Plug in estimator for variance from Wasserman

Question

In All of Statistics by Larry Wasserman, page 100, Example 7.11, the example is asking about the plug in estimator of the variance for a linear functional (i.e. the plug in estimator for $\sigma^2=T(F)=\mathbb{V}(X)=\int x^2 dF(x)-(\int xdF(x))^2$).

It says the plug in estimator is

\begin{align} \hat{\sigma}^2 &= \int x^2 d\hat{F}(x)-\left(\int xd\hat{F}(x)\right)^2\\ &=\frac{1}{n}\sum_{i=1}^n X_i^2 - \left(\frac{1}{n}\sum_{i=1}^n X_i\right)^2\\ &=\frac{1}{n}\sum_{i=1}^n (X_i-\bar{X}_n)^2 \end{align}

Here, $F$ is the distribution that we pick samples $X_1,\ldots,X_n$ from and $\hat{F}$ is the empirical distribution function generated from these sample.

I am confused about the algebra going from 2nd to 3rd line above. I feel like it should be \begin{align} \hat{\sigma}^2 &= \int x^2 d\hat{F}(x)-\left(\int xd\hat{F}(x)\right)^2\\ &=\frac{1}{n}\sum_{i=1}^n X_i^2 - \left(\frac{1}{n}\sum_{i=1}^n X_i\right)^2\\ &\color{red}{=\frac{1}{n}\sum_{i=1}^n X_i^2-(\bar{X}_n)^2} \end{align}

Can someone help me figure out what I am seeing wrong? Thanks a lot.

Answer 1

Both are correct.

\begin{align} \frac1n \sum_{i=1}^n X_i^2 - \left( \frac1n \sum_{i=1}^n X_i\right)^2 &= \frac1n \left( \sum_{i=1}^n X_i^2 - \frac1n\left(\sum_{i=1}^n X_i \right)^2\right)\\ &=\frac1n \left( \sum_{i=1}^n X_i^2 - \frac1n\left( n\bar{X}\right)^2\right)\\ &=\frac1n \left( \sum_{i=1}^n X_i^2 - n\left( \bar{X}\right)^2\right)\\ &=\frac1n \left( \sum_{i=1}^n X_i^2 - 2n\left( \bar{X}\right)^2+n\bar{X}^2\right)\\ &=\frac1n \left( \sum_{i=1}^n X_i^2 - 2\bar{X}\sum_{i=1}^n X_i+\sum_{i=1}^n\bar{X}^2\right)\\ &= \frac1n \sum_{i=1}^n (X_i - \bar{X})^2 \end{align}