Geometric Random Variable.

**simpleas123** · May 22nd 2010, 06:22 AM

Hey all i was stuck on this geometric random variable question.

Part (i)

(a) pmf = $p(k) = (1-p)^{k-1} \cdot p$
(b) is a simple proof.

Part (ii)

(a) $p(X \ge 4) = (1 -\dfrac{1}{3})^{4-1} = (\dfrac{2}{3})^3 = \dfrac {8}{27}$

(b) & (c) i dont have an idea of how to attempt, any help is greatly appricated.

Thanks guys

**matheagle** · May 22nd 2010, 07:33 AM

2b or not 2b, is that the question, just write out a few cases...

$P(X=2)=P(RR)=(1/3)^2$

$P(X=3)=P(BRR)+P(RBR)=2(1/3)^2(2/3)$

$P(X=4)=P(BBRR)+P(BRBR)+P(RBBR)=3(1/3)^2(2/3)^2$

It looks like $P(X=k)=(k-1)(1/3)^2(2/3)^{k-2}$ , $k=2,3,4....$

**matheagle** · May 22nd 2010, 05:35 PM

This is a valid distribution...

$p^2\sum_{k=2}^{\infty}(k-1)(1-p)^{k-2}$

$=-p^2{d\over dp}\left(\sum_{k=2}^{\infty}(1-p)^{k-1}\right)$

$=-p^2{d\over dp}\left(\sum_{j=0}^{\infty}(1-p)^{j+1}\right)$

$=-p^2{d\over dp}\left({1-p\over 1-(1-p))}\right)$

$=-p^2{d\over dp}\left(p^{-1}-1\right)$

$=-p^2\left(-p^{-2}\right)$

$=1$

**matheagle** · May 22nd 2010, 08:52 PM

I was waiting for someone to do 2c...
This is a finite rv, with W=1,2,...,11.

$P(W=1)=P(R)= {5\over 15}={1\over 3}$

$P(W=2)=P(BR)= \left({10\over 15}\right)\left({5\over 14}\right)$

$P(W=3)=P(BBR)= \left({10\over 15}\right) \left({9\over 14}\right)\left({5\over 13}\right)$

Thread: Geometric Random Variable.

LinkBack

Thread Tools

Display

Geometric Random Variable.

Similar Math Help Forum Discussions

Variance of a Poisson random variable whose parameter is also a random variable

Poisson and Geometric Random Variable questions

Geometric Random Variable

Finding the marginal distribution of a random variable w/ a random variable parameter

Geometric random variable

Search Tags