= Confidence interval from a relative normal pivot
{title2=$(T_n/(1+c/\sqrt n),\ T_n/(1-c/\sqrt n))$}
Suppose $\theta>0$ and a positive <estimator> $T_n$ satisfies $\sqrt n(T_n-\theta)/\theta\xrightarrow{d}N(0,1)$. Choose the <standard normal quantile> $c=\Phi^{-1}(1-\alpha/2)$. For $n>c^2$, an asymptotic <confidence interval> of level $1-\alpha$ is
$$
\boxed{\left(\frac{T_n}{1+c/\sqrt n},\frac{T_n}{1-c/\sqrt n}\right).}
$$
\b[Proof.] <Convergence in distribution> gives $\mathbb P(-c<\sqrt n(T_n-\theta)/\theta<c)\to1-\alpha$. Multiplication by the positive $\theta$ and rearrangement makes this event equivalent to $\theta(1-c/\sqrt n)<T_n<\theta(1+c/\sqrt n)$. Both factors are positive when $n>c^2$, so division yields exactly the displayed interval. Thus its coverage <probability> tends to $1-\alpha$. For $95\%$ coverage, $c\approx1.96$. This inversion retains the unknown scale $\theta$ in the normalized error, rather than replacing it by the observed <estimator> before inversion.
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