# A problem in convergence and limit

For any non-negative integer $$n$$ and some finite $$r$$, we introduce the notation $$n_k$$ which indicates the number of $$\{X_1, X_2, \cdots , X_n\}$$ belonging to the $$k$$-th distribution type, for $$k=1, 2, \cdots, r$$. Here, it is assumed that $$X_i$$'s are independent, and that $$E[X_i|S_n]$$ converges to $$E[X_i]$$, for $$i=1, 2, \cdots, n$$. Then, the following condition holds true.

There are at most $$r$$ different elements in the set of all distributions of the sequence $$X_1, X_2, \cdots,$$. Moreover, we have that

$$$$\lim_{n\rightarrow \infty} \min_{1\leq k\leq r} \frac{n_k}{ln \sigma[S_n]} = \infty,$$$$ and for $$i$$ given, there exists an $$n\geq i$$ such that

$$$$\sup_x \bar{f}_{S_{n/i}}(x) < \infty$$$$ where $$S_n = \sum_{i=1}^{n}X_i$$ and $$S_{n/i} = S_n - X_i$$, for $$i=1, 2, \cdots , n$$, and $$\bar{f}_{S_{n/i}}(x)$$ is the probability density function of the normalized version of $$S_{n/i}$$, i.e., $$\frac{S_{n/i} - E[S_{n/i}]}{\sigma[S_{n/i}]}$$ with expectation and standard deviation $$E[S_{n/i}]$$ and $$\sigma[S_{n/i}]$$, respectively.

My question is: Why $$r$$ should be a finite number, what if r is infinite (the number of different distributions goes to infinity)? Can we provide another condition that satisfies the above condition?

• When $r$ is infinite you can't even get started, because then potentially each $X_i$ has its own distribution.
– whuber
Commented Mar 10, 2022 at 12:55
• So, what happens if we say each $X_i$ has its own distribution. I mean, can we state something similar\equivalent to the above condition in case each $X_i$ has different distributions? Commented Mar 10, 2022 at 13:00
• It's impossible to say, because you haven't asserted anything: you have only listed a bunch of assumptions.
– whuber
Commented Mar 10, 2022 at 13:03
• I mean suppose that we have a sequence of independent, but not essentially identical distribution of $X_1$, $X_2, \cdots$. Then if it is possible to state a condition like above to make sure that $E[X_i| S_n]$ converges to $E[X_i]$? Commented Mar 10, 2022 at 13:09

$$r$$ need not be finite. Assuming your result is correct, it is sufficient that the condition on the $$n_k$$ holds for all $$r$$.
Proof by taking subsequences. Let $$i$$ and $$r<\infty$$ be given. Fix a set of $$r$$ distribution types including the type of $$X_i$$, and take the subsequence consisting only of those types. Write $$\tilde S_n$$ for the subsequence partial sums. Your condition on $$n_k$$ still applies to the subsequence (the numerator is the same and the denominator becomes smaller).
Your result thus says $$E[X_i|\tilde S_n]\to E[X_i]$$. But the full sequence partial sums $$S_n$$ cannot contain any more information about $$X_i$$ than $$\tilde S_n$$ do, because they are a function of $$\tilde S_n$$ and variables independent of the subsequence.
• Thank you. What do you mean by "each type appears infinitely often", you mean all $X_i$ has its own distribution? Commented Mar 11, 2022 at 10:46
• No, I mean there are infinitely many different types and for each type there are infinitely many $X_i$ of that type. Having every $X_i$ be a different type would contradict the assumption about $n_k$. Commented Mar 12, 2022 at 2:10