\documentclass[12pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[spanish]{babel}
\usepackage[margin=20mm]{geometry}
\usepackage{amsmath}
\usepackage{amssymb}            %para \leqslant
\usepackage{colonequals}
\renewcommand{\(}{\left(}
\renewcommand{\)}{\right)}
\renewcommand{\[}{\left[}
\renewcommand{\]}{\right]}
\renewcommand{\P}[1]{\Pr\[#1\]}
\newcommand{\sigue}{\hookrightarrow}
\newcommand{\gaus}[2]{\mathcal N\({#1},{#2}\)}
\newcommand{\<}{\leqslant}
\newcommand{\X}{\bar X}         %media muestral
\newcommand{\et}{\frac{\sigma}{\sqrt n}} %error típico
\newcommand{\pivote}{\frac{\X-\mu}{\sigma/\sqrt n}}
\newcommand{\Fi}[1]{\Phi\(#1\)}
\newcommand{\Fii}[1]{\Phi^{-1}\(#1\)}
\newcommand{\ffi}[1]{\varphi\(#1\)} %\fi cierra \if
\newcommand{\ffid}[1]{\varphi'\(#1\)} %derivada de \ffi
\renewcommand{\d}{\mathrm{d}\,}
\newcommand{\dd}[2]{\frac{\mathrm{d}^2\,#1}{\d #2^2}}
\begin{document}
\title{Minimizar longitud de intervalo}
\author{Población gausiana, $\sigma$ conocida}
\maketitle
\raggedright
\begin{itemize}
\item $X\sigue\gaus{\mu}{\sigma}$ $\implies$
  $\X\sigue\gaus{\mu}{\et}$$\implies$ pivote
  $\pivote\sigue\gaus{0}{1}$ $\equalscolon$ $\Phi$ $\equalscolon$
  $\int\varphi$
\item $\alpha_1+\alpha_2\colonequals\alpha$ \quad $\Fi{a}\colonequals\alpha_1$
  \quad $\Fi{b}\colonequals1-\alpha_2=1-\alpha+\alpha_1$
\item $1-\alpha$ $=$ $\P{a\<\pivote\<b}$
  $=$ $\P{\X-b\et\<\mu\<\X-a\et}$ $\implies$
  I.C.($\mu$) $=$ $\[\X-b\et,\X-a\et\]$
\item longitud $L$ $=$ $\X-a\et-\(\X-b\et\)$ $=$ $\et\(b-a\)$
  $=$ $\et\[\Fii{1-\alpha+\alpha_1}-\Fii{\alpha_1}\]$
\item si en $\alpha_1\in\(0,\alpha\)$ hay un extremo, entonces
  $0$ $=$ $\frac{\d L}{\d\alpha_1}$ $=$
  $\et\[\frac{\d \Fii{1-\alpha+\alpha_1}}{\d\alpha_1}
    -\frac{\d \Fii{\alpha_1}}{\d\alpha_1}\]$ $=$
  $\et\[\frac1{\ffi{\Fii{1-\alpha+\alpha_1}}}
    -\frac1{\ffi{\Fii{\alpha_1}}}\]$ $\iff$ 
  ${\ffi{\Fii{\alpha_1}}}={\ffi{\Fii{1-\alpha+\alpha_1}}}$ $\iff$
  $\Fii{\alpha_1}=-\Fii{1-\alpha+\alpha_1}$ por simetría de $\gaus01$ respecto
  a $0$ $\iff$ $\alpha_1=\alpha_2=\frac\alpha2$
\item se trata de un mínimo, pues $\dd{L}{\alpha_1}$ $=$
  $\et\frac{\d}{\d\alpha_1}\[\frac1{\ffi{\Fii{1-\alpha+\alpha_1}}}
    -\frac1{\ffi{\Fii{\alpha_1}}}\]$ $=$
  $\et\[\dfrac{-\overbrace{\ffid{\Fii{1-\alpha+\alpha_1}}}
      ^{<0\,\Leftarrow\,\Fii{1-\alpha+\alpha_1}>0}
      \overbrace{\dfrac1{\ffi{\Fii{1-\alpha+\alpha_1}}}}
      ^{>0\,\Leftarrow\,\varphi\text{ densidad}}}
    {\ffi{\Fii{1-\alpha+\alpha_1}}^2}
    -\dfrac{-\overbrace{\ffid{\Fii{\alpha_1}}}
      ^{>0\,\Leftarrow\,\Fii{\alpha_1}<0}
      \dfrac1{\ffi{\Fii{\alpha_1}}}}
    {\underbrace{\ffi{\Fii{\alpha_1}}^2}_{>0}}\]$ $>$ $0$
\item el intervalo más corto se obtiene con $\alpha_1=\alpha_2=\alpha/2$,
  es decir, con $a=-b=-z_{1-\alpha/2}=z_{\alpha/2}\colonequals\Fii{\alpha/2}=-\Fii{1-\alpha/2}$
\end{itemize}
\end{document}
