*Preface: I'm a statistician hoping to derive asymptotic normality for an estimator I'm proposing. TLDR: I'm trying to use the martingale CLT but I'm having trouble finding the right martingale for my situation.*

Suppose I have two iid data streams: $X = (X_1, X_2, \ldots)$ and $Y = (Y_1, Y_2, \ldots)$ independent of each other with $E X_1^2 < \infty$ and $E Y_1^2 < \infty$. I'm trying to estimate $\theta = E X_1 - E Y_1$ by sampling $X$ and $Y$ into sample means but with a wrinkle:

The entire timeline is divided up into slices, so that at the time when there are $n$ total points there are $k(n)$ slices, where in slice $i$ we have a difference-in-sample-means estimate $\hat{\theta}_i = \bar{X}_i - \bar{Y}_i$ calculated only from data within that slice. Also suppose that at time $n$, the number of points within slice $i$ (denoted $n_i$), is non-zero, known and not random for each $1 \leq i \leq k(n)$.

Assume there is dependence between the within-slice estimates $\{\hat{\theta}_i: 1 \leq i \leq n\}$ in the following way:

 - For each $i$, the probability $p_i$ that each of the $n_i$ total within-slice visitors will be sampled from stream $X$ (instead of $Y$) is a function of $\hat{\theta}_{i-1}$. Assume that $p_i \in (0,1)$ so that the $\hat{\theta}_i$'s are orthogonal.

I want to show that the following weighted quantity is asymptotically normal:
$$ T_{n,j} = \sum_{i=1}^{j} \frac{n_i}{n} (\bar{X}_i - \bar{Y}_i - \theta) $$
To ease notation let $E_{n,i} := \frac{n_i}{n} (\bar{X}_i - \bar{Y}_i - \theta)$ so that we can write simply $T_{n,j} = \sum_{i=1}^j E_{n,i}$ with each $E_{n,i}$ having zero mean. Because of orthogonality the variance of $T_{n,i}$ is just
$$ \textrm{Var}(T_{n,j}) = \sum_{i=1}^j \textrm{Var}(E_{n,i}) $$

Assuming that $k(n) \to \infty$ with $n$, I want to show a CLT for $T_{n,k(n)}$ as $n \to \infty$. I immediately reached for the martingale CLT, but I'm stuck on constructing a martingale array out of $T_{n,j}$ and $\textrm{Var}(T_{n,j})$.

Letting $\mathcal{F}_n$ be the sigma-algebra generated by all the data observed up to the $n$th point and $\mathcal{F}_{n,i}$ be the sigma-algebra generated by only the data up to and including all of slice $i$, one issue is that $\textrm{Var}(T_{n,i})$ is only $\mathcal{F}_{n,i}$-measurable so I can't build a martingale array in the typical way like:
$$ M_{n,i} = T_{n,i} \cdot \left( \textrm{Var}(T_{n,k(n)}) \right)^{1/2} $$
Since this would not be adapted to the filtration $\{\mathcal{F}_{n,i}: 1 \leq i \leq k(n)\}$. Instead, I tried to shoehorn something like
$$ \tilde{M}_{n,i} = T_{n,i} \cdot \left( \textrm{Var}(T_{n,i)}) \right)^{1/2} $$
but this isn't a martingale as written because
$$ E(\tilde{M}_{n,i} | \mathcal{F}_{n,i-1}) = T_{n,i-1} \cdot \left( \textrm{Var}(T_{n,i)}) \right)^{1/2} = T_{n,i-1} \cdot \left( \sum_{j=1}^i \textrm{Var}(E_{n,j}) \right)^{1/2} $$
Which is off by just a small yet annoying factor. Is there some easy way around this situation to get the martingale I need?

**EDIT**: I am making no assumptions on the manner of dependence of $p_i$ on $\hat{\theta}_{i-1}$ aside that $p_i$ is bounded away from 0 and 1. If it turns out that this is intractable then I'd love to investigate what conditions must be imposed to make this work!