What is the second equality in the bias of histogram estimate derived from?

[logarithmic]

Suppose we wish to estimate a probability density given the points {x_1, ..., x_n} using a histogram $$\hat{f}(x)$$.

I have a book that says $$Bias(\hat{f}(x))=E_f(\hat{f}(x))-f(x)=\frac{1}{2}f'(x)(h-2(x-b_j))+O(h^2)$$ for $$x\in(b_j,b_{j+1}]$$.

Can someone explain where the second equality comes from? I pretty sure it's a Taylor expansion, but I'm not sure how to Taylor expand the expected value.

The notation is as follows:

$$h$$ is the width of the histogram bins.
$$b_j$$ and $$b_{j+1}$$ are the boundaries of the j-th bin.
$$\hat{f}(x)=n_j/(nh)$$ for $$x\in(b_j,b_{j+1}]$$, where $$n_j$$ is the number of x points in the j-th bin. and n is the number of x points in total.

Any help is appreciated.

 

[lippyka]

The second equality comes from a Taylor expansion of the expected value of \hat{f}(x). The Taylor expansion is used to approximate the expected value of \hat{f}(x) for values of x near b_j.We begin by writing E_f(\hat{f}(x)) in terms of the expected value of f(x). Note that E_f(\hat{f}(x)) = E_f(n_j/(nh)) for x\in(b_j,b_{j+1}], where n_j is the number of x points in the j-th bin and n is the total number of x points.We then write f(x) as a Taylor expansion around b_j. Specifically,f(x) = f(b_j) + f'(b_j)(x-b_j) + O((x-b_j)^2).Substituting this into the equation for E_f(\hat{f}(x)), we getE_f(\hat{f}(x)) = f(b_j) + f'(b_j)(x-b_j) + O((x-b_j)^2).Now, we can write the bias as the difference between the expected value of \hat{f}(x) and f(x). Thus,Bias(\hat{f}(x))=E_f(\hat{f}(x))-f(x) = f'(b_j)(x-b_j) + O((x-b_j)^2).Finally, we can use the fact that h = (b_{j+1} - b_j) to rewrite the expression asBias(\hat{f}(x))=E_f(\hat{f}(x))-f(x)=\frac{1}{2}f'(x)(h-2(x-b_j))+O(h^2) for x\in(b_j,b_{j+1}].