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I’m studying a recursively defined estimator and want to prove consistency. The estimator is not merely computed by a recursive algorithm—the value on a sample is defined (also) through values on sub-samples.

Let's denote the estimator by $T_n$ and the parameter of interest by $\gamma$. So far, I've tried:

  • Induction on $n$ for $\mathbb{P}(|T_n - \gamma| > \varepsilon)$, but it doesn't close without additional structure
  • Covering $\{ |T_n - \gamma| > \varepsilon \} \subseteq E_n(\varepsilon)$ and bounding $\mathbb{P}(E_n(\varepsilon))$, but a tight enough cover is hard to quantify
  • Solving a recursion relation of tail bounds, which only yields decay for $\varepsilon$ bounded away from $0$

I've also tried to use asymptotic properties of some of the quantities involved in the definition of $T_n$, but it seems especially difficult to use those with recursion.

Are there any other common proof techniques? Pointers to standard references are very welcome.

Edit: I can't share specifics, but think of $n$ i.i.d observations, some "base" estimator $T'_{i:j}$ computed on observations $i$ through $j$ and an aggregation function $A$, then setting $T_{i:j} := A(\{ T'_{i:j} \} \cup \bigcup_{i \leq k < j} \{ T_{i:k}, T_{k+1:j}\})$ and $T_n := T_{1:n}$.

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    $\begingroup$ It is hard to get an intuition of what you would need without more details on your problem. Can you tell more about the specifics? $\endgroup$ Commented Sep 5 at 12:47
  • $\begingroup$ I've expanded the question with a high-level description of the estimator. $\endgroup$ Commented Sep 5 at 18:45
  • $\begingroup$ Thank you for expanding the description, but it is still too vague. We have no idea what the properties of the base estimator is (and why it is not sufficient for your purposes), or the aggregation function. Without details we will make assumptions which are not relevant to your problem, and a back-and-forth will waste time for everyone involved. I'd be happy to look at a more detailed description; however if you cannot share specifics on SE then might be best to turn to a colleague. $\endgroup$ Commented Sep 11 at 14:32

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