What about discrete distributions?

The K-S test is only appropriate for continuous distirbutions (the hypothesized distribution is continuous).

But what about if our hypothesis if for a discrete distribution, e.g.:

• Discrete Uniform
• Bernoulli
• Poisson

Setting

Population: $Y\sim$ some discrete population distribution with p.m.f $p\left(y\right)=P(Y=y)$

Sample: n i.i.d from population, $Y_1,\dots ,Y_n$

Parameter: Whole p.m.f

Hypotheses: $H_0:P(Y=y)=p_0\left(y\right)$, versus $H_A:P(Y=y)\ne p_0\left(y\right)$

Sample estimate of the p.m.f

The discrete sample based estimate of the probability mass function:

$$\hat{p}\left(y\right)=\frac{1}{n}\sum _{i=1}^{n}11\left\{Y_i=y\right\}$$

The Chi-square Goodness of Fit test

Chi-square goodness of fit compares the estimated p.m.f to the hypothesized one.

Pearson’s Chi-square statistic: $$X\left(p_0\right)=\sum _y\frac{n{\left(\hat{p}\left(y\right)-p_0\left(y\right)\right)}^2}{p_0\left(y\right)}$$

Under null hypothesis: $X\left(p_0\right)$ converges in distibrution (as n goes to infinity) to ${\chi }^2$ with $k-1$ degrees of freedom

$k=$ number of possible values for $Y$.

An alternative presentation

• $j=1,\dots ,n$ indexes possible values/categories for $Y$
• $O_j$ be the observed number of values in category $j$
• $E_j=n{p}_0\left(j\right)$ be the expected number of values in category $j$, based on the hypothesized distribution.

Pearson’s Chi-square statistic: $$X\left(p_0\right)=\sum _{j=1}^k\frac{(O_j-E_j{)}^2}{E_j}$$

Example: Dice rolling

I rolled a die 60 times and recorded how many times I got each side: 1, 2, 3, 4, 5, 6

Question: Is the die fair? That is, is $p_0\left(j\right)=1/6$ for $j=1,2,3,4,5,6$?

Rejection region

Reject $H_0$ for $X\left(p_0\right)>{\chi }_{(k-1)}^2(1-\alpha )$

Rejection region always to the right, p-values always area to right. Why? We usualy are only interested in evidence of poor fit (not evidence of extra good fit)

Example: Dice rolling cont.

Compare to ${\chi }_{\left(5\right)}^2$

qchisq(0.95, df = 5)

## [1] 11.0705

1 - pchisq(chi_sq, df = 5)

## [1] 0.01438768

In R:

rolls

##  [1] 6 5 2 2 6 2 1 1 5 1 2 2 1 1 3 4 5 1 2 4 1 6 6
## [24] 2 4 1 6 6 1 6 1 1 6 1 3 1 3 4 1 1 5 3 4 2 5 1
## [47] 6 3 1 4 3 2 5 1 6 2 1 4 1 2

chisq.test(table(rolls), p = rep(1/6, 6))

## 
##  Chi-squared test for given probabilities
## 
## data:  table(rolls)
## X-squared = 14.2, df = 5, p-value = 0.01439

Estimation of parameters

If the null hypothesis doesn’t completely specify the distribution $p_0$, but specifes a family of distributions, $p_0({\theta }_1,{\theta }_2,\dots ,{\theta }_d)$ where the $\theta$ are unknown parameters.

You can still use the Chi-square test with some modification

1. Estimate the parameters ${\theta }_1,{\theta }_2,\dots ,{\theta }_d$
2. Find $E_j$ based on estimated parameters and $p_0$.
3. Compute Pearson’s ${\chi }^2$ statistic as usual
4. Compare statistic to a ${\chi }^2$ with $k-d-1$ degrees of freedom, where $d$ is the number of parameters that were estimated.

Example: Poisson

I counted the number of passengers in $n=40$ vehicles passing through an intersection.

Question: Is the number of passengers per vehicle distributed according to a Poisson distribution?

mean(passengers)

## [1] 1.875

Example: Poisson

p_0 <- dpois(0:6, lambda = mean(passengers))
(E <- p_0 * n)

## [1]  6.1341987 11.5016225 10.7827711  6.7392319
## [5]  3.1590150  1.1846306  0.3701971

# For 7+ category
(E_7 <- (1 - sum(p_0)) * n)

## [1] 0.1283331

Example: Poisson

Test statistic:

$$\frac{(6-6.13{)}^2}{6.13}+\frac{(11-11.5{)}^2}{11.5}+\dots +\frac{(0-0.13{)}^2}{0.13}=2.66$$

Compare to ${\chi }_{k-d-1}^2$

1 - pchisq(X, df)

## [1] 0.8504434

Other points

• For binary data, the $X\left(p_0\right)$ statistic is equal to the square of the Z-statistic for testing a hypothesis regarding a binary proportion.

  Therefore, for two sided hypothesis testing $H_0:p=p_0$ vs $H_A:p\ne p_0$, the ${\chi }^2$ test and the z-test give the exact same result.

• The ${\chi }^2$ statistic has an asymptotic ${\chi }^2$ distribution.

  Therefore this test is approximate: the test is asymptotically exact.

  The approximation is generally considered appropriate when $E_j>5$ for all $j$.

Next time

Starting two sample inference…