You might want to explain the assumption that $\mathbb{E}[X_i] \in O(f(n))$ for all $i$.

@Did I don't understand the question. $f : \mathbb{N} \to \mathbb{R}$ and it should be continuous, I guess, if that's what you mean.

@AndrásSalamon The $X_i$ have the same distribution up to its parameters. Since their support is typically finite in my application, we can assume finite moments.

Let $x_i=E(X_i)$. I do not understand the assertion that $x_i\in O(f(n))$ for every $i$. You might want to translate it as a precise mathematical assertion: "there exists $C$ such that..." what?

@Did: What is wrong with the mathematical definition of $O$ (Landau notation)?

The statement $x_n\in O(f(n))$ means that there exists some finite $N$ and $C$ such that, for every $n\geqslant N$, $|x_n|\leqslant C|f(n)|$. Since the statement "$x_i\in O(f(n))$ for every $i$" does not enter this framework, you should explain its meaning. (Note that this is the third and last time I repeat the same comment.)

@Did: You miss that $x_i \in O(f(n))$ is not the same as $x_i \in O(f(i))$. I conciously wrote the former (which is well-defined). Since $n$ is bound in my setup, that should be clear; $x_i$ depends on $n$, not only on $i$. I tried to clarify that. (If you had repeated the same comment, I would have understood your misconception from the beginning.)

You added the crucial hypothesis that $E(X_i)$ depends on $i$ AND ON $n$ 52 minutes ago. Nothing before this moment indicated this was the case (and in fact everything pointed to some $E(X_i)$ INDEPENDENT of $n$). On the positive side, the question is clearer now (on the negative side, the steps needed to reach this result were unpleasant). However, the meaning of the new "we ignore $i$ asymptotically" is still mysterious--but frankly this does not concern me anymore.

To conclude, I wish to mention that your behaviour is becoming more and more unpleasant and that your recent "damage-control" comment to András Salamon's answer is quite unacceptable.

I'd like to understand your problem, both the mathematical and the personal one.

In my eyes (which may be clouded by the somewhat sloppy use of Landau notation in CS), $x_i \in O(f(n))$ directly implies that $x_i$ depends on $n$ (mathematically, it says "might depend on n" which makes the assumption "does not depend on n" unlikely).

Therefore, I honestly did not get your first two comments.

Why you choose to attack me personally, I don't understand.

I still disagree with "Nothing before this moment indicated this was the case", but I see how the notation could have been clearer. I adapted in a hopefully clarifying way.

Even though you say don't care and seem to have heated up quite a bit, thanks for pointing the notational shortcomings out to me.

@AlexanderGruber Hi. Someone called the cavalry? ;)

okay, i've cleaned this discussion from the comments in the question, now it may be continued in chat.

@AlexanderGruber Thanks.

yup, have fun kids

I think it was a simple misunderstanding. shrug