TLDR: I'm a statistician (bear with me!) trying to use the martingale CLT but I only can estimate the conditional variance instead of the unconditional one. Can I do anything to get a CLT with norming by the conditional variance?

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.

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. Fortunately because of orthogonality of the $E_{n,i}'s$ the variance of $T_{n,i}$ is just $$ \textrm{Var}(T_{n,j}) = \sum_{i=1}^j \textrm{Var}(E_{n,i}) $$ and from here it is straightforward enough to show that $T_{n,k(n)} / \textrm{Var}(T_{n,k(n)})^{1/2}$ has a limiting normal distribution.

My question: Unfortunately, in practice it's way easier to compute the conditional variances $E(E_{n,i}^2 | \mathcal{F}_{n,i-1})$ than the unconditional ones $\textrm{Var}(E_{n,i})$. Is it still possible to obtain a CLT using a norming by this conditional variance instead? For example:

$$ \frac{T_{n,k(n)}}{\sum_{i=1}^{k(n)} E(E_{n,i}^2 | \mathcal{F}_{n,i-1})^{1/2}} \to N(0,1) \textrm{ in distribution as } n \to \infty$$

Is this workable at all (maybe with mild assumptions) or is it totally missing an important point about how the CLT works?

UPDATE 1: 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!

UPDATE 2: I realized that my initial problem about constructing a martingale really boiled down to the question of the conditional variance vs. the unconditional one. Restructured the question to make it more concrete.

  • $\begingroup$ How does $p_i$ depend on $\hat{\theta}_{i-1}$? $\endgroup$ – Marcus M Oct 30 '18 at 17:36
  • $\begingroup$ @MarcusM Can we start by making no assumptions on this dependence and see if we can get something out of that? $\endgroup$ – gogurt Oct 30 '18 at 18:08
  • $\begingroup$ In my opinion, we should put more work on the statement of your problem. At this point, it is still unclear. $\endgroup$ – Taro NGUYEN Nov 3 '18 at 17:07
  • $\begingroup$ In first 2 papagraphs, do you want to say ? " Given 2 arrays of random variables $(X_n)$ and $(Y_n)$ such that i) $(X_n)$ are idd, (Y_n) are idd ii) $(X_n) \perp \!\!\! \perp (Y_n)$ iii) $X_1$ and $Y_1$ are square integrable. $\endgroup$ – Taro NGUYEN Nov 3 '18 at 17:13
  • $\begingroup$ ... Let $(n_i)_{i \ge 1} $ be an array of positive real numbers. Define $S_k = \sum_{i=1}^k n_i$ ; $\hat{\theta}_i= \frac{1}{n_i} \left( \sum_{m= S_{i-1}+1}^{S_i} X_m-Y_m\right)$ " ? $\endgroup$ – Taro NGUYEN Nov 3 '18 at 17:15

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