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更改
→一维函数映射
EI&=I(y;x|do(x\sim U[-L/2,L/2]))\\
EI&=I(y;x|do(x\sim U[-L/2,L/2]))\\
&=\int_{-L/2}^{L/2}\int_{f([-L/2,L/2])}p(x)p(y|x)\ln\frac{p(y|x)}{p(y)}dydx\\
&=\int_{-L/2}^{L/2}\int_{f([-L/2,L/2])}p(x)p(y|x)\ln\frac{p(y|x)}{p(y)}dydx\\
&=\int_{-L/2}^{L/2}\int_{f([-L/2,L/2])}\frac{1}{L}\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(y-f(x))^2}{\sigma^2}\right)\ln\frac{\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(y-f(x))^2}{\sigma^2}\right)}{\int_{-L/2}^{L/2}\frac{1}{L}\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(y-f(z))^2}{\sigma^2}\right)dz}dydx
&=\int_{-\frac{L}{2}}^{\frac{L}{2}}\int_{f([-\frac{L}{2},\frac{L}{2}])}\frac{1}{L}\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(y-f(x))^2}{\sigma^2}\right)\ln\left[\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(y-f(x))^2}{\sigma^2}\right)\right]dydx-\int_{-\frac{L}{2}}^{\frac{L}{2}}\int_{f([-\frac{L}{2},\frac{L}{2}])}\frac{1}{L}\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(y-f(x))^2}{\sigma^2}\right)\ln\left[\frac{1}{L}\int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(y-f(z))^2}{\sigma^2}\right)dz\right]dydx\\
&=\int_{-L/2}^{L/2}\int_{f([-L/2,L/2])}\frac{1}{L}\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(y-f(x))^2}{\sigma^2}\right)\ln\frac{\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(y-f(x))^2}{\sigma^2}\right)}{\int_{-L/2}^{L/2}\frac{1}{L}\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(y-f(z))^2}{\sigma^2}\right)dz}dydx\\
&\approx \ln(\frac{L}{\sqrt{2\pi e}})+\frac{1}{2L}\int_{-L/2}^{L/2}\ln \left(\frac{f'(x)}{\epsilon}\right)^2dx.
\end{aligned}
\end{aligned}
</math>
</math>
如果同时考虑两种噪声，并且如果干预空间大小为<math>L
如果同时考虑两种噪声，并且如果干预空间大小为<math>L