Simplifying this equation to get sum of squares

Question

$$\frac{1}{n\sum x_{i}^2-{(\sum x_{i}})^2}$$

I have this equation above. I am trying to simplify it such that I can get: $\dfrac{1}{nSS_{x}}$

Where $SS_{x}$ is the sum of squares of $x$. Any pointers on how to achieve this?

Answer 1

It's easier to work backwards, and only compare the denominators of your fractions. The sum of squared deviations of $x$ is

$$ SS_x = \sum_i {(x_i-\bar{x})^2} $$

$$ \Rightarrow nSS_x = n\sum_i {(x_i-\bar{x})^2} $$

$$ = n\sum_i \left({x_i^2 - 2x_i\bar{x} + \bar{x}^2}\right) $$

$$ = n\sum_i {x_i^2} - n\sum_i {2x_i\bar{x}} + n\sum_i {\bar{x}^2} . $$

Since $\bar{x}$ and $\bar{x}^2$ are constants, we can factor them out of the sums:

$$ = n\sum_i {x_i^2} - 2n\bar{x}\sum_i {x_i} + n^2\bar{x}^2 , $$

Note, the extra factor of $n$ in the third term comes from the fact that $\sum_i {1} = n$. In the second term and third terms, we can substitute $n\bar{x} = \sum_i x_i$:

$$ = n\sum_i {x_i^2} - 2 \left( \sum_i x_i \right)^2 + \left( \sum_i x_i \right)^2 $$

$$ = n\sum_i {x_i^2} - \left( \sum_i x_i \right)^2 . $$