Re: Expected value of the Geometric distribution
- To: mathgroup at smc.vnet.net
- Subject: [mg118497] Re: Expected value of the Geometric distribution
- From: Peter Breitfeld <phbrf at t-online.de>
- Date: Fri, 29 Apr 2011 07:36:12 -0400 (EDT)
- References: <ipbg1f$ahf$1@smc.vnet.net>
"Tonja Krueger" wrote: > Hi all, > I want to calculate expected value of diverse distributions like the > Geometric distribution (for example). > As I understand this, the expected value is the integral of the > density function *x. > But when I try to calculate this: > Integrate[(1-p)^k*p*k,k] > I get this as the answer: > ((1 - p)^k p (-1 + k Log[1 - p]))/Log[1 - p]^2 > Instead of: (1-p)/p. > I would be so grateful if someone could explain to me what I'm doing wrong. > Tonja > ___________________________________________________________ > Empfehlen Sie WEB.DE DSL Ihren Freunden und Bekannten und wir > belohnen Sie mit bis zu 50,- Euro! https://freundschaftswerbung.web.de > The Geometric Distribution is discrete, so you must use Sum instead of Integrate: Sum[(1-p)^k p k, {k,1,Infinity}] Out= (1-p)/p This is also the result of Mean[GeometricDistribution[p]] as well as of ExpectedValue[#&,GeometricDistribution[p]] -- _________________________________________________________________ Peter Breitfeld, Bad Saulgau, Germany -- http://www.pBreitfeld.de