Suppose I have a sequence of $n$ i.i.d. random variables $X_1,X_2,\ldots,X_n$. Each $X_i$ is positive and has variance $\sigma(n)$ that is an increasing function of the number of variables in the sequence, i.e. $\sigma(n)=\omega(1)$. The mean $\mu$ of each $X_i$ is a constant. I am trying to lower-bound the probability that the maximum of this sequence $X_\max$ exceeds a threshold $T(n)$ (which may either be a constant or a function of $n$ that grows much slower that $\sigma(n)$, i.e. $T(n)=o(\sigma(n))$).

Since $X_i$'s are i.i.d., by elementary probability we have:

$$P(X_\max> T(n))=1-(P(X_1\leq T(n)))^n=1-(1-P(X_1>T(n)))^n$$

Thus, I want to lower-bound the probability that a positive random variable $X_1$ with variance $\sigma(n)$ exceeds a threshold $T(n)=o(\sigma(n))$. Unfortunately, the only variance-based bound that I know is Chebyshev's and it's an upper bound. I also know that most extreme value theory results rely on the structure of the particular distribution function...

However, intuitively, it seems that, since $X_i$'s are positive and since their variance is growing faster than the threshold, $X_\max$ should exceed $T(n)$ with high probability... But I am having hard time proving this... can anyone help?

8more comments