# Does tightness imply exponential tightness?

I just learned about the notion of exponential tightness which is described as a generalisation of tightness. The first question that pops into my head is: So does tightness imply exponential tightness?

I haven't quickly found any answer for that, so I guess the answer must be obvious. Still, I'm heaving trouble verifying if it actually is true.

Let $X$ be a topological space equipped with its Borel-$\sigma$-algebra $\Sigma$ (the one generated by all open subsets). A series $(\mu_n)_n$ of probability measures on $(X,\Sigma)$ is called tight if for each $\epsilon>0$ there exists a compact subset $K_\epsilon\subset X$ (which is of course in $\Sigma$) such that $\mu_n(K_\epsilon^c)<\epsilon$ is true for all $n$ ($K_\epsilon^c$ means $X\setminus K_\epsilon$). $(\mu_n)_n$ is called exponentially tight on a scale $(\gamma_n)_n$ if for each $s>0$ there exists a compact subset $K_s\subset X$ such that $$\limsup_{n\rightarrow\infty} \frac{1}{\gamma_n} \log \mu_n(K_s^c) < -s.\tag{1}$$ I should add that a scale is a series $(\gamma_n)_n$, $\gamma_n\in(0,\infty)$, such that $\gamma_n\rightarrow \infty$ if $n\rightarrow \infty$.

Now I can put my question a bit more accurately: If $(\mu_n)_n$ is tight, is it also exponentially tight on every scale? Or can we at least impose some condition on the decay of the scale in order to conclude exponential tightness?

If I take $(\mu_n)$ to be tight and look at (1), I can see that we can choose $K_s$ such that $\mu_n(K_s^c)$ is bounded by an arbitrarily small $\epsilon>0$, so $\log \mu_n(K_s^c)$ gets (negatively) large and competes with the decay of $\frac{1}{\gamma_n}$ which by assumption converges to $0$.

Here I'm a little bit confused about how to proceed and conclude anything meaningful about the relation of tightness and exponential tightness. Can someone help me clear that up?

No, tightness does, in general, not imply exponential tightness. Exponential tightness means that

$$\mu_n(K_s^c) \leq e^{-s \gamma_n} \qquad \text{for n \gg 1};$$

this is a much stronger assumption than tightness.

Example: Let $\mu$ be an arbitrary finite measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ and consider the sequence $\mu_n := \mu$, $n \in \mathbb{N}$. Then the sequence $(\mu_n)_{n \in \mathbb{N}}$ is tight (this follows e.g. from the continuity of the measure from below). On the other hand,

$$\lim_{n \to \infty} \frac{1}{\gamma_n} \underbrace{\mu_n(K^c)}_{\mu(K^c)} = 0$$

for any scale $(\gamma_n)_{n \in \mathbb{N}}$. Consequently, $(\mu_n)_{n \in \mathbb{N}}$ is not exponentially tight.

• Thank you so much! I hope it's appropriate to ask another question by comment (otherwise I might open a new question). I'm thinking about "easy ways" to tell that a sequence or, more generally, a set $M$ of probability measures is exponentially tight. For tightness there is the famous Prokhorov theorem that gives the tightness of $M$ if its closure is sequentially compact with respect to weak convergence, provided that the measures are defined on a polish space. Is there an "exponential version" of this statement? At least $M$ should be exponentially tight if $M$ itself is compact, I think. Feb 25 '16 at 18:00
• @Amarus Yes, there is an analogue of Prokhorov's theorem in this setting; it was proved by Pukhalskii in 1991 ("On functional principle of large deviations"). It states the following: If $X$ is a complete separable metric space, then $(\mu_n)_{n \in \mathbb{N}}$ is exponentially tight if, and only if, any subsequence of $(\mu_n)_n$ has a further subsequence which satisfies a large deviation principle.
– saz
Feb 25 '16 at 18:34