Proving that the bias of the derivative of Parzen-Rosenblatt (kernel density) estimator is of order $O(h^2) $ and $O(h)$ when $h$ approaches $0$

Question

I came across this property that I don't get and I couldn't find the proof anywhere:

Suppose we have a density $K$ of the standard normal distribution and $K'$ its derivative. Suppose that the density $f$ is of class $C^4$ in $\Bbb{R}$ .

Let $\hat{f'}_{h,n} = \cfrac{1}{nh^2}\sum_{j=1}^n K'(\cfrac{x-X_j}{h})$ be the estimator of the density $f'$ with $h > 0 $.

For a small $h$, as in if $h$ approaches $0$ the bias verifies :

$B(\hat{f'}_{h,n}) = O(h)$ and $B(\hat{f'}_{h,n}) = O(h^2)$.

Can someone explain to me how did we get to that result?

EDIT: I found this theorem:

But here we are discussing the density $f$ and not $f'$. How can we extend the proposition?

EDIT: Here is my attempt :

$$E(\hat{f'}_{h,n}(x)) = \int\cfrac{1}{h^2}K'(\cfrac{u-x}{h})f'(u)$$

Let $u = x +hv$ so we get : $$ E(\hat{f'}_{h,n}(x)) = \cfrac{1}{h}\int f'(hv+x) K'(v)dv $$

Now I'm thinking about using taylor's expansion for f because 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?

Answer 1

To answer your question, I'm following this question: Expectation of derivative of kernel density estimator.

With some abuse in notation, we can write, for any fixed $x,$ \begin{align} h E(\hat{f}'(x)) &= \frac{1}{h} \mathbb E\left[K'\left(\frac{X-x}{h} \right) \right]\\ &=\int f(t) \cdot \frac{1}{h} K'\left(\frac{t-x}{h} \right) dt \\ &=\Bigg[f(t) \cdot K\left(\frac{t-x}{h}\right)\Bigg]_{-\infty}^{\infty} - \int K\left(\frac{t-x}{h} \right) \cdot f'(t)dt \\ &= -\int K\left(\frac{t-x}{h} \right) \cdot f'(t)dt \end{align} using integration by parts, and both $f$ and $K$ are valid pdf.

Now, take $u=(t-x)/h$ and $du=dt/h,$ gives \begin{align} E(\hat{f}'(x)) - f'(x) &= \frac{-1}{h}\int K\left(\frac{t-x}{h} \right) \cdot f'(t)dt - f'(x)\\ &= -\int K(u) \cdot \left[f'(x+uh)-f'(x)\right] du \\ &= -\int K(u) \cdot \left[f''(x)(uh)+f'''(x)(uh)^2/2 + o(h^2)\right] du \end{align} which you can check is $O(h)$.

I think derivative in both $K$ and $f$ is not possible in your approach.