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--- deriviation_of_a_and_b_in_a_simple_regression [2024/05/23 08:27] – hkimscil
+++ deriviation_of_a_and_b_in_a_simple_regression [2025/08/05 06:24] (current) – hkimscil
@@ Line 1: / Line 1: @@
+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 \\
+\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{dv}} \sum{(Y_i - (a + bX_i))^2} & = & \sum \dfrac{\text{d}}{\text{dv}} {(Y_i - (a + bX_i))^2} \\
+\dfrac{\text{d}}{\text{da}} \sum{(Y_i - (a + bX_i))^2}
-& = & \sum{2 (Y_i - (a + bX_i))} * (-1) \;\;\;\; \because \dfrac{\text{d}}{\text{dv for a}} (Y_i - (a+bX_i)) = -1 \\
+& = & \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))} \\
 \\
@@ Line 24: / Line 38: @@
 a & = & \overline{Y} - b \overline{X} \\
 \end{eqnarray*}
+</WRAP>
+<WRAP box>
 \begin{eqnarray*}
 \text{for b, (coefficient)} \\
 \\
-\dfrac{\text{d}}{\text{dv}} \sum{(Y_i - (a + bX_i))^2}  & = & \sum \dfrac{\text{d}}{\text{dv}} {(Y_i - (a + bX_i))^2} \\
+\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 \\
+& = & \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))} \\
 \\
@@ Line 45: / Line 60: @@
 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}} \\
+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?
-{{:pasted:20240522-084708.jpeg?400}}
-{{:pasted:20240522-084738.jpeg?400}}