# Geometric Random Variable.

• May 22nd 2010, 06:22 AM
simpleas123
Geometric Random Variable.
Hey all i was stuck on this geometric random variable question.

http://img541.imageshack.us/img541/9366/39409785.jpg

Part (i)

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

Part (ii)

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

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

Thanks guys
• May 22nd 2010, 07:33 AM
matheagle
2b or not 2b, is that the question, just write out a few cases...

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

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

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

It looks like $\displaystyle P(X=k)=(k-1)(1/3)^2(2/3)^{k-2}$, $\displaystyle k=2,3,4....$
• May 22nd 2010, 05:35 PM
matheagle
This is a valid distribution...

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

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

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

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

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

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

$\displaystyle =1$
• May 22nd 2010, 08:52 PM
matheagle
I was waiting for someone to do 2c...
This is a finite rv, with W=1,2,...,11.

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

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

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