Tests for goodness-of-fit

Activity 14.1   Show that the two forms for $X^2$ are equivalent.
$\blacksquare$

Answer 14.1  

$$X^2 = \sum \frac{(O_i-E_i)^2}{E_i}$$

$$=\sum \frac{O_i^2-2O_iE_i+E_i^2}{E_i}$$

$$=\sum\left[\frac{O_i^2}{E_i}-2O_i +E_i\right]$$

$$=\sum\frac{O_i^2}{E_i}-2\sum O_i + \sum E_i$$

$$=\sum\frac{O_i^2}{E_i} -2n +n.$$

The fact that $\sum E_i=n$ follows from $E_i=n\pi_i$ and $\sum \pi_i=1$.
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Activity 14.2   Show from results for the binomial distribution that $EX^2=k-1$. This is exactly the mean of a $\chi^2_{(k-1)}$ distribution.
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Answer 14.2   We have

$$X^2 = \sum \frac{(O_i-E_i)^2}{E_i}$$

and so

$$E[X^2] = \sum \frac{E[(O_i-E_i)^2]}{E_i}$$

$$=\sum \frac{\operatorname{var}O_i}{E_i}$$

$$=\sum \frac{n\pi_i(1-\pi_i)}{n\pi_i}$$

$$=\sum (1-\pi_i)$$

$$=k-\sum\pi_i$$

$$= k-1.$$

$\blacksquare$