Computing the plug in estimator of the variance

Question

My professor wants us to show that the plug in estimator of $var(X)$ is $$\frac{1}{n}\sum_{i=1}^n(x_i-\overline{x})^2$$ where $x_i$ are observations of the random variable $X$. But when I compute I get the following, $$\int x^2dF_n-\left(\int xdF_n\right)^2=\frac{1}{n}\sum_{i=1}^nx_i^2-\left(\frac{1}{n}\sum_{i=1}^nx_i\right)^2=\frac{1}{n}\sum_{i=1}^nx_i^2-\overline{x}^2=\frac{1}{n}\sum_{i=1}^n(x_i^2-\overline{x}^2).$$ Is my computation wrong somewhere?

Answer 1

your last passage is not useful to prove what you are requesting to do.

You get

$$\frac{1}{n}\Sigma_i x_i^2-\overline{x}^2=\frac{1}{n}\Sigma_i(x_i-\overline{x})^2$$

the simplest way to verify it is to expand RHS finding

$$\frac{1}{n}\Sigma_i(x_i-\overline{x})^2=\frac{1}{n}\left[ \Sigma_i x_i^2+n\overline{x}^2-2\overline{x} \Sigma_ix_i \right]=\frac{1}{n}\left[ \Sigma_i x_i^2+n\overline{x}^2-2n\overline{x}^2\right]=$$

$$=\frac{1}{n}\left[ \Sigma_i x_i^2-n\overline{x}^2\right]=\frac{1}{n}\Sigma_ix_i^2-\overline{x}^2$$