= Bayes factor for the sign of a normal mean
{c}
{title2=$B_{01}=\Phi(-z)/\Phi(z)$}
For a <random sample> $X_i\mid\theta\sim N(\theta,\sigma^2)$ with known $\sigma>0$ and <prior distribution> $\theta\sim N(0,\tau^2)$ with $\tau>0$, compare $H_0:\theta<0$ to $H_1:\theta\geq0$. The <prior odds> are one by symmetry. <Normal-normal conjugacy> gives a normal <posterior distribution> with
$$
\mu_n=\frac{n\tau^2\overline X}{\sigma^2+n\tau^2},\qquad
v_n=\frac{\sigma^2\tau^2}{\sigma^2+n\tau^2}.
$$
Its negative and nonnegative masses are $\Phi(-z)$ and $\Phi(z)$, where
$$
z=\frac{\mu_n}{\sqrt{v_n}}=
\frac{n\tau\overline X}{\sigma\sqrt{\sigma^2+n\tau^2}}.
$$
Since the <Bayes factor> is the <posterior odds> divided by the <prior odds>,
$$
\boxed{B_{01}=\frac{\Phi(-z)}{\Phi(z)}.}
$$
Here $\Phi$ is the <cumulative distribution function> of the <standard normal distribution>. The ratio decreases with the <sample mean>. For any threshold $b>0$, the inequality $B_{01}<b$ is equivalent to $\Phi(z)>1/(1+b)$ and hence to
$$
\overline X>
\frac{\sigma\sqrt{\sigma^2+n\tau^2}}{n\tau}
\Phi^{-1}\!\left(\frac1{1+b}\right).
$$
For $b=1/10$, the <quantile> is $\Phi^{-1}(10/11)\approx1.33518$, and the equivalent <posterior probability> threshold for $H_0$ is $1/11$. The symmetry and proper normalization of the two hypothesis-conditional priors are essential to replacing the <Bayes factor> by the <posterior odds> directly.
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