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\subsubsection*{Testing independence in contingency tables}

$N_{ij}$: number of samples of type $(i, j)$.
\[ (N_{ij}) \sim \Multinomial(n, (p_{ij})) \]
$H_0$: $p_{ij} = p_{i+} \times p_{+j}$ \\
$H_1$: $(p_{ij})$ unconstrained. \\
Found $2\log\Lambda$, which has asymptotic
$\chi_{(r - 1)(c - 1)}^2$ distribution.

\begin{example*}[COVID-19 deaths]
    Problems with $\chi^2$ independence test:
    \begin{enumerate}[(1)]
        \item $\chi^2$ approximation can be bad
            when we have large tables. Rule of
            thumb: Need $N_{ij} \ge 5$ for all $i,
            j$. \myskip
            Solution (non-examinable): exact
            testing.  Idea: under $H_0$, the
            margins of $N$ $(N_{i+})$, $(N_{+j})$
            are sufficient statistics for $p$.
            therefore 2 tables $N$, $\tilde{N}$
            with the same margins are equally
            likely under $H_0$. An exact test
            contrasts the test statistic observed
            $2\log\Lambda(N)$ with the
            distribution of this statistic for the
            set of tables with the same margins as
            $N$. This gives a test of \emph{exact}
            size $\alpha$.
        \item $2\log\Lambda$ can detect deviations
            from $H_0$ in any direction.
            $\implies$ Low power, especially when
            $r$, $c$ is large. This is why $H_0$
            is not rejected in a test of size
            $1\%$ in COVID-19 example. Solutions:
            \begin{enumerate}[(1)]
                \item Define a parametric
                    alternative $H_1$ with fewer
                    degrees of freedom.
                \item Lump categories in the
                    table.
            \end{enumerate}
    \end{enumerate}
\end{example*}

\subsubsection*{Tests of Homogeneity}

Instead of assuming $\sum_{i, j} N_{ij}$ fixed, we
assume row totals are fixed.

\begin{example*}
    $150$ patients, split into groups of $50$ for
    placebo, half-dose, full-dose. We record
    whether each patient improved, showed no
    difference or got worse.
    \begin{center}
        \begin{tabular}{p{4em}|c|c|c}
             & I & N.D. & W \\
             \hline
            Placebo & \hspace{4em} & \hspace{4em} & \hspace{4em} \\
            Half & & & \\
            Full & & &
        \end{tabular}
    \end{center}
    Now row totals are fixed. Null of homogeneity:
    probability of each outcome is the same in
    each treatment group. \myskip
    Model:
    \[ (N_i1, \ldots, N_{ic}) \sim
    \Multinomial(n_{i+}, p_{i1}, \ldots, p_{ic}) \]
    independent for $i = 1, \ldots, r$. Paramters
    satisfy $\sum_j p_{ij} = 1$ for all $i$.
    $H_0$: $p_{1j} = p_{2j} = \cdots p_{rj}$ for
    all $j = 1, \ldots, c$. $H_1$: $(p_{i1},
    \ldots, p_{ic})$ is a probability vector for
    all $i$.
    \begin{align*}
        L(p)
        &= \prod_{i = 1}^r \frac{n_{i+}!}{N_{i1}!
        \cdots N_{ic}!} p_{i1}^{N_{i1}} \cdots
        p_{ic}^{N_{ic}} \\
        l(p)
        &= \text{const} + \sum_{i, j} N_{ij} \log
        p_{ij}
    \end{align*}
    To find $2\log\Lambda$ we need to maximise
    $l(p)$ over $H_0$, $H_1$. $H_1$: use Lagrange
    multipliers with constraints $\sum_j p_{ij} =
    1$ for all $i$. Then the mle is
    \[ \hat{p}_{ij} = \frac{N_{ij}}{n_{i+}} \]
    $H_0$: let $p_j = p_{1j} = \cdots = p_{+j}$.
    \[ l(p) = \text{const} + \sum_{j = 1}^c N_{+j}
    \log p_j \]
    hence the mle is $\hat{p}_j =
    \frac{N_{+j}}{n_{++}}$, $n_{++} = \sum_i
    n_{i+}$. Thus
    \[ 2\log\Lambda = 2 \sum_{i, j} N_{ij} \log
    \left( \frac{N_{ij}}{n_{i+} N_{+j} / n_{++}} \right) \]
    This is exactly the same statistic as
    $2\log\Lambda$ for the independence test. Let
    $o_{ij} = N_{ij}$, $e_{ij} = n_{i+} \hat{p}_j
    = n_{i+} \frac{N_{+j}}{n_{++}}$
    \begin{align*}
        \implies 2\log\Lambda
        &= 2\sum_{i, j} o_{ij} \log \left(
        \frac{o_{ij}}{e_{ij}} \right) \\
        &\approx \sum_{i, j} \frac{(o_{ij} -
        e_{ij})^2}{e_{ij}}
    \end{align*}
    This is also the same as Pearson's statistic
    for independence test.
\end{example*}

\noindent
Wilk's implies $2\log\Lambda$ is approximately
$\chi_d^2$,
\begin{align*}
    d
    &= \dim(\Theta_1) - \dim(\Theta_0) \\
    &= (c - 1)r - (c - 1) \\
    &= (c - 1)(r - 1)
\end{align*}
Asymptotic distribution of $2\log\Lambda$ is also
the same as in the independence test. \myskip
Testing independence or homogeneity with size
$\alpha$ always has the same conclusion.

\subsubsection*{Relationship between tests and confidence sets}

Define the \emph{acceptance ragion} $A$ of a test
to be the complement of the critical region. Let
$X \sim f_X(\bullet \mid \theta)$ for some $\theta
\in \Theta$.

\begin{theorem*}
    \begin{enumerate}[(1)]
        \item Suppose that for each $\theta_0 \in \Theta$
            there is a test of $H_0$: $\theta =
            \theta_0$ of size $\alpha$ with
            acceptance region $A(\theta_0)$. Then,
            the set
            \[ I(X) = \{\theta : X \in A(\theta)\} \]
            is a $100(1 - \alpha)\%$ confidence
            set.
        \item Suppose $I(X)$ is a $100(1 -
            \alpha)\%$ confidence set for
            $\theta$. Then
            \[ A(\theta_0) = \{x : \theta_0 \in
            I(X)\} \]
            is the acceptance region of a size
            $\alpha$ test for $H_0$: $\theta =
            \theta_0$.
    \end{enumerate}
\end{theorem*}

\begin{proof}
    In each part:
    \[ \theta_0 \in I(X) \iff X \in A(\theta_0) \]
    For part (1), we calculate:
    \begin{align*}
        \PP_{\theta_0}(I(X) \ni \theta_0)
        &= \PP_{\theta_0}(x \in A(\theta_0)) \\
        &= 1 - \PP_{\theta_0}(x \in C(\theta_0))
        \\
        &= 1 - \alpha
    \end{align*}
    as desired. For part (2):
    \begin{align*}
        \PP_{\theta_0}(X \in C(\theta_0))
        &= \PP_{\theta_0}(X \not\in A(\theta_0))
        \\
        &= \PP_{\theta_0}(I(X) \not\ni \theta_0)
        \\
        &= 1 - \PP_{\theta_0}(I(x) \ni \theta_0)
        \\
        &= 1 - (1 - \alpha) \\
        &= \alpha
    \end{align*}
    as desired.
\end{proof}

\begin{example*}
    $X_1, \ldots, X_n \iidsim \normaldist(\mu,
    \sigma_0^2)$, $\sigma^2$ known.
    \[ I(X) = \left( \ol{X} \pm \frac{z_{\alpha/2}
    \sigma_0}{\sqrt{n}} \right) \]
    confidence interval. Test: $H_0$: $\mu =
    \mu_0$, $H_1$: $\mu \neq \mu_0$. Critical
    region:
    \[ \left\{ x : \left| \sqrt{n} \frac{(X -
    \ol{X}}{\sigma_0} \right| > z_{\alpha/2} \right\} \]
\end{example*}