# distribution discretization

Let $\mu$ be a distribution on $\mathbb{R}^n$. We partition $\mathbb{R}^n$ into small cubes congruent with $[0,\delta)^n$, parallel to the axes. In each cube, pick a point $x$ (for instance, the center of the cube). Let $\Lambda$ be the point set of all such centres. We will define a discrete probability $\tilde\mu$ on $\Lambda$ as follows: the density function $\tilde p(y) = \mu(B(y))$ ($y\in\Lambda$), where $B(y)$ denotes the cube that contains $y$. Basically $\tilde\mu$ is like a discretized version of $\mu$. For instance, using computer to generate random variables, which truncates the random number at a certain precision, actually gives such a discretized distribution over a lattice.

Let $X\sim \mu$ and $\tilde X\sim \tilde\mu$. Let $f:\mathbb{R}^n\to\mathbb{R}^k$ be some function. I wish to know under what assumptions (on $f$, $\mu$ and $\delta$) we can say that the distribution of $f(\tilde X)$ approximates a discretized version of the distribution of $f(X)$. our daily use of numerical simulation does seem to suggest some kind of this result holds -- when $\tilde X$ is truncated version of $X$, we all take the histogram of samples of $f(\tilde X)$ to be an approximation to the density function of $f(X)$.

I would like to formalize this but don't have a clear, formal definition of what 'approximate' means here, probably it has small total variation distance to some discretized version of $\mathcal{L}(f(X))$ with support on $f(\Lambda)$. Here the underlying partition of $\mathbb{R}^k$ does not need to have congruent parts, I would only expect that the diameter of the parts to be small. I don't quite see how to argue this though, because for two elements $B(x_1)$ and $B(x_2)$ in the original partition of $\mathbb{R}^n$, $f(B(x_1))$ and $f(B(x_2))$ may overlap partially but not completely.

Can anyone give me some references?

• I suspect that you will not have approximation in the total variation distance but only weak approximation. Probably this can be quantified by Prokhorov or Wasserstein metrics but I don't know a good pointer... – Dirk Jan 4 '15 at 21:42
• $\tilde \mu$ converges weakly to $\mu$ as $\delta \to 0$ without any additional assumption, so the law of $f(\tilde X)$ converges weakly to that of $f(X)$ as soon as $f$ is continuous. This follows immediately from the fact that pointwise convergence when testing against a smooth test function implies weak convergence on $R^n$. – Martin Hairer Jan 9 '15 at 9:36
• Three unrelated thoughts: (1) Probably this depends what you want to use it for. Perhaps there is a hidden assumption in your setting that $\mu$ is "nice" in some way? (2) Maybe some good pointers would come from research in "learning a distribution" or "density estimation" from i.i.d. samples. But I would guess that usually continuous distributions are used to approximate continuous distributions. (3) One thing you could do is let $\bar{\mu}$ be uniform over each given cube, rather than discrete on some point in the cube. – usul Jan 9 '15 at 11:50