Verify that $(n-1)s^2(x) = \sum_{k=1}^{n}x_2^k − n\bar{x}^2$

Question

Let $ θ$ be the mean of a normal distribution and $ aθ^2$ be the variance where $a$ in a known constant. That is $normal~ ( θ, a θ^2)$ How do i verify that:

$( − 1)^2 () = $$\sum_{k=1}^{n} _^2 − \bar{x}^2$

I know that $s^2(x) = 1/(n-1) \sum_{k=1}^{n} (X_ − \bar{X})^2$ which is the sum of squares. But im not sure where the $\sum_{k=1}^{n}2x_k\bar{x}$ goes when expanding this.

Furthermore, how does $exp(\sum_{k=1}^{n}(x_k - θ)^2)$ expand to becomes $exp((n-1)s^2(x) + n(\bar{x} - θ)^2)$.

I tried $exp( \sum_{k=1}^{n}((x_k - \bar{x} + \bar{x} - θ)^2) $ and that is suppose to become $exp(\sum_{k=1}^{n}(x_k - \bar{x})^2 + \sum_{k=1}^{n}(\bar{x} - θ)^2) $ but is that even possible without a transition step or something; is this correct?

Would really appreciate your help.

Answer 1

As you stated in the question text (I assume the $X$ were supposed to be $x$),

$$s^2(x) = 1/(n-1) \sum_{k=1}^{n} (x_k − \bar{x})^2 \tag{1}\label{eq1}$$

Thus,

$$\begin{equation}\begin{aligned} (n-1)s^2(x) & = \sum_{k=1}^{n} (x_k − \bar{x})^2 \\ & = \sum_{k=1}^{n} (x_k^2 - 2x_k\bar{x} + \bar{x}^2) \\ & = \sum_{k=1}^{n} x_k^2 - \sum_{k=1}^{n}2x_k\bar{x} + \sum_{k=1}^{n}\bar{x}^2 \\ & = \sum_{k=1}^{n} x_k^2 - 2\bar{x}\sum_{k=1}^{n}x_k + n\bar{x}^2 \\ & = \sum_{k=1}^{n} x_k^2 - 2\bar{x}(n\bar{x}) + n\bar{x}^2 \\ & = \sum_{k=1}^{n} x_k^2 - n\bar{x}^2 \end{aligned}\end{equation}\tag{2}\label{eq2}$$

Note in the second last line I used that

$$\bar{x} = \frac{\sum_{k=1}^{n} x_k}{n} \implies \sum_{k=1}^{n} x_k = n\bar{x} \tag{3}\label{eq3}$$

Regarding your additional question of

Furthermore, how does $exp(\sum_{k=1}^{n}(x_k - θ)^2)$ expand to becomes $exp((n-1)s^2(x) + n(\bar{x} - θ)^2)$.

With the exponent expressions, you get that

$$\begin{equation}\begin{aligned} \sum_{k=1}^{n}(x_k - \theta)^2) & = \sum_{k=1}^{n}\left(\left(x_k - \bar{x}\right) + \left(\bar{x} - \theta\right)\right)^2 \\ & = \sum_{k=1}^{n}\left(x_k - \bar{x}\right)^2 + 2\sum_{k=1}^{n}\left(x_k - \bar{x}\right)\left(\bar{x} - \theta\right) + \sum_{k=1}^{n}\left(\bar{x} - \theta\right)^2 \\ & = (n-1)s^2(x) + 2\left(\bar{x} - \theta\right)\sum_{k=1}^{n}\left(x_k - \bar{x}\right) + n\left(\bar{x} - \theta\right)^2 \\ & = (n-1)s^2(x) + 2\left(\bar{x} - \theta\right)\left(\sum_{k=1}^{n}x_k - n\bar{x}\right) + n\left(\bar{x} - \theta\right)^2 \\ & = (n-1)s^2(x) + n\left(\bar{x} - \theta\right)^2 \end{aligned}\end{equation}\tag{4}\label{eq4}$$

I used \eqref{eq3} to get that the middle term in the second last line above is $0$.