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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?

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Amarus
    Feb 25 '16 at 18:00
  • $\begingroup$ @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. $\endgroup$
    – saz
    Feb 25 '16 at 18:34

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