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!



Averaging function of sum of variables using central limit theorem
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>


