I'm trying to calculate the bias of this estimator of $f$ a $C^4$ mesurable function:
$$\hat{f'}_{h,n} = \cfrac{1}{nh^2}\sum_{j=1}^n K'\left(\cfrac{x-X_j}{h}\right) =\cfrac{1}{h^2}K'\left(\cfrac{x-X_1}{h}\right)$$
with $K$ being the standard normal distribution and $K'$ its derivative.
The formula for the bias is this :
$$B\left(\hat{f'}_{h,n}\right)= E\left(\hat{f'}_{h,n}(x)\right) - f'(x)$$
My goal is to prove that this bias is of order $O(h^2)$ and $O(h)$ when $h$ approaches $0$.
Here is my attempt :
$$E\left(\hat{f'}_{h,n}(x)\right) = \int\cfrac{1}{h^2}K'\left(\cfrac{u-x}{h}\right)f'(u)$$
Let $u = x +hv$ so we get : $$ E\left(\hat{f'}_{h,n}(x)\right) = \cfrac{1}{h}\int f'(hv+x) K'(v)dv $$
Now I'm thinking about using taylor's expansion for f when $ h $ approaches $0$:
$$f'(hv+x) = f^{(2)}hv +\cfrac{(hv)^2}{2} f^{3}(x)+\cfrac{(hv)^3}{3!}f^{4}(x)+ o(h^3)$$
but seeing this calculation, I don't know whether it will get me to prove that the order is indeed $O(h^2)$ and $O(h)$.
How should I proceed?
