functions over dependent random variables Say we have a set of identically distributed integer-valued random variables: $\{ A_i \}_{i=1}^n$, such that they are not independent. Say we have another set of identically distributed integer-valued random variables $\{ B_i \}_{i=1}^n$, such that they are not independent (different dependence than that of the $A_i$'s).
If we have $A_i \sim B_i$, this is, $\lim A_i / B_i = 1$, what can we say about a function of the random variables such as:
$f(A_1, \dots, A_n) = \max_i A_i$ and $f(B_1, \dots, B_n) = \max_i B_i$
Does it follow that $f(A_1, \dots, A_n) \sim f(B_1, \dots, B_n)$?
Thanks!
 A: The limit, $\lim \frac{A_{i}}{B_{i}} = 1$ is ill defined, but seems to mean that $A_{i}$ and $B_{i}$ are essentially the same till the first degree in the exponent. 
I am assuming the limit $\displaystyle\lim_{n \to\infty} \frac{A_{i}}{B_{i}} = 1$ holds $\forall i = 1, \dots, n$, and that the random variable $A_{i}$ and $B_{i}$ depend on $n$. 
The question breaks down to the following: 
Is $\mathbb{P}\left(f(A_1, \dots, A_n) = f(B_1, \dots, B_n)\right) = 1$? (almost sure convergence)
For the given functions, $f(A_1, \dots, A_n) = \max_i A_i$ and $f(B_1, \dots, B_n) = \max_i B_i$, we have:
Is $\mathbb{P}\left(\max_i A_i = \max_i B_i\right) = 1$? 
Intuitively, yes. Let me try to give a (semi)-rigorous proof.

Proof:
Let us explicitly represent $A_i$'s as a function of $n$, i.e. $A_i(n)$. Also, without loss of generality, let us take these random variables to be positive. 
Let $\max_i A_i(n) = A_j(n)$ for some $j \in \{1, \dots n\}$. Then, by the relationship between $A_j(n)$ and $B_j(n)$ in the form of the above limit, for every $\epsilon > 0$, there exists a large enough $n_0$, such that $n \geq n_0$ and
$$A_j(n)/\left(1 - \epsilon\right) \geq B_j(n) \geq A_j(n)/\left(1 + \epsilon\right)\hspace{20 mm}(1)$$
Hence, if $A_j(n)$ is the maximum, as assumed above, then it stands to reason that $B_j(n)$ is the maximum among $B_i(n)$'s. Let me prove this rigorously. 
Assume that $B_j(n)$ is not the maximum among $B_i(n)$'s. Let us also assume that there is only one $A_j(n)$ satisfying $\max_i A_i(n) = A_j(n)$. Consider $\max_i B_i(n) = B_k(n)$ for some $k \in \{1, \dots n\} \neq j.$ Then, for every $\epsilon > 0$, there exists some $n_1$ such that $n > n_1$ and:
$$ A_k(n)/\left(1 - \epsilon\right) \geq B_k(n) \geq A_k(n)/\left(1 + \epsilon\right) \hspace{20 mm} (2)  $$
This implies that $B_k(n)$ is arbitrarily close to $A_k(n)$ as $ n \to \infty$. Similarly, eq. (1) tells us that $B_j(n)$ is arbitrarily close to $A_j(n)$ as $n \to \infty$. However, $max_i A_i(n) = A_j(n)$ and hence $A_j(n) \geq A_k(n)$. We, however, assumed that there exists only one $j$ that satisfies $max_i A_i(n) = A_j(n)$. Hence, clearly, $B_j(n) > B_k(n)$ for sufficiently large $n  > \max(n_0, n_1)$. Clearly, we have a contradiction and therefore, $B_j(n)$ is indeed the maximum among $B_i(n)$ in the limit.
The case, where $A_j(n) = A_k(n)$ (for sufficiently large $n > \max(n_0, n_1)$) is trivially taken care of, since in this case $B_j(n) = B_k(n)$ and hence, in either case the limit,
$$\lim_{n \to\infty} \frac{\max_i A_{i}}{\max_i B_{i}} = \lim_{n \to\infty} \frac{A_{j}}{B_{j}} = 1 \hspace{20 mm} (3)$$
where $j$ is as defined above.
To finish it off, note that the limits mentioned in eq. (3) are the definition of almost sure convergence, and hence: 
$\displaystyle \mathbb{P}\left(A_i = B_i\right) = 1$, $\forall i$. 
Therefore, 
$\displaystyle \mathbb{P}\left(\max_i A_i = \max_i B_i\right) = 1$, where $j$ is such that $\max_i A_i = A_j$ and $\max_i B_i = B_j$. 
Q.E.D.
By the way, not quite sure if the same reasoning can lead to a similar outcome for any general functions, $f(A_1, \dots, A_n)$ and $f(B_1, \dots, B_n)$.
