MathGroup Archive 2011

[Date Index] [Thread Index] [Author Index]

Search the Archive

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


  • Prev by Date: Re: Expected value of the Geometric distribution
  • Next by Date: Re: and color via PlotStyle
  • Previous by thread: Re: Expected value of the Geometric distribution
  • Next by thread: Re: Expected value of the Geometric distribution