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I don't know why the order of error term is $O(n^{-1})$ or more high in asymptotically computing integral using Laplace's method? For instance, the following examples:

Suppose that $h(\theta)$ is a real function, has a unique manimum at $\hat{\theta}$ and has
continuously second derivative, then we get $\int_{-\infty}^{+\infty}e^{-nh(\theta)}d\theta=e^{-nh(\hat{\theta})}(\frac{2\pi}{nh^{''}(\hat{\theta})})^{\frac{1}{2}}(1+O(n^{-1})),$ where $h^{''}(\hat{\theta})\neq 0.$

But, I have computed this asymptotic expression many times, I still can not get the order of error term is $O(n^{-1}).$ On contrary, I can only get the order is $O(n).$ I do not know why.

There is another question on this expression, which is what is the $n$ that is arbitrary nature number. A lot of literature only use this result give some illustrations and do not give some proofs.

Hence, I consult everyone and wish to get your answers about this questions. Thanks a lot!

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migrated from Dec 19 '11 at 16:25

This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.

Thank very much to Zarrax and Wesley for their unusually helpful answers! – user21535 Dec 20 '11 at 14:23

In 1989, Tierney, Kass, and Kadane answered this question far better than I could:

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This is the basic idea... I think I'm assuming $h(\theta)$ is $C^4$, but anyhow:

First note you can factor out $e^{-nh(\hat{\theta})}$, so you can replace $h(\theta)$ by $h(\theta) - h(\hat{\theta})$ and assume $h(\hat{\theta}) = 0$. Then pick $\epsilon > 0$ such that $h(\theta)$ is of the form $c(\theta - \hat{\theta})^2 + O((\theta- \hat{\theta})^3)$ for $|\theta - \hat{\theta}| <\epsilon$, where ${\displaystyle c = {h''(\hat{\theta}) \over 2}}$. You only have to worry about the integral for $|\theta - \hat{\theta}| <\epsilon$ since the rest of the integral decays much faster than the inside.

You can always make a change of variables to $y = (\theta - \hat{\theta})+ O(\theta - \hat{\theta})^2$ so that the integral becomes that of $e^{-ncy^2}$. However you get a Jacobian factor of the form $1 + ay+ O(y^2)$. So your integral is $$\int_{-\epsilon}^{\epsilon} e^{-ncy^2}(1 + ay + O(y^2))\,dy$$ The first term tends to the main asymptotic term as $n$ goes to infinity. The second is $$a\int_{-\epsilon}^{\epsilon} ye^{-ncy^2}\,dy$$ Being the integral of an odd function, you just get 0. The final term is bounded in absolute value by $$C\int_{-\epsilon}^{\epsilon} y^2e^{-ncy^2}\,dy$$

If you change variables to $z = \sqrt{n}y$, you get a term asymptotic to ${C' \over n^{3 \over 2}}$ for some $C'$ as $n \rightarrow \infty$.

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