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We have $E[\hat{f_n}(x)]=\frac{F(x+h)-F(x-h)}{2h}$, $h\downarrow0$. In order to compute this expectation I need to use a Taylor expansion, under the assumption that f' and f'' exists:

$F(x+h)=F(x)+hf(x)+\frac{h^2}{2}f'(x)+\frac{h^3}{6}f''(x)+o(h^3)$, $F(x-h)=F(x)-hf(x)+\frac{h^2}{2}f'(x)-\frac{h^3}{6}f''(x)+o(h^3)$.

Now plugging this into the expectation I find:

$E[\hat{f_n}(x)]=\frac{f(x)+\frac{2h^3}{6}f''(x)}{2h}=f(x)+\frac{h^2}{6}f''(x)$.

So the error terms cancel out, but the actual expectation should be $E[\hat{f_n}(x)]=f(x)+\frac{h^2}{6}f''(x)+o(h^2)$. So obviously I'm making a mistake, how do I handle the error term?

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    $\begingroup$ $o(x) - o(x) = o(x)$, not $0$ $\endgroup$ – Michael Galuza Aug 12 '15 at 10:14
  • $\begingroup$ Note that $O(h^3)$ and $o(h^3)$is different. Make sure you are correct about this notation. $\endgroup$ – Rowan Aug 12 '15 at 10:23

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