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| Started by | DrMajorBob <btreat1@austin.rr.com> |
|---|---|
| First post | 2011-04-29 11:33 +0000 |
| Last post | 2011-04-29 11:33 +0000 |
| Articles | 1 — 1 participant |
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Re: Expected value of the Geometric distribution DrMajorBob <btreat1@austin.rr.com> - 2011-04-29 11:33 +0000
| From | DrMajorBob <btreat1@austin.rr.com> |
|---|---|
| Date | 2011-04-29 11:33 +0000 |
| Subject | Re: Expected value of the Geometric distribution |
| Message-ID | <ipe7m9$r1f$1@smc.vnet.net> |
The geometric distribution is discrete (of course), so...
Sum[(1 - p)^k*p*k, {k, 1, Infinity}]
(1 - p)/p
Bobby
On Thu, 28 Apr 2011 05:37:38 -0500, Tonja Krueger <tonja.krueger@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
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--
DrMajorBob@yahoo.com
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