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I'm trying to evaluate an integral of the following form

$$\int \prod_i \left[ dx_i \,P(x_i) \right] \; f \Big( \frac{1}{N} \! \sum_{i=1}^N x_i \Big)$$

and I know that the distribution of $x$ is such that the central limit theorem is applicable, i.e., $\frac{1}{N} \sum_{i = 1}^N x_i \sim \mathcal{N} \big( \mathbb{E} (x), \frac{1}{N} \! \text{Var}(x) \big)$. Am I allowed to say the equation above is well approximated by

$$\int dz\, \mathcal{N} \big(z; \mathbb{E} (x), \frac{1}{N} \! \text{Var}(x) \big) \; f(z)$$

Thanks!

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  • $\begingroup$ Some properties of $f$ are required, otherwise "no". $\endgroup$ Apr 28, 2014 at 1:11
  • $\begingroup$ Hmm, specific properties? I'd say $f$ is some well-behaved function, i.e. analytic/smooth, but I wouldn't know much more about it... $\endgroup$ Apr 29, 2014 at 1:48
  • $\begingroup$ The usual condition for convergence in distribution implying convergence of moments is uniform integrability, here for the sequence f(∑X_i/N) in N. You'll get the convergence in distribution from the smoothness of f. However uniform integrability will take some work on your part, as it involves the interplay between f and the tails of the distribution of X. $\endgroup$
    – guest
    Apr 29, 2014 at 10:35

1 Answer 1

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I assume that $\mathbb{E}|X|<\infty$ and $\mathrm{Var}(X)<\infty.$

Let $a_N=\int \Pi_{i=1}^N [dx_i P(x_i)] f\left(\frac{1}{N}\sum_{i=1}^N x_i\right)$ and $b_N = \int dz \mathcal{N}\left(z;\mathbb{E}X,\frac{1}{N}\mathrm{Var}(X)\right)f(z).$

We know from the weak law of large numbers that if $X_1,X_2,\ldots$ are independent and identically distributed, then the sequence of random variables $\frac{1}{N}\sum_{i=1}^N X_i$ converges in probability and hence in distribution to the constant random variable $\mathbb{E}X.$ It is easy to see that a sequence of random variables distributed as $\mathcal{N}\left(\mathbb{E}X,\frac{1}{N}\mathrm{Var}(X)\right)$ converges in probability and hence in distribution to the constant random variable $\mathbb{E}X.$

If $f$ is known to be continuous and bounded, we have by the portmanteau lemma that $a_N\to f(\mathbb{E}X)$ and $b_N\to f(\mathbb{E}X)$ as $N\to\infty$ and so $a_N$ and $b_N$ must be close for sufficiently large $N.$

References:

http://en.wikipedia.org/wiki/Convergence_of_random_variables http://en.wikipedia.org/wiki/Portmanteau_lemma http://en.wikipedia.org/wiki/Law_of_large_numbers#Weak_law

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