Suppose we want to approximate a real-valued random variable $X$ by a discrete random variable $Z$ with finitely many atoms. Suppose all moments of $X$ is finite. We want to match the moments of $X$ up to the $m^{\rm th}$ order:

$$\mathbb{E}[X^k] = \mathbb{E}[Z^k] \text{ for } k = 1, \ldots, m.\tag{1}\label{1}$$

Here is a positive result, which is a simple consequence of convex analysis (Carathéodory's theorem): there exists $Z$ with at most $m+1$ atoms such that \eqref{1} holds.

Here are my questions:

Is there a

**converse**result about this? Say $X$ has an absolutely continuous distribution supported on $\mathbb{R}$ (e.g. Gaussian). When $m$ is large, given that $Z$ has only $m$ atoms, can we conclude that we cannot approximate all $2m$ moments of $X$ well, i.e., can we lower bound the error $\max_{1 \leq k \leq 2m}|\mathbb{E}[X^k] - \mathbb{E}[Z^k]|$? My intuition is the following: for a Gaussian $X$, $\mathbb{E}[X^k]$ grows like $k^{\frac{k}{2}}$ superexponentially. When we find a $Z$ who matches all moments of $X$ up to $m$, it cannot catch up with higher-order moments $X$; if $Z$ matches all moments from $m+1$ up to $2m$, then its low-order moments will be quite different from $X$.Is there an

**efficient algorithm**to compute the location and weights of the approximating discrete distribution? Does there exist a table to record these for approximating common distribution (e.g. Gaussian) for each fixed $m$? It could be very handy...I heard from folklore that when \eqref{1} holds, the total variation distance between their distributions can be upper bounded by, say, $e^{-m}$ or $1/m!$. Of course, this won't be true for a discrete $Z$. But let's say $X$ and $Z$ both has smooth and bounded density on $\mathbb{R}$. Could this be true? Now two characteristic functions matches at $0$ up to $m^{\rm th}$ derivatives. They should be pretty close?