Tightness Condition

Helly’s selection theorem produces a subsequential limit $F$ that is right-continuous and non-decreasing, but the total mass of $F$ may be strictly less than $1$ (mass can escape to $\pm\infty$). Tightness is the condition that rules this out and promotes Helly’s conclusion to weak convergence to a probability measure.

Definition

Definition: Tightness

A sequence of probability measures $\{\mu_n\}_{n \ge 1}$ on $(\mathbb{R}, \mathcal{B})$ is called tight if for every $\epsilon > 0$, there exists $M > 0$ such that

$$\mu_n((-M, M]) \ge 1 - \epsilon \quad \text{for all } n \ge 1.$$

A sequence of distribution functions is tight if the corresponding sequence of measures is tight.

Written in terms of distribution functions, this is equivalent to

$$1 - F_n(M) + F_n(-M) \le \epsilon \quad \text{for all } n.$$

Some equivalent formulations of the same condition:

1. $\liminf_{n \to \infty} \mu_n((-M, M]) \ge 1 - \epsilon$ for some $M$,
2. $\limsup_{n \to \infty} \big(1 - F_n(M) + F_n(-M)\big) \le \epsilon$ for some $M$,
3. $\mu_n((-M, M]) \ge 1 - \epsilon$ for all $n$ (after enlarging $M$).

The equivalence between holding for all $n$ and holding eventually is immediate: any finite collection of measures can be absorbed by taking $M$ large enough, since each individual $\mu_n$ is a probability measure and so $\mu_n((-M, M]) \to 1$ as $M \to \infty$.

Tightness ⟺ subsequential weak convergence

Theorem

Statement

A sequence $\{\mu_n\}_{n \ge 1}$ of probability measures on $(\mathbb{R}, \mathcal{B})$ is tight if and only if every subsequence has a further subsequence that converges weakly to a probability measure.

Proof

(⟹) Tightness implies subsequential weak convergence.

Fix any subsequence $\{F_{n_k}\}$. By Helly’s selection theorem, there is a further subsequence (which we relabel $\{F_{n_m}\}$) and a right-continuous, non-decreasing function $F$ with $F_{n_m}(y) \to F(y)$ at every continuity point $y$ of $F$.

It remains to show $\lim_{y \to -\infty} F(y) = 0$ and $\lim_{y \to \infty} F(y) = 1$, since then $F$ is a distribution function and $\{F_{n_m}\}$ converges to it weakly. Both limits exist because $F$ is monotone and bounded in $[0,1]$, and they satisfy $\lim_{y \to -\infty} F(y) \ge 0$ and $\lim_{y \to \infty} F(y) \le 1$. So it suffices to show

$$\lim_{y \to \infty} F(y) - \lim_{y \to -\infty} F(-y) = 1.$$

1. Pick $M_\epsilon$ from tightness.

   For $\epsilon > 0$, choose $M_\epsilon$ such that

   $$\limsup_{n \to \infty} \big(1 - F_n(M_\epsilon) + F_n(-M_\epsilon)\big) \le \epsilon. \quad\textcolor{gray}{\text{---(1.1)}}$$

2. Compare $F$ to $F_{n_m}$ at $M_\epsilon$ via continuity points.

   Pick continuity points $r < -M_\epsilon$ and $s > M_\epsilon$ of $F$. Such points exist since $F$ has at most countably many discontinuities. By monotonicity of each $F_{n_m}$,

   $$F_{n_m}(s) \ge F_{n_m}(M_\epsilon), \quad F_{n_m}(r) \le F_{n_m}(-M_\epsilon),$$

   so

   $$1 - F_{n_m}(s) + F_{n_m}(r) \le 1 - F_{n_m}(M_\epsilon) + F_{n_m}(-M_\epsilon).$$

3. Pass to the limit.

   Since $r, s$ are continuity points of $F$, $F_{n_m}(r) \to F(r)$ and $F_{n_m}(s) \to F(s)$. Taking $\limsup$ on the right and using $(1.1)$,

   $$1 - F(s) + F(r) \le \epsilon.$$

   Equivalently, $F(s) - F(r) \ge 1 - \epsilon$.

4. Send $r \to -\infty$ and $s \to \infty$.

   Choosing continuity points $r_k \to -\infty$ and $s_k \to \infty$,

   $$\lim_{s \to \infty} F(s) - \lim_{r \to -\infty} F(r) \ge 1 - \epsilon.$$

   Since $\epsilon > 0$ was arbitrary, the difference is at least $1$. Combined with $\lim_{y \to \infty} F(y) \le 1$ and $\lim_{y \to -\infty} F(y) \ge 0$, both limits must take their extreme values: $\lim_{y \to \infty} F(y) = 1$ and $\lim_{y \to -\infty} F(y) = 0$.

Thus $F$ is a distribution function, and the corresponding measure $\mu$ is a probability measure with $\mu_{n_m} \Rightarrow \mu$.

(⟸) No tightness implies no convergent sub-subsequence.

We prove the contrapositive: if $\{\mu_n\}$ is not tight, then some subsequence has no further subsequence converging weakly to a probability measure.

Non-tightness gives an $\epsilon > 0$ such that for every $M$, some $n$ satisfies $1 - F_n(M) + F_n(-M) \ge \epsilon$. Building this index by index, there exist sequences $M_k \to \infty$ and indices $n_k$ with

$$1 - F_{n_k}(M_k) + F_{n_k}(-M_k) \ge \epsilon \quad \text{for all } k. \quad\textcolor{gray}{\text{---(2.1)}}$$

By Helly’s selection theorem, $\{F_{n_k}\}$ has a further subsequence (still denoted $\{F_{n_k}\}$) with $F_{n_k}(y) \to F(y)$ at continuity points of a right-continuous, non-decreasing $F$. We show $F$ is not a distribution function, hence does not correspond to a probability measure.

1. Bound $F$ at any pair of continuity points $r < 0 < s$.

   Since $M_k \to \infty$, for $k$ large enough, $M_k > s$ and $-M_k < r$. By monotonicity of $F_{n_k}$,

   $$F_{n_k}(s) \le F_{n_k}(M_k), \quad F_{n_k}(r) \ge F_{n_k}(-M_k),$$

   so

   $$1 - F_{n_k}(s) + F_{n_k}(r) \ge 1 - F_{n_k}(M_k) + F_{n_k}(-M_k) \ge \epsilon$$

   using $(2.1)$.

2. Pass to the limit and send $r \to -\infty, s \to \infty$.

   Taking $k \to \infty$ at continuity points $r, s$,

   $$1 - F(s) + F(r) \ge \epsilon.$$

   Letting $s \to \infty$ and $r \to -\infty$ along continuity points,

   $$\lim_{s \to \infty} F(s) - \lim_{r \to -\infty} F(r) \le 1 - \epsilon < 1.$$

   So $F$ does not satisfy property 2 of a distribution function: either $\lim_{y \to \infty} F(y) < 1$ or $\lim_{y \to -\infty} F(y) > 0$ (or both).

Therefore the subsequence $\{F_{n_k}\}$ has no further subsequence converging weakly to a probability measure: any such sub-sub-limit must agree with $F$ at continuity points, but $F$ is not a distribution function.

Why both directions matter

The forward direction is the workhorse: in practice, one shows a sequence of laws is tight (often via a moment bound such as $\sup_n \mathbb{E}[|X_n|] < \infty$, which controls the tails through Markov’s inequality), then invokes the theorem to extract a weakly convergent subsequence. If a uniqueness argument pins down all possible subsequential limits to the same $\mu$, the full sequence converges weakly to $\mu$.

The reverse direction says tightness is not just sufficient but necessary: any sequence enjoying this subsequential compactness must control its tails uniformly.