Re: Expected value of the Geometric distribution
- To: mathgroup at smc.vnet.net
- Subject: [mg118476] Re: Expected value of the Geometric distribution
- From: Stefan <wutchamacallit27 at gmail.com>
- Date: Fri, 29 Apr 2011 07:32:22 -0400 (EDT)
- References: <ipbg1f$ahf$1@smc.vnet.net>
On Apr 28, 6:37 am, "Tonja Krueger" <tonja.krue... at web.de> 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 Tonja, Two things to be considered. First you are on the right track regarding the definition of expected value of a random variable. Note that the integral though should not be an indefinite integral, but one over whatever domain the variable takes its values from. In this case, 0 to Infinity. Your second mistake though, was to use an *integral* to compute the expected value of a *discrete* random variable. The geometric distribution is discrete and so any expected values should be computed using sums, in this case from 0 to Infinity. The line you're looking for is Sum[(1 - p)^k*p*k, {k, 0, Infinity}] = (1-p)/p Hope this helps. -Stefan S