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Let $(X_i)_{i\in\mathbb{N}}$ be an iid sequence of random variables and $S_n:=\sum_{i=1}^n X_i$. Moreover, let $M_{X_1}$ denote the moment-generating function and $\Lambda:=\log M_{X_1}$. Define $$ I(x):=\sup_{\theta\geqslant 0}\left\{\theta x-\Lambda(\theta)\right\}. $$ Last, but not least $$ D:=\left\{x: I(x)<\infty\right\}, E:=\left\{\theta: \Lambda(\theta)<\infty\right\}. $$ Cramér's Theorem says that for each $x>\mathbb{E}(X_1)=:z$ with $x\in\text{int}(D)$, we have $$ \lim_{n\to\infty}\frac{1}{n}\ln P(S_n\geqslant nx)=-I(x). $$

Intuitively, I would think that this implies that $$ P(S_n\geqslant nx)\sim e^{-nI(x)}. $$ Indeed, I have read this in some books and papers.

Others say that this is false and say that $$ P(S_n\geqslant nx)\sim e^{-nI(x)+o(n)} $$ and again others write that $$ P(S_n\geqslant nx)=\Phi(n)e^{-nI(x)}\text{ with }\log\Phi(n)\in o(n). $$

I am a bit confused. Which version is correct?

Rhjg
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    You mean $$\lim_{n \to \infty} \dfrac{1}{n} \ln P(S_n \ge n x) = -I(x)$$ – Robert Israel Aug 26 '15 at 16:53
  • Yes, thanks. I changed it. – Rhjg Aug 26 '15 at 17:09
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    The asymptotic that you want is only "morally" correct, while both of the alternative formulations toward the bottom of your question are truly correct. Similarly, under some circumstances the limit involving the logarithm only "morally" exists; in these cases you have to fall back on the general large deviation principle for iid variables, as we discussed at some length in your other recent question. – Ian Aug 26 '15 at 17:50
  • For $x>E(X_1)$ and $x\in int(D)$, I have that $P(S_n\geqslant nx)=-I(x)$ by Cramér. We discussed that. Then, when I got it right, I have, $P(S_n\geqslant nx)\sim e^{-nI(x)+o(n)}$. Choosing $1<d<e^{I(x)}$, I get the estimation $P(S_n\geqslant nx)\leqslant d^{-n}e^{o(n)}$ for large $n$, right? Moreover, $d^{-n}e^{o(n)}\to 0$, ya? – Rhjg Aug 26 '15 at 18:28

1 Answers1

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$$\lim_{n \to \infty} \dfrac{1}{n} \ln a_n = c$$ says that for any $\epsilon > 0$, we eventually have $$c - \epsilon < \dfrac{1}{n} \ln a_n < c + \epsilon$$ and thus $$ e^{n c - n \epsilon} < a_n < e^{nc + n \epsilon} $$ It is equivalent to $$ a_n = e^{nc + o(n)}$$ and if you define $\Phi(n) = a_n e^{-nc}$ you have $\Phi(n) = e^{o(n)}$, i.e. $\ln \Phi(n) = o(n)$. So your two "others" are equivalent and correct.

But it's not necessarily true that $a_n \sim e^{nc}$, which would imply that the $o(n)$ is $O(1)$.

For example, suppose $X_n \sim \mathcal N(0,1)$, so $S_n \sim \mathcal N(0, n)$. Then

$$P(S_n \ge n x) = \dfrac{1 - \text{erf}(x \sqrt{n/2})}{2}$$ and using Watson's lemma, I get $$ P(S_n \ge n x) \sim e^{-n x^2/2} \left( \dfrac{1}{x \sqrt{2\pi n}} - \dfrac{1}{x^3 \sqrt{2\pi n^3}} + \dfrac{3}{x^5 \sqrt{2\pi n^5}}\right)$$

So in this case it's certainly not $\sim e^{-n I(x)}$, but it is $e^{-n I(x) + o(n)}$ where $I(x) = -x^2/2$. Note that $$ \Phi(n) = \ln\left( \dfrac{1}{x \sqrt{2\pi n}} + \ldots \right) = - \ln(x \sqrt{2\pi}) - \dfrac{1}{2} \ln(n) + \ldots = o(n)$$

Robert Israel
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  • I do not understand why $e^{nc-n\varepsilon} < a_n < e^{nc+n\varepsilon}$ is equivalent to $a_n=e^{nc+o(n)}$. – Rhjg Aug 26 '15 at 17:23
  • Write $a_n = e^{nc + b_n}$. Then for every $\epsilon > 0$ we have, for $n$ sufficiently large, $|b_n| < n \epsilon$. That says $\lim_{n \to \infty} |b_n|/n = 0$, which is exactly the definition of $b_n = o(n)$. – Robert Israel Aug 26 '15 at 17:33
  • What is your $b_n$? $b_n=\varepsilon n$? – Rhjg Aug 26 '15 at 17:40
  • No, $b_n = \ln(a_n e^{-nc}) = \ln(a_n) - n c$. – Robert Israel Aug 26 '15 at 17:42
  • Ok, got it. So $P(S_n\geqslant nx)\sim e^{-n I(x)+o(n)}$ is correct. Is it possible to say in which cases this reduces to $\sim e^{-n I(x)}$, i.e. in which cases the $o(n)$ is indeed $O(1)$? – Rhjg Aug 26 '15 at 17:51
  • How do you get $\lvert b_n\rvert < n\varepsilon$? – Rhjg Aug 26 '15 at 19:39
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    $|b_n| < n \epsilon$ when $\left| \dfrac{\ln(a_n)}{n} - c \right| < \epsilon$ – Robert Israel Aug 27 '15 at 00:28
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    I don't know if it's possible in general to say in which cases the $o(n)$ is $O(1)$, but the fact that it doesn't happen for the normal distribution makes me guess that this will be extremely rare, if indeed it ever occurs. – Robert Israel Aug 27 '15 at 00:33