Prove that the mean integrated squared error can be written as the sum of the integrated square bias and the integrated variance. That is, prove that \(\operatorname{MISE}\left(\bar{f}_{n}, fight)=\operatorname{ISB}\left(\bar{f}_{n}, fight)+\operatorname{IV}\left(\bar{f}_{n}ight)\), where

\[\operatorname{ISB}\left(\bar{f}_{n}, fight)=\int_{-\infty}^{\infty} \operatorname{Bias}^{2}\left(\bar{f}_{n}(x), f(x)ight) d x=\int_{-\infty}^{\infty}\left[E\left[\bar{f}_{n}(x)ight]-f(x)ight]^{2} d x\]

and

\[\operatorname{IV}\left(\bar{f}_{n}ight)=\int_{-\infty}^{\infty} E\left\{\left[\left(\bar{f}_{n}(x)-E\left(\bar{f}_{n}(x)ight)ight]^{2}ight\} d xight.\]