Let $\mathcal{D}(m)$ be the set of [phase-type distributions](http://en.wikipedia.org/wiki/Phase-type_distribution) constructed from $m+1$-state Markov chains. Recall that the *coefficient of variation* of a distribution $D$ is the ratio of the standard deviation to the mean, which we will denote by $C(D)$.
Let
$$ F(m)=\inf\_{D\in\mathcal{D}} C(D)$$


What is $F(m)$, and what Markov chains (if any) achieve the infimum?


Note: One candidate distribution for the minimum might be a one-way chain of states with identical transition probabilities. The corresponding phase-type distribution has an Erlang distribution, and the coefficient of variation is $1/\sqrt{m}$. (So, $F(m)\leq 1/\sqrt{m}$.)


Motivation: I'm modeling a stochastic process with a Markov chain and want to match a particular (measured) mean and variance, but for computational reasons I want to use the fewest states possible. It is easy to make the variance unboundedly large (for a fixed mean), but making the variance small seems to require more care.



Minimal variance for phase-type distributions?
I haven't checked the details, but Aldous and Shepp's paper ["The least variable phase-type distribution is Erlang"](http://ftp.stat.berkeley.edu/~aldous/Papers/me32-scan.PDF) seems to show that $F(m)=\frac{1}{\sqrt{m}}$ is the minimum. The standard deviation over expectation is the square root of the coefficient of variation, which they show cannot be smaller than $\frac{1}{m}$.


