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\begin{lemma*}
    If $C$ is a LRT with size $\alpha$, and $C^*$
    is another test of size $\le \alpha$, then $C$
    is more powerful than $C^*$, i.e.
    \[ \beta = \PP_{H_1}(x \not\in C) \le
    \PP_{H_1}(x \not\in C^*) = \beta^* \]
\end{lemma*}

\begin{example*}
    $X_1, \ldots, X_n \iidsim \normaldist(\mu,
    \sigma_0^2)$, $\sigma_0^2$ is known. Want the
    best size $\alpha$ test for $H_0$: $mu =
    \mu_0$, $H_1$: $\mu = \mu_1$ for some fixed
    $\mu_1 > \mu_0$
    \begin{align*}
        \Lambda_x(H_0; H_1)
        &= \frac{\cancel{(2\pi \sigma^2)^{-\half}}
        \exp \left( -\frac{1}{2\sigma_0^2} \sum
        (x_i - \mu_1)^2 \right)}
        {\cancel{(2\pi \sigma_0^2)^{\half}} \exp
        \left( -\frac{1}{2\sigma_0^2} \sum (x_i -
        \mu_0)^2 \right)} \\
        &= \exp \left( \frac{(\mu_1 -
        \mu_0)}{\sigma_0^2} n\ol{x} +
        \frac{n(\mu_0^2 - \mu_1^2)}{2\sigma_0^2} \right)
    \end{align*}
    $\Lambda_x(H_0; H_1)$ is monotone increasing
    in $\ol{x} = \frac{1}{n} \sum x_i$. Hence, for
    any $k$, there is a $c$, such that
    $\Lambda_x(H_0; H_1) > k \iff \ol{x} > c$.
    Thus the LRT critical region is $\{x : \ol{x}
    > a\}$ for some constant $c$. By the same
    logic the LRT is of the form
    \[ C = \{\sqrt{n} \frac{(\ol{x} -
    \mu_0)}{\sigma_0} < c'\} \]
    want to pick $c'$ such that
    \[ \PP_{H_0} \left( \sqrt{n} \frac{(\ol{x} -
    \mu_0)}{\sigma_0} > c' \right) = \alpha \]
    But $\sqrt{n} \frac{(\ol{x} -
    \mu_0)}{\sigma_0} \sim \normaldist(0, 1)$
    (this is a pivot). So if we take $c' =
    \Phi^{-1}(1 - \alpha) \cdot z_\alpha$. Finally
    the LRT has critical region
    \[ \left\{ x : \frac{\sqrt{n}(\ol{x} -
    \mu_0)}{\sigma_0} > z_\alpha \right\} \]
    By N-D lemma, this is the most powerful test
    of size $\alpha$. This is called a
    ``$z$-test'' because we use a $z$ statistic $z
    = \sqrt{n} \left( \frac{\ol{x} -
    \mu_0}{\sigma_0} \right)$ to define the
    critical region.
\end{example*}

\subsubsection*{P-value}

For any test with critical region of the form $\{x
: T(x) > k\}$ for some statistic $T$, a
\emph{$p$-value} or observed significance level is
\[ p = \PP_{H_0}(T(X) > T(X^*)) \]
where $x^*$ is the observed date. In example we
just saw, let $\mu_0 = 5$, $\mu_1 = 6$, $\sigma_0
= 1$, $\alpha = 0.05$, observe
\[ x^* = (5.1, 5.5, 4.9, 5.3) \]
$\ol{x^*} = 5.2$, $z^* = 0.4$. $z_\alpha =
\Phi^{-1}(1 - \alpha) = 1.645$
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/21f565e8adee11ed.png}
\end{center}
Here, we fail to reject $H_0$: $\mu_0 = 5$, $p =
0.35$.

\begin{proposition*}
    Under $H_0$, $p$ has a $\Unif(0, 1)$
    distribution. $p$ is a function of $x^*$; null
    distribution assumes $x^* \sim \PP_{H_0}$.
\end{proposition*}

\begin{proof}
    \[ \PP_{H_0}(p < u) = \PP_{H_0}(1 - F(T) < u) \]
    where $F$ is the cdf of $T$.
    \begin{align*}
        &= \PP_{H_0}(F(T) > 1 - u)) \\
        &= \PP_{H_0} (T > F^{-1}(1 - u)) \\
        &= 1 - F(F^{-1}(1 - u)) \\
        &= u
    \end{align*}
    for all $u \in [0, 1]$. Thus $p \sim \Unif(0,
    1)$.
\end{proof}

\subsubsection*{Composite Hypotheses}

$X \sim f_X(\bullet \mid \theta)$, $\theta  \in
\Theta$. $H_0$: $\theta \in \Theta_0 \subset
\Theta$, $H_1$: $\theta \in \Theta_1 \subset
\Theta$. Type I, II error probabilities depend on
the value of $\theta$ within $\Theta_0$ or
$\Theta_1$ respectively. Let $C$ be some critical
region.

\begin{flashcard}[power-func-and-ump-test]
\begin{definition*}[Power Function and UMP test]
    The \emph{power function} of the test $C$
    is\cloze{
    \[ W(\theta) = \PP_\theta(\ub{x \in
    C}_{\text{$H_0$ rejected}}) \]}
    The \emph{size} of $c$ is \cloze{the worst case Type
    I error probability:
    \[ \alpha=  \sup_{\theta \in \Theta} W(\theta)
    \]}
    We say that $C$ is \emph{uniformly most
    powerful} (UMP) \cloze{of size $\alpha$ for
    $H_0$ against $H_1$ if:}
    \begin{enumerate}[(1)]
        \item \cloze{$\sup_{\theta \in \Theta_0}
            W(\theta) = \alpha$}
        \item \cloze{For any other test $C^*$ of size
            $\le \alpha$, with power function
            $W^*$, we have $W(\theta) \ge
            W^*(\theta)$ for all $\theta \in
            \Theta_1$.}
    \end{enumerate}
\end{definition*}
\end{flashcard}

\begin{note*}
    UMP test need not exist. But, in some simple
    cases, the LRT is UMP.
\end{note*}

\begin{example*}
    $X_1, \ldots, X_n \iidsim \normaldist(\mu,
    \sigma_0^2)$: $\sigma_0^2$ known. We wish to
    test $H_0$: $\mu \le \mu_0$ against $H_1$:
    $\mu > \mu_0$ for some fixed $\mu_0$. We just
    studied the simple hypothesis:
    \[ H'_0: \mu = \mu_0, \qquad H'_1 : \mu =
    \mu_1 \qquad (\mu_1 > \mu_0) \]
    LRT was:
    \[ C = \left\{ x : z = \frac{\sqrt{n} (\ol{x}
    - \mu_0)}{\sigma_0} > z_\alpha \right\} \]
    Claim: the same test $C$ is UMP for $H_0$
    against $H_1$. The power function for $C$ is
    \begin{align*}
        W(\mu)
        &= \PP_\mu(X \in C) = \PP_\mu \left(
        \frac{\sqrt{n} (\ol{x} - \mu_0)}{\sigma_0} >
        z_\alpha \right) \\
        &= \PP_\mu \left( \frac{\sqrt{n}(\ol{x} -
        \mu)}{\sigma_0} > z_\alpha +
        \frac{\sqrt{n}(\ol{x} - \mu)}{\sigma_0}
        \right) \\
        &= 1 - \Phi \left( z_\alpha +
        \frac{\sqrt{n} (\mu_0 - \mu)}{\sigma_0}
        \right)
    \end{align*}
    This is monotone increasing in $\mu \in
    (-\infty, \infty)$
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/63609b52adf211ed.png}
    \end{center}
    The test has size $\alpha$ as $\sup_{\mu \in
    \Theta_0} W(\mu) = \alpha$. It remains to show
    that if $C^*$ is another test of size $\le
    \alpha$ with power function $W^*$ then
    $W(\mu_1) \ge W^*(\mu_1)$ for all $\mu_1 >
    \mu_0$. Main observation: critical region only
    depends on $\mu_0$. And $C$ is the LRT for the
    simple hypothesis $H_0'$: $\mu = \mu_0$,
    $H_1'$: $\mu = \mu_1$. Any test $C^*$ of $H_0$
    vs $H_1$ of size $\le \alpha$ also has size
    $\le \alpha$ for $H_0'$ vs $H_1'$.
    \[ W^*(\mu_0) \le \sup_{\mu \in \Theta_0}
    W^*(\mu) \le \alpha \]
    Hence by N-D lemma, we know $W(\mu_1) \ge
    W(\mu_2)$. As we can apply this argument for
    any $\mu_1 > \mu_0$, we have
    \[ W^*(\mu_1) \le W(\mu_1) \qquad \forall
    \mu_1 > \mu_0 \]
\end{example*}
