$ is trying to estimate.

The identity you refer to concerning sums of squares turns out just to be some arithmetic.

\begin{align} \sum_{i=1}^{n} (X_i - \bar{X})^2 &= \sum_{i=1}^{n} (X_i^2 + \bar{X}^2 - 2 X_i \bar{X}) \\ &= \sum_{i=1}^{n} X_i^2 + n \bar{X}^2 - 2 n \bar{X}^2 \\ &= \sum_{i=1}^{n} X_i^2 - n \bar{X}^2 \end{align}

where all we've used is that $\sum_{i=1}^{n} \bar{X}^2 = n \bar{X}^2$ and $\sum_{i=1}^{n} X_i = n \bar{X}$. You shouldn't be taking expectations at any point here.