Convergence of maximum of random variables converging to zero

Question

Suppose that for a sample of identically distributed but non-independent uniformly bounded non-negative random variables we can show that $X_i = O_P(b_n)$ with $b_n$ a deterministic sequence converging to zero. Would it be possible to say something about the maximum of these random variables, namely $Y_n = \max_{i \leq n} X_i$? Intuitively one would expect that $Y_n = O_P(b_n)$ but with such things the devil is always in the details. Thank you.

Answer 1

Still not completely sure about your question setting but here's a shot:

For any $n$, the following are true:

• There are random variables $X_{i,n}$ for $i \in \{1,2,...,n\}$.

• Every $X_{i,n}$ is uniformly bounded and non-negative.

• $\forall i \neq j: X_{i,n}$ and $X_{j,n}$ are identically distributed, but not independent.

• $X_{i,n} = O_P(b_n)$ or in other words, ${X_{i,n} \over b_n}$ is probabilistically bounded for large enough $n$.

Am I correct above? If so, your conjecture is false, with this counterexample: For any $n$, imagine there are $n$ balls and you pick one of them uniformly at random, and $X_{i,n}$ is the indicator for picking the $i$th ball. I.e., $X_{i,n} = 1$ with probabililty $1/n$ and $= 0$ otherwise. These satisfy the first 3 bullets above.

For the 4th bullet, pick $b_n = 1/n$. Then $X_{i,n}/b_n$ is either $n$ (with prob $1/n$) or $0$ (otherwise). Thus $\forall \epsilon>0, \delta>0: \exists N: \forall n > N: Prob(X_{i,n}/b_n > \delta) < \epsilon$.

If these $X_{i,n}$ satisfy your prerequisites, then it is a counterexample because $\forall n: Y_n \equiv 1$ since exactly one of the $X_{i,n} = 1$.

== ADDENDUM ==

In fact, a slightly modified example shows that independence wouldn't help. Consider $X_{i,n}$ binary with $Prob(X_{i,n} = 1) = 1/n$, and all the $X_{i,n}$ are independent. You still have $Y_n = 0 \iff \forall i: X_{i,n} = 0$, and so $Prob(Y_n=0) = (1 - {1\over n})^n \approx {1 \over e}$ and $Prob(Y_n=1) \approx 1 - {1 \over e}$ and does not tend to $0$.