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I am a physicist caught in the following situation: I have two probability measures $\mathbb{P}_1$ and $\mathbb{P}_2$ and have to deal with the following integral where $X_i$ are random iid:

$$\int_{B} \mathbb{P}_1\left(\frac{1}{N} \sum_{i=1}^N X_i \ge x\right) d\mathbb{P}_2(x).$$

I was able to obtain a large deviation principle for $X$, i.e. I was able to show that

$$\frac{1}{N} \log\left( \mathbb{P}_1\left(\frac{1}{N} \sum_{i=1}^N X_i \ge x\right)\right) = -I(x).$$

Now it is tempting to replace $\mathbb{P}_1\left(\frac{1}{N} \sum_{i=1}^N X_i \ge x\right) $ in the above integral by $e^{-N I(x)}$ and compute $$\int_{B} e^{-NI(x)} d\mathbb{P}_2(x).$$

However, I do not know in which sense this now close to the expression I am looking for?

Are there any standard bounds to estimate the error between my approximation and the object I am interested in?

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  • $\begingroup$ The function I has to be monotone and the rate on the integral ought to be the smallest x in the support of $P_2$. $\endgroup$
    – user83457
    Commented Dec 18, 2017 at 11:39
  • $\begingroup$ @michael The function $I$ is indeed monotone. What precisely does the second part of your comment mean? $\endgroup$
    – Thorben Driggs
    Commented Dec 18, 2017 at 11:57
  • $\begingroup$ That if V is the value of your integral $\frac {log(V)} N \rightarrow -\inf \lbrace I(x) \rbrace$ where the inf is over x in the support of $P_2$. If $P_2$ is really nice, like uniform on an interval, I think it is easy to write down, but if it has 0 or no density at the lower limit it is probably tedious. $\endgroup$
    – user83457
    Commented Dec 18, 2017 at 13:34
  • $\begingroup$ On second thought, If $P_2$ is really nasty I don't know how bad it can get. What do you have in mind for $P_2$? $\endgroup$
    – user83457
    Commented Dec 18, 2017 at 13:42
  • $\begingroup$ I am not quite sure you understood my question. I am interested in comparing the value of $$\int_{B} \mathbb{P}_1\left(\frac{1}{N} \sum_{i=1}^N X_i \ge x\right) d\mathbb{P}_2(x)$$ to the value of $$\int_{B} e^{-NI(x)} d\mathbb{P}_2(x).$$ $\endgroup$
    – Thorben Driggs
    Commented Dec 18, 2017 at 14:02

1 Answer 1

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It all depends on the shape of $P_2$ and on the assumptions you put on $X_i$. In what follows I'll assume that $\Lambda(\lambda)=\log E_1 e^{\lambda X_1}$ is finite for all $\lambda$. I will also assume that $E_1X_i=0$. Further I will assume that $P_2$ is supported on $R$ with density $f$.

Case 1: $B\cap (-\infty,0]>0$. In that case your formula may be false, and the answer is essentially $P_2(B\cap (-\infty,0))$.

Case 2: (which is what I guess you had in mind): $B\cap (-\infty,a)=\emptyset$ for some $a>0$. In that case, use Bahadur-Rao to approximate, for $x\in B$, $P_1(N^{-1}\sum_{i=1}^N X_i>x)\sim C(x)e^{-NI(x)}/\sqrt{N}$, with explicit $C=C(x)$. So the expression you write is correct at the exponential scale but if you meant up to precise asymptotics, you are missing a constant multiple and a factor $\sqrt{N}$. The constant $C(x)$ depends on whether the law of $X_1$ is lattice or not. See the original paper of Bahadur-Rao or Theorem 3.7.4 in Dembo-Zeitouni's large deviations book.

One can also deal with other cases ($B$ touching $0$ with density of $P_2$ vanishing there, etc.) but I'll stop here.

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  • $\begingroup$ thank you. I wonder where the the first case comes from though. Let us say if $\mathbb{P}_2$ is Gaussian. Then, we are in Case $1$, right? So how does this one come about then? $\endgroup$
    – Thorben Driggs
    Commented Dec 20, 2017 at 6:22
  • $\begingroup$ Depends what B is. if B charges the negative axis, the probability involving $P_1$ is essentially 1! $\endgroup$
    – ofer zeitouni
    Commented Dec 20, 2017 at 7:24

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