Lema de Neyman-Pearson

Sea ${\displaystyle X_{1},X_{2},\dots ,X_{n}}$ una muestra aleatoria de una población con función de densidad ${\displaystyle f(x;\theta )}$ donde ${\displaystyle \theta \in \Theta =\{\theta _{0},\theta _{1}\}}$ y sean ${\displaystyle 0<\alpha <1}$, ${\displaystyle k\in \mathbb {R} ^{+}}$ y ${\displaystyle {\mathcal {C}}}$ tales que

1. ${\displaystyle \operatorname {P} [\mathbf {X} \in {\mathcal {C}}|H_{0}]=\alpha }$
2. ${\displaystyle \lambda ={\frac {{\mathcal {L}}(\theta _{0})}{{\mathcal {L}}(\theta _{1})}}={\frac {\prod _{i=1}^{n}f(x_{i};\theta _{0})}{\prod _{i=1}^{n}f(x_{i};\theta _{1})}}\leq k}$ si ${\displaystyle \mathbf {x} \in {\mathcal {C}}}$.
3. ${\displaystyle \lambda >k}$ si ${\displaystyle \mathbf {x} \in {\mathcal {C}}^{c}}$.

entonces la prueba asociada a ${\displaystyle {\mathcal {C}}}$ es una prueba más potente para probar ${\displaystyle H_{0}:\theta =\theta _{0}}$ contra ${\displaystyle H_{1}:\theta =\theta _{1}}$, es decir, ${\displaystyle {\mathcal {C}}}$ es la mejor región crítica.

Sea ${\displaystyle X_{1},X_{2},\dots ,X_{n}}$ una muestra aleatoria de una población con distribución ${\displaystyle N(\mu ,\sigma _{0}^{2})}$ donde ${\displaystyle \sigma _{0}^{2}}$ es conocida. Considere

${\displaystyle {\begin{aligned}H_{0}&:\mu =\mu _{0}\\H_{1}&:\mu =\mu _{1}\\\alpha \end{aligned}}}$

siendo ${\displaystyle \mu _{0}<\mu _{1}}$.

En esta caso la función de verosimilitud es

${\displaystyle {\begin{aligned}{\mathcal {L}}(x_{1},\dots ,x_{n};\mu ,\sigma _{0}^{2})&=\prod _{i=1}^{n}{\frac {1}{\sqrt {2\pi \sigma _{0}^{2}}}}\exp \left(-{\frac {(x_{i}-\mu )^{2}}{2\sigma _{0}^{2}}}\right)\\&=\left({\frac {1}{\sqrt {2\pi \sigma _{0}^{2}}}}\right)^{n}\exp \left(-{\frac {1}{2\sigma _{0}^{2}}}\sum _{i=1}^{n}(x_{i}-\mu )^{2}\right)\end{aligned}}}$

por el lema de Neyman-Pearson

${\displaystyle {\begin{aligned}{\frac {{\mathcal {L}}_{0}}{{\mathcal {L}}_{1}}}&={\frac {\left({\frac {1}{\sqrt {2\pi \sigma _{0}^{2}}}}\right)^{n}\exp \left(-{\frac {1}{2\sigma _{0}^{2}}}\sum _{i=1}^{n}(x_{i}-\mu _{0})^{2}\right)}{\left({\frac {1}{\sqrt {2\pi \sigma _{0}^{2}}}}\right)^{n}\exp \left(-{\frac {1}{2\sigma _{0}^{2}}}\sum _{i=1}^{n}(x_{i}-\mu _{1})^{2}\right)}}\\&=\exp \left(-{\frac {1}{2\sigma _{0}^{2}}}\sum _{i=1}^{n}(x_{i}-\mu _{0})^{2}+{\frac {1}{2\sigma _{0}^{2}}}\sum _{i=1}^{n}(x_{i}-\mu _{1})^{2}\right)\end{aligned}}}$

pero

${\displaystyle {\begin{aligned}\sum _{i=1}^{n}(x_{i}-\mu )^{2}&=\sum _{i=1}^{n}(x_{i}^{2}-2\mu x_{i}+\mu ^{2})\\&=\sum _{i=1}^{n}x_{i}^{2}-2\mu \sum _{i=1}^{n}x_{i}+n\mu ^{2}\\&=\sum _{i=1}^{n}x_{i}^{2}-2\mu n{\bar {x}}+n\mu ^{2}\end{aligned}}}$

por lo que

${\displaystyle {\begin{aligned}{\frac {{\mathcal {L}}_{0}}{{\mathcal {L}}_{1}}}&=\exp \left[-{\frac {1}{2\sigma _{0}^{2}}}\left(\sum _{i=1}^{n}x_{i}^{2}-2\mu _{0}n{\bar {x}}+n\mu _{0}^{2}-\sum _{i=1}^{n}x_{i}^{2}+2\mu _{1}n{\bar {x}}-n\mu _{1}^{2}\right)\right]\\&=\exp \left[-{\frac {1}{2\sigma _{0}^{2}}}\left(2n{\bar {x}}(\mu _{1}-\mu _{0})+n(\mu _{0}^{2}-\mu _{1}^{2})\right)\right]\\&=\exp \left[{\frac {n{\bar {x}}(\mu _{0}-\mu _{1})}{\sigma _{0}^{2}}}-{\frac {n(\mu _{0}^{2}-\mu _{1}^{2})}{2\sigma _{0}^{2}}}\right]\leq k_{1}\end{aligned}}}$

lo anterior implica

${\displaystyle {\begin{aligned}&{\frac {n{\bar {x}}(\mu _{0}-\mu _{1})}{\sigma _{0}^{2}}}-{\frac {n(\mu _{0}^{2}-\mu _{1}^{2})}{2\sigma _{0}^{2}}}\leq k_{2}=\ln(k_{1})\\&{\frac {n{\bar {x}}(\mu _{0}-\mu _{1})}{\sigma _{0}^{2}}}\leq k_{3}=k_{2}+{\frac {n(\mu _{0}^{2}-\mu _{1}^{2})}{2\sigma _{0}^{2}}}\end{aligned}}}$

como ${\displaystyle \mu _{1}>\mu _{0}}$ entonces ${\displaystyle \mu _{0}-\mu _{1}<0}$ luego

${\displaystyle {\bar {x}}\geq k={\frac {k_{3}\sigma _{0}^{2}}{n(\mu _{0}-\mu _{1})}}}$

por lo tanto se rechaza ${\displaystyle H_{0}}$ si ${\displaystyle {\bar {x}}\geq k}$, es decir la región de rechazo ${\displaystyle {\mathcal {C}}}$ queda descrita como

${\displaystyle {\mathcal {C}}=\{(X_{1},X_{2},\dots ,X_{n}):{\bar {X}}\geq k\}}$