derivate of a and b in regression
dv for a
dv for b
to understand [[:gradient descent]]
[{{:r.regressionline3.png}}]

\begin{eqnarray*}
\sum{(Y_i - \hat{Y_i})^2} 
& = & \sum{(Y_i - (a + bX_i))^2}  \;\;\; \because \hat{Y_i} = a + bX_i \\
& = & \text{SSE or SS.residual} \;\;\; \text{(and this should be the least value.)}
\end{eqnarray*}

<WRAP box>
\begin{eqnarray*}
\text{for a (constant)} \\ 
\\
\dfrac{\text{d}}{\text{da}} \sum{(Y_i - (a + bX_i))^2} 
& = & \sum \dfrac{\text{d}}{\text{da}} {(Y_i - (a + bX_i))^2} \\
& & \because {(Y_i - (a + bX_i))^2} = \text{residual}^2 \\
& & \therefore{} \\
& = & \sum \dfrac{\text{dresidual}^2} {da}  \\
& = & \sum \dfrac{\text{dresidual}^2}{\text{dresidual}} * \dfrac{\text{dresidual}}{\text{da}} \\
& = & \sum{2 * \text{residual}} * {\dfrac{\text{dresidual}}{\text{da}}} \;\;\;\; \\
& = & \sum{2 * \text{residual}} * {\dfrac{d{(Y_i - (a + bX_i))}}{\text{da}}} \;\;\;\; \\
& = & \sum{2 * \text{residual}} * (0 - 1 - 0) \;\;\;\; \\
& & \because{Y_i = 0; \;\;\; a = 1; \;\;\; bX_i = 0} \\
& = & \sum{2 * \text{residual}} * (-1) \;\;\;\; \\
& = & -2 \sum{(Y_i - (a + bX_i))} \\ 
\\
\text{in order to have the least value, the above should be zero} \\ 
\\
-2 \sum{(Y_i - (a + bX_i))} & = &  0 \\
\sum{(Y_i - (a + bX_i))} & = & 0 \\ 
\sum{Y_i} - \sum{a} - b \sum{X_i} & = & 0 \\
\sum{Y_i} - n*{a} - b \sum{X_i} & = & 0 \\
n*{a} & = & \sum{Y_i} - b \sum{X_i} \\
a & = & \dfrac{\sum{Y_i}}{n} - b \dfrac{\sum{X_i}}{n} \\
a & = & \overline{Y} - b \overline{X} \\
\end{eqnarray*} 
</WRAP>

<WRAP box>
\begin{eqnarray*}
\text{for b, (coefficient)} \\ 
\\
\dfrac{\text{d}}{\text{db}} \sum{(Y_i - (a + bX_i))^2}  & = & \sum \dfrac{\text{d}}{\text{db}} {(Y_i - (a + bX_i))^2} \\ 
& = & \sum{2 (Y_i - (a + bX_i))} * (-X_i) \;\;\;\; \\
& \because & \dfrac{\text{d}}{\text{dv for b}} (Y_i - (a+bX_i)) = -X_i \\
& = & -2 \sum{X_i (Y_i - (a + bX_i))} \\
\\
\text{in order to have the least value, the above should be zero} \\ 
\\
-2 \sum{X_i (Y_i - (a + bX_i))} & = & 0 \\
\sum{X_i (Y_i - (a + bX_i))} & = & 0 \\ 
\sum{X_i (Y_i - ((\overline{Y} - b \overline{X}) + bX_i))} & = & 0 \\ 
\sum{X_i ((Y_i - \overline{Y}) - b (X_i - \overline{X})) } & = & 0 \\ 
\sum{X_i (Y_i - \overline{Y})} - \sum{b X_i (X_i - \overline{X}) } & = & 0 \\ 
\sum{X_i (Y_i - \overline{Y})} & = &  b \sum{X_i (X_i - \overline{X})} \\ 
b & = & \dfrac{\sum{X_i (Y_i - \overline{Y})}}{\sum{X_i (X_i - \overline{X})}} \\
b & = & \dfrac{\sum{(Y_i - \overline{Y})}}{\sum{(X_i - \overline{X})}} \\
b & = & \dfrac{ \sum{(Y_i - \overline{Y})(X_i - \overline{X})} } {\sum{(X_i - \overline{X})(X_i - \overline{X})}} \\
b & = & \dfrac{ \text{SP} } {\text{SS}_\text{x}} = \dfrac{\text{Cov(X, Y)}} {\text{Var(X)}} = \dfrac{\text{Cov(X, Y)}} {\text{Cov(X, X)}}\\
\end{eqnarray*} 
</WRAP>
리그레션 라인으로 예측하고 틀린 나머지 error의 제곱의 합을 (ss.res) 최소값으로 만드는 선의 기울기와 절편값은 위와 같다 (a and b).

위는 증명을 통해서 a와 b값을 알아낸 것이고, [[:gradient descent|R과 같은 어플리케이션에서 a와 b를 알아내는 방법은]] 없을까?