Talk:Phase-type distribution

WikiProject Mathematics (Rated C-class, Low-importance)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
 C Class
 Low Importance
Field: Probability and statistics
WikiProject Statistics (Rated C-class, Mid-importance)

This article is within the scope of the WikiProject Statistics, a collaborative effort to improve the coverage of statistics on Wikipedia. If you would like to participate, please visit the project page or join the discussion.

C  This article has been rated as C-Class on the quality scale.
Mid  This article has been rated as Mid-importance on the importance scale.

This Phase-type distribution article is in serious need of revision, or perhaps even reversion, or both! It currently sounds like it has being copied from a text book or an original work. Earlier versions are somewhat more readable. Perhaps it has been over-constructed. -- Cameron Dewe 10:05, 7 October 2006 (UTC)

There is a mention of 'hyperexponential distribution' that should probably read 'hypoexponential distribution' in 3.1.2. —Preceding unsigned comment added by 134.221.191.89 (talk) 13:27, 15 October 2007 (UTC)

Limit vs special case

Is the "constant" distribution really a special case? If so, the definition needs to be changed. The limiting distribution requires an infinite-dimensional matrix ($m=\infty$), but then the absorbing state can't be state $\infty+1$. It should be state 1.

It would be much easier to say that the constant distribution is not phase-type. (Note that all distributions are the limit of phase type distributions, and so this might server as a useful conter-example, rather than an example. LachlanA (talk) 03:45, 17 November 2008 (UTC)

Easy fix. 129.127.252.4 (talk) 07:49, 20 September 2011 (UTC)

Things to check

In the box the formula stated for the variance seems to be instead the formula for the mean square. Also much of the article aims to address the case where $\alpha_{m+1}$ is not necessarily zero, but in that case surely the p.d.f. needs to have $\alpha_{m+1}\delta(x)$ adding to it, both in the box and in the main text.Fathead99 (talk) 12:47, 11 November 2009 (UTC)