Does the probability of obtaining a sample for which a new element will be larger than the sample approaches 0 as the sample size increases?

Question

Choose arbitrary $\epsilon > 0$ and arbitrary probability distribution $D$ over $[0, 1]$.

For a given natural number $m$, sample $S_m = (x_1, x_2, ..., x_m)$ as indepentent identically distributed random variables from $D$. Let $\overline{S_m} = max ({x_1, ... x_m})$.

Now consider the probability that a newly sampled random variable $x$ will be larger than $\overline{S_m}$, that is consider $P({S_m})$ = $P_{x \sim D}${x > $\overline{S_m}$}

We are interested in the probability $Q(m)$ of obtaining a sample $S_m$, for which $P(S_m) > \epsilon$, that is:

$Q(m)$ = $P_{S_m \sim D^m}${$P(S_m) > \epsilon$}.

a) Is it true that $Q(m)\rightarrow 0$ as $m$ tends to infinity, regardless of the initial choice of $\epsilon$ and $D$?

b) If a) is true, can one state an explicit bound function (parameterised by $\epsilon$ but independent of $D$) $F_\epsilon : N \rightarrow [0,1]$ such that $Q(m) < F_\epsilon(m)$ and $F_\epsilon(m) \rightarrow 0$ for large m?

Answer 1

Let $F$ be the CDF for the distribution $D$, and $y = \inf \{v: F(v) \ge 1-\epsilon\}$. Thus $F(y) \ge 1-\epsilon$ but $F(t) < 1-\epsilon$ for $t < y$.
If $\max(S_m) \ge y$, $P(S_m) = 1-F(y)\le \epsilon$ but if $\max(S_m) < y$, $P(S_m) > \epsilon$. Thus $Q(m) = P(\max(S_m) < y) = F(y-)^m$ (i.e. $\lim_{v \to y-} F(y)^m$). In particular $Q(m) \le (1-\epsilon)^m \to 0$ as $m \to \infty$.