Forma de calcular el método de los mínimos cuadrados
SSE={{\sum\limits_{i=1}^{n}{\left( {{y}_{i}}-{{{\hat{y}}}_{i}} \right)}}^{2}}\text{ sea m }\!\!\acute{\mathrm{i}}\!\!\text{ nima}
Como:
\overset{\hat{\ }}{\mathop{{{y}_{i}}}}\,=a+b{{x}_{i}}
Remplazando:
SSE=\underset{i=1}{\overset{n}{\mathop \sum }}\,{{\left( {{y}_{i}}-\left( a+b{{x}_{i}} \right) \right)}^{2}}
SSE=\underset{i=1}{\overset{n}{\mathop \sum }}\,{{\left( {{y}_{i}}-a-b{{x}_{i}} \right)}^{2}}
Para minimizar se debe derivar parcialmente e igualar a cero:
\frac{\partial SSE}{\partial a}=0$ $\frac{\partial SSE}{\partial b}=0
La derivada con respecto de «a»:
2\underset{i=1}{\overset{n}{\mathop \sum }}\,\left( {{y}_{i}}-a-b{{x}_{i}} \right)\left( -1 \right)=0
Distribuyendo la sumatoria:
\underset{i=1}{\overset{n}{\mathop \sum }}\,\left( {{y}_{i}} \right)-\underset{i=1}{\overset{n}{\mathop \sum }}\,a-b\underset{i=1}{\overset{n}{\mathop \sum }}\,\left( {{x}_{i}} \right)=0
Como:
La derivada con respecto de «b»:
2\underset{i=1}{\overset{n}{\mathop \sum }}\,\left( {{y}_{i}}-a-b{{x}_{i}} \right)\left( -{{x}_{i}} \right)=0
Distribuyendo la sumatoria:
\underset{i=1}{\overset{n}{\mathop \sum }}\,{{x}_{i}}{{y}_{i}}-a\underset{i=1}{\overset{n}{\mathop \sum }}\,{{x}_{i}}-b\underset{i=1}{\overset{n}{\mathop \sum }}\,{{x}_{i}}^{2}=0
a\underset{i=1}{\overset{n}{\mathop \sum }}\,{{x}_{i}}+b\underset{i=1}{\overset{n}{\mathop \sum }}\,{{x}_{i}}^{2}=\underset{i=1}{\overset{n}{\mathop \sum }}\,{{x}_{i}}{{y}_{i}}~ecuaci\acute{o}n~2
De la ecuación 2, si se despeja «a»:
a=\frac{\mathop{\sum }_{i=1}^{n}{{x}_{i}}{{y}_{i}}-b\mathop{\sum }_{i=1}^{n}{{x}_{i}}^{2}}{\mathop{\sum }_{i=1}^{n}{{x}_{i}}}
Remplazando lo anterior en la ecuación 1:
n\left( \frac{\mathop{\sum }_{i=1}^{n}{{x}_{i}}{{y}_{i}}-b\mathop{\sum }_{i=1}^{n}{{x}_{i}}^{2}}{\mathop{\sum }_{i=1}^{n}{{x}_{i}}} \right)+b\underset{i=1}{\overset{n}{\mathop \sum }}\,\left( {{x}_{i}} \right)=\underset{i=1}{\overset{n}{\mathop \sum }}\,\left( {{y}_{i}} \right)
\frac{n\left( \mathop{\sum }_{i=1}^{n}{{x}_{i}}{{y}_{i}}-b\mathop{\sum }_{i=1}^{n}{{x}_{i}}^{2} \right)}{\mathop{\sum }_{i=1}^{n}{{x}_{i}}}+b\underset{i=1}{\overset{n}{\mathop \sum }}\,\left( {{x}_{i}} \right)=\underset{i=1}{\overset{n}{\mathop \sum }}\,\left( {{y}_{i}} \right)
\frac{n\left( \mathop{\sum }_{i=1}^{n}{{x}_{i}}{{y}_{i}}-b\mathop{\sum }_{i=1}^{n}{{x}_{i}}^{2} \right)+b\mathop{\sum }_{i=1}^{n}\left( {{x}_{i}} \right)\mathop{\sum }_{i=1}^{n}\left( {{x}_{i}} \right)}{\mathop{\sum }_{i=1}^{n}{{x}_{i}}}=\underset{i=1}{\overset{n}{\mathop \sum }}\,\left( {{y}_{i}} \right)
n\underset{i=1}{\overset{n}{\mathop \sum }}\,{{x}_{i}}{{y}_{i}}-nb\underset{i=1}{\overset{n}{\mathop \sum }}\,{{x}_{i}}^{2}+b\underset{i=1}{\overset{n}{\mathop \sum }}\,{{\left( {{x}_{i}} \right)}^{2}}=\underset{i=1}{\overset{n}{\mathop \sum }}\,\left( {{x}_{i}} \right)\underset{i=1}{\overset{n}{\mathop \sum }}\,\left( {{y}_{i}} \right)
b\left( \underset{i=1}{\overset{n}{\mathop \sum }}\,{{\left( {{x}_{i}} \right)}^{2}}-n\underset{i=1}{\overset{n}{\mathop \sum }}\,{{x}_{i}}^{2} \right)=~\underset{i=1}{\overset{n}{\mathop \sum }}\,\left( {{x}_{i}} \right)\underset{i=1}{\overset{n}{\mathop \sum }}\,\left( {{y}_{i}} \right)-n\underset{i=1}{\overset{n}{\mathop \sum }}\,{{x}_{i}}{{y}_{i}}
b=\frac{\mathop{\sum }_{i=1}^{n}\left( {{x}_{i}} \right)\mathop{\sum }_{i=1}^{n}\left( {{y}_{i}} \right)-n\mathop{\sum }_{i=1}^{n}{{x}_{i}}{{y}_{i}}}{\mathop{\sum }_{i=1}^{n}{{\left( {{x}_{i}} \right)}^{2}}-n\mathop{\sum }_{i=1}^{n}{{x}_{i}}^{2}}
Para encontrar «a» se despeja de la ecuación 1:
a=\frac{\mathop{\sum }_{i=1}^{n}\left( {{y}_{i}} \right)-b\mathop{\sum }_{i=1}^{n}\left( {{x}_{i}} \right)}{n}
a=\frac{\mathop{\sum }_{i=1}^{n}\left( {{y}_{i}} \right)}{n}-\frac{b\mathop{\sum }_{i=1}^{n}\left( {{x}_{i}} \right)}{n}
Como:
\bar{x}=\frac{\mathop{\sum }_{i=1}^{n}\left( {{x}_{i}} \right)}{n} y \bar{y}=\frac{b\mathop{\sum }_{i=1}^{n}\left( {{y}_{i}} \right)}{n}
a=\bar{y}-b\bar{x}
