# What happens in the martingale CLT if I norm by the conditional variance instead?

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.

• How does $p_i$ depend on $\hat{\theta}_{i-1}$? – Marcus M Oct 30 '18 at 17:36
• @MarcusM Can we start by making no assumptions on this dependence and see if we can get something out of that? – gogurt Oct 30 '18 at 18:08
• In my opinion, we should put more work on the statement of your problem. At this point, it is still unclear. – Taro NGUYEN Nov 3 '18 at 17:07
• 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. – Taro NGUYEN Nov 3 '18 at 17:13
• ... Let $(n_i)_{i \ge 1}$ be an array of positive real numbers. Define $S_k = \sum_{i=1}^k n_i$ ; $\hat{\theta}_i= \frac{1}{n_i} \left( \sum_{m= S_{i-1}+1}^{S_i} X_m-Y_m\right)$ " ? – Taro NGUYEN Nov 3 '18 at 17:15