线性回归、Lasso回归、岭回归、逻辑回归的损失函数
线性回归:
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J(\theta)=\frac{1}{2m}\sum_{i=1}^m(h(x^{(i)})-y^{(i)})^2
J(θ)=2m1?∑i=1m?(h(x(i))?y(i))2 Lasso回归:
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J(\theta)=\frac{1}{2m}\sum_{i=1}^m(h(x^{(i)})-y^{(i)})^2+\lambda\sum_{j=1}^{n}|\theta|
J(θ)=2m1?∑i=1m?(h(x(i))?y(i))2+λ∑j=1n?∣θ∣ 岭回归:
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J(\theta)=\frac{1}{2m}\sum_{i=1}^m(h(x^{(i)})-y^{(i)})^2+\lambda\sum_{j=1}^{n}\theta^2
J(θ)=2m1?∑i=1m?(h(x(i))?y(i))2+λ∑j=1n?θ2 LR:
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J(\theta)=-\frac{1}{m}\sum_{i=1}^m[(1-y^{(i)})log(1-h(x^{(i)}))+y^{(i)}log(h(x^{(i)}))]
J(θ)=?m1?∑i=1m?[(1?y(i))log(1?h(x(i)))+y(i)log(h(x(i)))]
推导LR
LR的损失函数推导
根据sigmoid函数的定义,
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P(y=1|x,\theta)=h(x)
P(y=1∣x,θ)=h(x),
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P(y=0|x,\theta)=1-h(x)
P(y=0∣x,θ)=1?h(x) 因此,
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P(y|x,\theta)=h(x)^y[1-h(x)]^{1-y}
P(y∣x,θ)=h(x)y[1?h(x)]1?y。 目标是最大化
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P(y|x)
P(y∣x),即最大化其对数。 令似然函数L=
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lnL=ylog(h(x)+(1-y)log(1-h(x))
lnL=ylog(h(x)+(1?y)log(1?h(x))。 损失函数求最小化,注意加负号:
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J(\theta)=-\frac{1}{m}\sum_{i=1}^m[y^{(i)}log(h(x^{(i)}))+(1-y^{(i)})log(1-h(x^{(i)}))]
J(θ)=?m1?∑i=1m?[y(i)log(h(x(i)))+(1?y(i))log(1?h(x(i)))]
LR的导数推导
对损失函数
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J(\theta)=-\frac{1}{m}\sum_{i=1}^m[y^{(i)}log(h(x^{(i)}))+(1-y^{(i)})log(1-h(x^{(i)}))]
J(θ)=?m1?∑i=1m?[y(i)log(h(x(i)))+(1?y(i))log(1?h(x(i)))]求导:
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\begin{aligned} J'(\theta_j)&=-\frac{1}{m}\sum_{i=1}^m[y^{(i)}\frac{h'_{\theta}(x^{(i)})}{h(x^{(i)})}+(1-y^{(i)})\frac{h'_{\theta}(x^{(i)})}{1-h(x^{(i)})}]\\[2ex] &=-\frac{1}{m}\sum_{i=1}^m[y^{(i)}\frac{h(x^{(i)})[1-h(x^{(i)})](x_j^{(i)})}{h(x^{(i)})}+(1-y^{(i)})\frac{h(x^{(i)})[1-h(x^{(i)})](x_j^{(i)})}{1-h(x^{(i)})}]\\[2ex] &=-\frac{1}{m}\sum_{i=1}^m[(y^{(i)}-h(x^{(i)}))x_j^{(i)}] \end{aligned}
J′(θj?)?=?m1?i=1∑m?[y(i)h(x(i))hθ′?(x(i))?+(1?y(i))1?h(x(i))hθ′?(x(i))?]=?m1?i=1∑m?[y(i)h(x(i))h(x(i))[1?h(x(i))](xj(i)?)?+(1?y(i))1?h(x(i))h(x(i))[1?h(x(i))](xj(i)?)?]=?m1?i=1∑m?[(y(i)?h(x(i)))xj(i)?]?
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\theta_j=\theta_j+\alpha\frac{1}{m}\sum_{i=1}^m[(y^{(i)}-h(x^{(i)}))x_j^{(i)}]
θj?=θj?+αm1?∑i=1m?[(y(i)?h(x(i)))xj(i)?] 对比线性回归对参数的导数:
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J'(\theta_j)=\frac{1}{m}\sum_{i=1}^m(h(x^{(i)})-y^{(i)})x_j^{(i)}
J′(θj?)=m1?∑i=1m?(h(x(i))?y(i))xj(i)?
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\theta_j=\theta_j+\alpha\frac{1}{m}\sum_{i=1}^m[(y^{(i)}-h(x^{(i)}))x_j^{(i)}]
θj?=θj?+αm1?∑i=1m?[(y(i)?h(x(i)))xj(i)?] 可以发现二者虽然损失函数不同,但导数和梯度下降的公式却是相同的(神奇)
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