Is $\sum_{i=1}^n(x_i - \bar{x})x_i = \sum_{i=1}^n (x_i - \bar{x})^2$?

Question

I am proving a statement about the expectation of $\hat{\beta}$ in a linear regression and the following statement could help in my proof.

Assuming $X\sim N(\mu, \sigma^2)$ can we show that

$$\sum_{i=1}^n(x_i - \bar{x})x_i = \sum_{i=1}^n (x_i - \bar{x})^2$$

Thanks.

Answer 1

The summand in the LHS is $x_i^2 - \bar{x} x_i$ and the summand in the RHS is $x_i^2 - 2 \bar{x} x_i + \bar{x}^2$, so you're asking whether $$\sum_{i=1}^n (\bar{x}^2 - \bar{x}x_i) = 0.$$ The LHS can be rearranged as $$n \bar{x}^2 - \bar{x} \sum_{i=1}^n x_i$$ which equals zero because by definition, $\bar{x} = \frac{1}{n} \sum x_i$.

Answer 2

Essentially you need $\bar{x}\sum_{i=1}^n (x_i-\bar{x})=0$ This true from the definition of $\bar{x}$ as the average of $x_i$. Note that this does not depend on the distribution function.

Answer 3

With $E X= {1 \over n} \sum_k X_k$, we have $E((X -\overline{X})X ) = E X^2 - \overline{X}^2 = E (X-\overline{X})^2$.