Talk:Hyperexponential distribution

Latest comment: 11 years ago by Jafeluv in topic Requested move

E[x^2]= sum((1/lambda^2)p). — Preceding unsigned comment added by 165.132.58.112 (talkcontribs)

I believe the formula in the article is correct with the 2 upstairs. You might be thinking of variance, but that's still not quite right. I added formulae I found in a book, and cited it. Dicklyon (talk) 03:20, 3 May 2013 (UTC)Reply

Requested move

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The following discussion is an archived discussion of a requested move. Please do not modify it. Subsequent comments should be made in a new section on the talk page. Editors desiring to contest the closing decision should consider a move review. No further edits should be made to this section.

The result of the move request was: Move. Jafeluv (talk) 07:56, 10 May 2013 (UTC)Reply


Hyper-exponential distributionHyperexponential distribution – There is no apparent reason to separate the prefix "hyper" with a hyphen here; most sources don't do that, and we don't do that with Hypoexponential distribution or Hypergeometric distribution. Dicklyon (talk) 03:28, 3 May 2013 (UTC)Reply

The above discussion is preserved as an archive of a requested move. Please do not modify it. Subsequent comments should be made in a new section on this talk page or in a move review. No further edits should be made to this section.