# What is the density of $X$ under fixed design?

We observe an i.i.d. sample $(X_1, Y_1), \ldots (X_n, Y_n).$ Let $m(x) = E(Y|X=x),$ $\sigma^2(x) = \operatorname{Var}(Y|X=x)$ and let $f(\cdot)$ be the density of $X.$

Under some regularity conditions, the conditional variance of the local linear estimator $\hat m(\cdot)$ of $m(\cdot)$ at $x_0$ is $$\operatorname{Var}\{\hat m(x_0)|X_1,\ldots,X_n\} = \int\!\! K^2(u) du \, \frac{\sigma^2(x_0)}{f(x_0)} \, \frac{1}{nh} + o_P\left(\frac{1}{nh}\right),\tag{1}$$ as $h\to0$ and $nh \to \infty,$ where $K$ is the employed kernel.

Note that this expression involves the density $f$ of $X.$ Yet the expression is supposed to remain valid under fixed design.*

My understanding is that under fixed design, $X$ is not random but deterministic. But if $X$ is not random, what is its density? How can $(1)$ remain true under fixed design?

*My source for this is page 68 of Fan, J. and Gijbels, I. (1996). Local Polynomial Modelling and its Applications

## 1 Answer

If $$x_0\in [0,1]$$ and $$X_i=i/n$$, it is equivalent to $$X_i\sim i.i.d. Uniform[0,1]$$. So I believe that, in this case, $$f(x_0)=1$$.