回归模型中对数变换的含义

2024-07-11 13:50| 来源: 网络整理| 查看: 265

1 精确解释 1.1 因变量采用对数变换

l n ( y ^ ) = β 0 + β 1 × x ln(\hat y)=\beta_0 +\beta_1 \times x ln(y^)=β0+β1×x x → x + 1 ; y ^ 1 → y ^ 2 x \to x+1; \hat y_1 \to \hat y_2 x→x+1;y^1→y^2

{ y ^ 1 = e β 0 + β 1 × x y ^ 2 = e β 0 + β 1 × ( x + 1 ) \begin{cases} \hat y_1=e^{\beta_0 +\beta_1 \times x}\\ \hat y_2=e^{\beta_0 +\beta_1 \times (x+1)} \end{cases} {y^1=eβ0+β1×xy^2=eβ0+β1×(x+1)

y ^ 2 y ^ 1 = e β 0 + β 1 × ( x + 1 ) e β 0 + β 1 × x = e [ β 0 + β 1 × ( x + 1 ) ] − [ β 0 + β 1 × x ] = e β 1 \begin{aligned} &\frac{\hat y_2}{\hat y_1}\\ &=\frac{e^{\beta_0+\beta_1 \times (x+1)}}{e^{\beta_0+\beta_1 \times x}}\\ &=e^{[\beta_0+\beta_1 \times (x+1)]-[\beta_0+\beta_1 \times x]}\\ &=e^{\beta_1} \end{aligned} y^1y^2=eβ0+β1×xeβ0+β1×(x+1)=e[β0+β1×(x+1)]−[β0+β1×x]=eβ1

结论

x x x每增加一个单位变为 x + 1 x+1 x+1， y ^ \hat y y^变为原来的 e β 1 e^{\beta_1} eβ1倍 x x x每增加一个单位变为 x + 1 x+1 x+1， y ^ \hat y y^相比原来增加 [ e β 1 − 1 ] × 100 % [e^{\beta_1}-1]\times 100\% [eβ1−1]×100% 1.2 自变量采用对数变换

y ^ = β 0 + β 1 × l n ( x ) \hat y=\beta_0+\beta_1 \times ln(x) y^=β0+β1×ln(x) x → e × x ; y ^ 1 → y ^ 2 x \to e\times x; \hat y_1 \to \hat y_2 x→e×x;y^1→y^2 { y ^ 1 = β 0 + β 1 × l n ( x ) y ^ 2 = β 0 + β 1 × l n ( e × x ) \begin{cases} \hat y_1=\beta_0+\beta_1 \times ln(x)\\ \hat y_2=\beta_0+\beta_1 \times ln(e \times x) \end{cases} {y^1=β0+β1×ln(x)y^2=β0+β1×ln(e×x)

y ^ 2 − y ^ 1 = [ β 0 + β 1 × l n ( e × x ) ] − [ β 0 + β 1 × l n ( x ) ] = [ β 0 + β 1 × ( l n ( e ) + l n ( x ) ) ] − [ β 0 + β 1 × l n ( x ) ] = [ β 0 + β 1 × l n ( e ) + β 1 × l n ( x ) ] − [ β 0 + β 1 × l n ( x ) ] = β 1 × l n ( e ) = β 1 \begin{aligned} &\hat y_2-\hat y_1 \\ &=[\beta_0+\beta_1 \times ln(e \times x)]-[\beta_0+\beta_1 \times ln(x)]\\ &=[\beta_0+\beta_1 \times (ln(e)+ln(x))]-[\beta_0+\beta_1 \times ln(x)]\\ &=[\beta_0+\beta_1 \times ln(e)+\beta_1 \times ln(x)]-[\beta_0+\beta_1 \times ln(x)]\\ &=\beta_1 \times ln(e)\\ &=\beta_1 \end{aligned} y^2−y^1=[β0+β1×ln(e×x)]−[β0+β1×ln(x)]=[β0+β1×(ln(e)+ln(x))]−[β0+β1×ln(x)]=[β0+β1×ln(e)+β1×ln(x)]−[β0+β1×ln(x)]=β1×ln(e)=β1 结论

x x x变为原来的 e e e倍后，则 y ^ \hat y y^增加 β 1 \beta_1 β1如果使用以2为底的对数变换，则 x x x变为原来的2倍后， y ^ \hat y y^增加 β 1 \beta_1 β1 1.3 因变量和自变量同时采用对数变换

l n ( y ^ ) = β 0 + β 1 × l n ( x ) ln(\hat y)=\beta_0 +\beta_1\times ln(x) ln(y^)=β0+β1×ln(x) x → k × x ; y ^ 1 → y ^ 2 x \to k \times x; \hat y_1 \to \hat y_2 x→k×x;y^1→y^2 { y ^ 1 = e β 0 + β 1 × l n ( x ) y ^ 2 = e β 0 + β 1 × l n ( k × x ) \begin{cases} \hat y_1=e^{\beta_0+\beta_1\times ln(x)}\\ \hat y_2=e^{\beta_0+\beta_1\times ln(k \times x)} \end{cases} {y^1=eβ0+β1×ln(x)y^2=eβ0+β1×ln(k×x)

y ^ 2 y ^ 1 = e β 0 + β 1 × l n ( k × x ) e β 0 + β 1 × l n ( x ) = e [ β 0 + β 1 × l n ( k × x ) ] − [ β 0 + β 1 × l n ( x ) ] = e [ β 0 + β 1 × l n ( k ) + β 1 × l n ( x ) ] − [ β 0 + β 1 × l n ( x ) ] = e β 1 × l n ( k ) = [ e l n ( k ) ] β 1 = k β 1 \begin{aligned} &\frac{\hat y_2}{\hat y_1}\\ &=\frac{e^{\beta_0+\beta_1\times ln(k \times x)}}{e^{\beta_0+\beta_1\times ln(x)}}\\ &=e^{[\beta_0+\beta_1\times ln(k \times x)]-[\beta_0+\beta_1\times ln(x)]}\\ &=e^{[\beta_0+\beta_1 \times ln(k)+\beta_1 \times ln(x)] - [\beta_0+\beta_1\times ln(x)]}\\ &=e^{\beta_1 \times ln(k)}\\ &=[e^{ln(k)}]^{\beta_1}\\ &=k^{\beta_1} \end{aligned} y^1y^2=eβ0+β1×ln(x)eβ0+β1×ln(k×x)=e[β0+β1×ln(k×x)]−[β0+β1×ln(x)]=e[β0+β1×ln(k)+β1×ln(x)]−[β0+β1×ln(x)]=eβ1×ln(k)=[eln(k)]β1=kβ1 结论

x x x变为原来的 k k k倍， y ^ \hat y y^变为原来的 k β 1 k^{\beta_1} kβ1倍 x x x变为原来的 k k k倍， y ^ \hat y y^增加 [ k β 1 − 1 ] × 100 % [k^{\beta_1} - 1]\times 100\% [kβ1−1]×100% 2 粗略解释 2.1 e β − 1 e^{\beta} - 1 eβ−1与 β \beta β的关系 library(ggplot2) library(latex2exp) x

【本文地址】

公司简介

联系我们