We have $(X_i)_{i \in \mathbb Z}$ iid random variables with $1\le X_i \le2$ almost surely.

We define $X(x,\omega) \equiv X_i (\omega)$ if $x\in [i,i+1[$ and $X_\epsilon (x, \omega) \equiv X(x/\epsilon, \omega)$.

A post that solves a very similar problem: Show that those random quantities converge in distribution to a normal variable (hard analysis problem)


$$u'_\epsilon(x,\omega)= \frac {c_\epsilon(\omega) - F(x)}{X_\epsilon(x,\omega)}$$

where $F$ is an $L^1([0,1])$ function and $c(\omega)$ is defined by

$$c_\epsilon(\omega)\equiv \frac{\int_0^1 \frac{F(y)}{X_\epsilon(y,\omega)} \, dy}{\int_0^1\frac 1 {X_\epsilon(y,\omega)}\, dy}$$

Show that $$\epsilon^\alpha\int_0^1 u'_\epsilon(x,\omega) g(x)\, dx \to \mathcal N(?, ?)$$

for a certain $\alpha\in\mathbb R$, in distribution when $\epsilon \downarrow 0$ for any sufficiently smooth function $g$, where we need to characterize the expectation and the variance.

A hint says for this problem that we should notice that $u'_\epsilon(x,\omega)$ can be written as a product of a random part and a deterministic part + an error that can be controlled in $L^2(]0,1[ \times \Omega)$. We can easily notice that

$$u'_\epsilon(x,\omega)= \frac {F(x)}{X_\epsilon(x,\omega)} + err_\epsilon(x,\omega)$$


$$err_\epsilon(x,\omega)=\frac {\int_0^1 \frac {F(y)}{X_\epsilon(y,\omega)} dy}{X_\epsilon(x,\omega)\int_0^1 \frac {1}{X_\epsilon(y,\omega)} dy}$$

I have no idea how to control this and how will this help (maybe this is not the intended form). The question related to this control can be found here : Controlling this function in $L^2$ norm

Another post that brings more information about $u'_\epsilon$ is the following: Showing an $L^2$ convergence (with convergence rate)

EDIT: we can suppose that $F$ is continuous on $[0,1]$ if it helps.

EDIT2: Bonus rep will be awarded if, in addition to the above, the convergence is shown using quantitative argument (total variation distance, Wasserstein distance, Kolmogorov distance, Zolotarev distance...)


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