Mean, variance of non-parametric estimator

[rayge]

Homework Statement

For the nonparameteric estimator $\hat{f}(x)=\frac{1}{2hn}\sum\limits_{i=1}^n I_i(x)$ of a pdf,
(a) Obtain its mean and determine the bias of the estimator
(b) Obtain its variance

Homework Equations

The Attempt at a Solution

For (a), I think it goes like this:
$$E[\hat{f}(x)] = E[\frac{1}{2hn}\sum\limits_{i=1}^n I_i(x)]=\frac{1}{2hn}nE[I_1(X)]=\frac{1}{2h}P(x_h<X<x+h)=\frac{1}{2h}\int_{x-h}^{x+h}f(x)dx=f(\xi_n)$$

However my professor stated he wants the answer in terms of integrals of f. He doesn't want the answer in terms of $\xi$, as the function $f(\xi)$ is existential. I'm confused about what he is asking for. (I should probably ask him for clarification rather than strangers, but maybe someone sees something in this that I'm missing.) Maybe it has something to do with taking the limit as h goes to 0; but in this case, I think we get $f(x)$, which is somewhat trivial and not really helpful.

As for obtaining the bias, I think I need to find $E[\hat{f}(x)] - \mu$. Which I also think becomes $f(x) - \mu$, which is also trivial.

I think once I understand the key to this I'll understand how to get part (b), but I'm kind of lost. Thanks for any pointers.

 

[BruceW]

hmm. I don't understand your working. Each $I_i$ is going to be centred at a different place, corresponding to each of your data points, right? So each $I_i$ is a different function, so the expectation is not going to be the same for all of them.