Using the fact that the pointwise bias of the histogram is given by, [operatorname{Bias}left[bar{f}_{n}(x)ight]=frac{1}{2} f^{prime}(x)left[h-2left(x-g_{i}ight)ight]+Oleft(h^{2}ight),] as (h
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Using the fact that the pointwise bias of the histogram is given by,
\[\operatorname{Bias}\left[\bar{f}_{n}(x)ight]=\frac{1}{2} f^{\prime}(x)\left[h-2\left(x-g_{i}ight)ight]+O\left(h^{2}ight),\]
as \(h ightarrow 0\), prove that the square bias is given by
\[\operatorname{Bias}^{2}\left[\bar{f}_{n}(x)ight]=\frac{1}{4}\left[f^{\prime}(x)ight]^{2}\left[h-2\left(\left(x-g_{i}ight)ight]^{2}+O\left(h^{3}ight) .ight.\]
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